trig without tears

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Trig without Tears

or, How to Remember Trigonometric Identities

revised 25 Jan 2012

Copyright © 1997–2012 Stan Brown, Oak Road Systems

Faced with the large number of trigonometric identities, students tend to try tomemorize them all. That way lies disaster. When you memorize a formula by rote, youhave no way to know whether you’re remembering it correctly. I believe it is much more

effective (and, in the long run, much easier) to understand thoroughly how the trigfunctions work, memorize half a dozen formulas, and work out the rest as needed.That’s what these pages show you how to do.

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Contents

Trig without TearsIntroduction

About TrigonometryAbout Trig without Tears

Bonus TopicsWhat You Won’t Find Here

NotationInterval NotationDegrees and Radians

The Six FunctionsThe Basic Two: Sine and Cosine

Expressions for Lengths of SidesThe Other Four: Tangent, Cotangent, Secant, CosecantSix Functions in One PictureWhy Call It Sine?

Functions of Special Angles

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Functions of 45°Functions of 30° and 60°Mnemonic for All Special Angles

Functions of Any AngleNot Just Triangles Any More

Why Bother?Reference AnglesSigns of Function ValuesExamples: Function Values

Identities for Related AnglesPeriodic Functions

Solving TrianglesLaw of SinesLaw of CosinesDetective Work: Solving All Types of Triangles

The CasesSpecial Note: Side‐Side‐AngleSolving Triangles on the TI‐83 or TI‐84

The “Squared” IdentitiesSum and Difference Formulas

Sine and Cosine of A±BEuler’s FormulaSine and Cosine of a SumSine and Cosine of a DifferenceSome Geometric Proofs

Tangent of A±BProduct‐Sum Formulas

Product to SumSum to Product

Double Angle and Half Angle FormulasSine or Cosine of a Double AngleTangent of a Double AngleSine or Cosine of a Half AngleTangent of a Half Angle

Inverse FunctionsPrincipal ValuesFunctions of Arcfunctions

Example 1: cos(Arctan x)Example 2: cos(Arcsin x)Example 3: cos(Arctan 1/x))

Arcfunctions of FunctionsExample 4: Arccos(sin u)Example 5: Arcsec(cos u)Example 6: Arctan(sin u)

Notes and DigressionsThe Problem with Memorizing

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On the Other Hand ...Proof of Euler’s FormulaPolar Form of a Complex NumberPowers and Roots of a Complex Number

Square Root of iPrincipal Root of Any NumberMultiple Roots

Logarithm of a Negative NumberCool Proof of Double‐Angle FormulasMultiple‐Angle Formulas

Functions of 3AFunctions of 4A

Great Book on Problem SolvingWhat’s New

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Trig without Tears Part 1:

Introduction

About Trigonometry

Trigonometry is fascinating! It started as the measurement (Greek metron) of triangles (Greek trigonon),but now it has been formalized under the inFluence of algebra and analytic geometry and we talk oftrigonometric functions. not just sides and angles of triangles.

Trig is almost the ideal math subject. Big and complex enough to have all sorts of interesting oddcorners, it is still small and regular enough to be taught thoroughly in a semester. (You can master theessential points in a week or so.) It has lots of obvious practical uses, some of which are actually taughtin the usual trig course. And trig extends plenty of tentacles into other Fields like complex numbers,logarithms, and calculus.

If you’d like to learn some of the history of trigonometry and peer into its dark corners, Irecommend Trigonometric Delights by Eli Maor (Princeton University Press, 1998).

The computations in trigonometry used to be a big obstacle. But now that we have calculators, that’s nolonger an issue.

Would you believe that when I studied trig, back when dinosaurs ruled the earth (actually, in the1960s), to solve any problem we had to look up function values in long tables in the back of the book, andthen multiply or divide those Five‐place decimals by hand? The “better” books even included tables oflogs of the trig functions, so that we could save work by adding and subtracting Five‐place decimalsinstead of multiplying and dividing them. My College Outline Series trig book covered all of plane andspherical trigonometry in 188 pages—but then needed an additional 138 pages for the necessary tables!

Though calculators have freed us from tedious computation, there’s still one big stumbling block in theway many trig courses are taught: all those identities. They’re just too much to memorize. (Manystudents despair of understanding what’s going on, so they just try to memorize everything and hope forthe best at exam time.) Is it tan²A + sec²A = 1 or tan²A = sec²A + 1? (Actually, it’s neither—see equation39!)

Fortunately, you don’t need to memorize them. This paper shows you the few that you do need tomemorize, and how you can produce the others as needed. I’ll present some ideas of my own, and awonderful insight by W.W. Sawyer.

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About Trig without Tears

I wrote Trig without Tears to show that you need to memorize very little. Instead, you learn how all thepieces of trigonometry hang together, and you get used to combining identities in different ways so thatyou can derive most results on the )ly in just a couple of steps.

You might like to read some ideas of mine on the pros and cons of memorizing.

To help you Find things, I’ll number the most important equations and other facts. (Don’t worry about thegaps in the numbering. I’ve left those to make it easier to add information to these pages.)

A very few of those, which you need to memorize, will be marked “memorize:”. Please don’t

memorize the others. The whole point of Trig without Tears is to teach you how to derive them asneeded without memorizing them. If you can’t think how to derive one, the boxes should make it easy toFind it. But then please work through the explanation. I truly believe that if you once thoroughlyunderstand how all these identities hang together, you’ll never have to memorize them again. (It’sworked for me since I First studied trig in 1965.)

Bonus Topics

By the way, I love explaining things but sometimes I go on a bit too long. So I’ve put some interesting butnonessential notes in a separate page and inserted hyperlinks to them at appropriate points. If youfollow them (and I hope you will), use your browser’s “back” command to return to the main text.

Much as it pains me to say so, if you’re pressed for time you can still get all the essential points byignoring those side notes. But you’ll miss some of the fun.

What You Won’t Find Here

Trig without Tears concerns itself exclusively with plane trigonometry, which is what’s taught today innearly every First course. Spherical trigonometry is not dealt with.

I’m also restricting myself to real arguments to the functions and real values of the functions. I haveto draw the line somewhere! (I do use real‐valued functions with the polar form of complex numbers inthe Notes.) For complex trigonometric functions, see chapter 14 of Eli Maor’s Trigonometric delights(Princeton University Press, 1998).

In Trig without Tears we’ll work with identities and solve triangles. A separate (and much shorter)page of mine explains how to solve trigonometric equations.

Finally, there are a lot of good math sites out there: some are good, some less good. I’ve collected adozen or so that you may Find useful.

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Notation

I’ve made some compromises since many common math characters can’t be displayed in a standardway:

sqrt is the square‐root function in formulas. (I use the radical sign √ for a square root of anumber.)Fractions other than ½ are written using the slash, such as a/b for a over b.

Interval Notation

In talking about the domains and ranges of functions, it is handy to use interval notation. Thus insteadof saying that x is between 0 and π, we can use the open interval (0;π) if the endpoints are not included,or the closed interval [0;π] if the endpoints are included.

You can also have a half-open interval. For instance, the interval [0;2π) is all numbers ≥ 0 and < 2π.You could also say it’s the interval from 0 (inclusive) to 2π (exclusive).

Degrees and Radians

These pages will show examples with both radians and degrees. The same theorems apply to either wayof measuring an angle, and you need to practice with both.

A lot of students seem to Find radians terrifying. But measuring angles in degrees and radians is no worsethan measuring temperature in °F and °C. In fact, angle measure is easier because 0°F and 0°C are not thesame temperature, but 0° and 0 radians are the same angle.

Just remember that a complete circle is 2π radians, and therefore half a circle is π radians:π radians (or just π) = 180°

You can convert between degrees and radians exactly the same way you convert between inches andfeet, or between centimeters and meters. (If conversions in general are a problem for you, you might liketo consult my page on that topic.)

But even though you can convert between degrees and radians, it’s probably better to learn to think in

both. Here’s an analogy.When you learn a foreign language L, you go through a stage where you mentally translate what

someone says in L into your own language, formulate your answer in your own language, mentallytranslate it into L, and then speak. Eventually you get past that stage, and you carry on a conversation inthe other language without translating. Not only is it more fun, it’s a heck of a lot faster and easier.

One easy way to visualize the special angles in radians is to think of the twelve hours numbered

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around the circumference of a clock face. One hour of travel by the hour hand is π/6, two hours is π/3,three hours is π/2, six hours is π, and so on. (Thanks are due to Jeffrey T. Birt for this suggestion.)

You want to train yourself to work with radians for the same reason: it’s more efFicient, and savingwork is always good. Practice visualizing an angle of π/6 or 3π/4 or 5π/3 directly, without translating

to degrees. You’ll be surprised how quickly it will become second nature!

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Summary:

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The Six Functions

Every one of the six trig functions is just one side of a right triangle divided by another

side. Or if you draw the triangle in a unit circle, every function is the length of one linesegment. The easy way to remember all six deFinitions: memorize the de)initions of

sine and cosine and then remember the other four as combinations of sine and

cosine, not as independent functions.

The Basic Two: Sine and Cosine

A picture is worth a thousand words (which is why it takes a thousand times as long to download). Thetrig functions are nothing more than lengths of various sides of a right triangle in various ratios.Since there are three sides, there are 3×2 = 6 different ways to make a ratio (fraction) of sides. That’swhy there are six trig functions, no more and no less.

Of those six functions, three—sine, cosine, and tangent—get the lion’s share of the work. (The othersare studied because they can be used to make some expressions simpler.) We’ll begin with sine andcosine, because they are really basic and the others depend on them.

Here is one of the conventional ways of showinga right triangle. A key point is that thelower‐case letters a, b, c are the sides opposite tothe angles marked with the correspondingcapital letters A, B, C. Most books use thisconvention: lower-case letter for side

opposite upper-case angle.The two fundamental deFinitions are

marked in the diagram. You must commit them

to memory. In fact, they should become secondnature to you, so that you recognize them nomatter how the triangle is turned around.Always, always, the sine of an angle is equal tothe opposite side divided by the hypotenuse(opp/hyp in the diagram). The cosine is equal to the adjacent side divided by the hypotenuse (adj/hyp).

memorize:

sine = (opposite side) / hypotenuse

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cosine = (adjacent side) / hypotenuse

What is the sine of B in the diagram? Remember opp/hyp: the opposite side is b and the hypotenuse is c,so sin B = b/c. What about the cosine of B? Remember adj/hyp: the adjacent side is a, so cos B = a/c.

Do you notice that the sine of one angle is the cosine of the other? Since A+B+C = 180° for any triangle,and C is 90°, A+B must equal 90°. Therefore A = 90°−B and B = 90°−A. When two angles add to 90°, eachangle is the complement of the other, and the sine of each angle is the cosine of the other. These are thecofunction identities:

sin A = cos(90°−A) or cos(π/2−A)

cos A = sin(90°−A) or sin(π/2−A)

Expressions for Lengths of Sides

The deFinitions of sine and cosine can be rearranged a little bit to let you write down the lengths of thesides in terms of the hypotenuse and the angles. For example, when you know that b/c = cos A, you canmultiply through by c and get b = c×cos A. Can you write another expression for length b, one that uses asine instead of a cosine? Remember that opposite over hypotenuse equals the sine, so b/c = sin B.Multiply through by c and you have b = c×sin B.

Can you see how to write down two expressions for the length of side a? Please work from thedeFinitions and verify that a = c×sin A = c×cos B.

Example: Given a right triangle with angle A =52° and hypotenuse c = 150 m. What is the length of sideb?

Solution: b = c×cos A = 150×cos 52° = about 92.35 m.

Example: A guy wire is anchored in the ground and attached to the top of a 45‐foot Flagpole. If it meetsthe ground at an angle of 63°, how long is the guy wire?

Solution: This is a right triangle, with A = 63°, a = 45 ft, and hypotenuse c unknown. But you doknow that a = c×sin A, and therefore c = a/sin A = 45/sin 63° = about 50.5 ft.

You may be wondering how to Find sides or angles of triangles when there is no right angle. We’ll get tothat, under the topic of Solving Triangles.

Sine and Cosine in the Unit Circle

One important special case comes up frequently. Suppose thehypotenuse c = 1; then we call the triangle a unit right triangle.You can see from the paragraphs just above that if c = 1 thena = sin A and b = cos A. In other words, in a unit right triangle the

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opposite side will equal the sine and the adjacent side will equal thecosine of the angle.

The triangle is often drawn in a unit circle, a circle of radius 1, asshown at right. The angle A is at the center and the adjacent sidelies along the x axis. If you do this, the hypotenuse is the radius,which is 1. The (x,y) coordinates of the outer end of the hypotenuse are the x and y legs of the triangle(cos A,sin A). The unit circle is your friend: it can help you visualize lots of trig identities.

The Other Four: Tangent, Cotangent, Secant, Cosecant

The other four functions have no real independent life of their own; they’re just combinations of the

)irst two. You could do all of trigonometry without ever knowing more than sines and cosines. Butknowing something about the other four, especially the tangent, can often save you some steps in acalculation—and your teacher will expect you to know about them for exams.

I Find it easiest to memorize (sorry!) the deFinition of the tangent in terms of the sine and cosine:

memorize:

tan A = (sin A) / (cos A)

You’ll use the tangent (tan) function very much more than the last three functions. (I’ll get to them in aminute.)

There’s an alternative way to remember the meaning of the tangent. Remember from thediagram that sin A = opposite/hypotenuse and cos A = adjacent/hypotenuse. Plug those into equation 3,the deFinition of the tan function, and you have tan A = (opposite/hypotenuse) / (adjacent/hypotenuse)or

tangent = (opposite side) / (adjacent side)

Notice this is not marked “memorize”: you don’t have to memorize it because it Flows directly from thedeFinition equation 3, and in fact the two statements are equivalent. I’ve chosen to present them in thisorder to minimize the jumble of opp, adj, and hyp among sin, cos, and tan. However, if you prefer you canmemorize equation 4 and then derive the equivalent identity equation 3 whenever you need it.

The other three trig functions—cotangent, secant, and cosecant—are de)ined in terms of the

)irst three. They’re much less often used, but they do simplify some problems in calculus.

memorize:

cot A = 1 / (tan A)

sec A = 1 / (cos A)

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csc A = 1 / (sin A)

Guess what! That’s the last trig identity you have to memorize.(You’ll probably Find that you end up memorizing certain other identities without even intending to,

just because you use them frequently. But equation 5 makes the last ones that you’ll have to sit down andmake a point of memorizing just on their own.)

Unfortunately, the deFinitions in equation 5 aren’t the easiest thing in the world to remember. Does thesecant equal 1 over the sine or 1 over the cosine? Here are two helpful hints: Each of those deFinitionshas a co‐function on one and only one side of the equation, so you won’t be tempted to think that sec Aequals 1/sin A. And secant and cosecant go together just like sine and cosine, so you won’t be tempted tothink that cot A equals 1/sin A.

For an alternative approach to remembering the above identities, you might like this 24‐secondvideo, suggested by Gene Laratonda: <http://www.youtube.com/watch?v=McIbWHgjQng>

You can immediately notice an important relation between tangent and cotangent. Each is theco‐function of the other, just like sine and cosine:

tan A = cot(90°−A) or cot(π/2−A)

cot A = tan(90°−A) or tan(π/2−A)

If you want to prove this, take the deFinition of tan and use equation 2 to substitute cos(90°−A) for sin Aand sin(90°−A) for cos A. Tangent and cotangent are co‐functions just like sine and cosine. By doing thesame sort of substitution, you can show that secant and cosecant are also co‐functions:

sec A = csc(90°−A) or csc(π/2−A)

csc A = sec(90°−A) or sec(π/2−A)

The formulas for negative angles of tangent and the other functions drop right out of the deFinitionsequation 3 and equation 5, since you already know the formulas equation 22 for sine and cosine ofnegative angles. For instance, tan(−A) = sin(−A)/cos(−A) = −sin A/cos A.

tan(−A) = −tan A

cot(−A) = −cot A

sec(−A) = sec A

csc(−A) = −csc A

You can reason out things like whether sec(180°−A) equals sec A or −sec A: just apply the deFinition anduse what you already know about cos(180°−A).

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unit circle (radius = AB = 1)sin θ = BC; cos θ = AC; tan θ = EDcsc θ = AG; sec θ = AE; cot θ = FG

graphic courtesy of TheMathPage

Six Functions in One Picture

You saw earlier how the sine and cosine of an angle are the sides of a triangle in a unit circle. It turns outthat all six functions can be shown geometrically in this way.

In the illustration at right, triangle ABC has angle θ at thecenter of a unit circle (AB = radius = 1). You already knowthat BC = sin θ and AC = cos θ.

What about tan θ? Well, since DE is tangent to the unitcircle, you might guess that its length is tan θ, and you’d beright. Triangles ABC and AED are similar, and therefore

ED / AD = BC / ACED / 1 = sin θ / cos θED = tan θ

More information comes from the same pair of similartriangles:

AE / AB = AD / ACAE / 1 = 1 / cos θAE = sec θ

The lengths that represent cot θ and csc θ will come from the other triangle, GAF. That triangle is alsosimilar to triangle AED. (Why? FG is perpendicular to FA, and FA is perpendicular to AD; therefore FG andAD are parallel. In beginning geometry you learned that when parallel lines are cut by a third line, thecorresponding angles—marked θ in the diagram—are equal. Thus FG is a tangent to the unit circle, andtherefore angles G and θ are equal. )

Using similar triangles GAF and AED,FG / FA = AD / EDFG / 1 = 1 / tan θFG = cot θ

That makes sense: FG is tangent to the unit circle, and is the tangent of the complement of angle θ,namely angle GAF, and therefore it is the cotangent of the original angle θ (or angle GAD).

Finally, using the same pair of similar triangles again, you can also say thatAG / FA = AE / EDAG / 1 = sec θ / tan θAG = [ 1 / cos θ ] / [ sin θ / cos θ ]AG = 1 / sin θAG = csc θ

This one diagram beautifully depicts the geometrical meaning of all six trig functions when the angle θ isdrawn at the center of a unit circle:

sin θ = BC; cos θ = AC; tan θ = EDcsc θ = AG; sec θ = AE; cot θ = FG

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Why Call It Sine?

From the picture, it’s obvious why the name “tangent” makes sense: the tangent of an angle is the lengthof a segment tangent to the unit circle. But what about the sine function? How did it get its name?

Please look at the picture again, and notice that sin θ = BC is half a chord of the circle. The Hindumathematician Aryabhata the elder (about 475–550) used the word “jya” or “jiva” for this half‐chord. InArabic translation the word was unchanged, but in the Arabic system of writing “jiva” was written thesame way as the Arabic word “jaib”, meaning bosom, fold, or bay. The Latin word for bosom, bay, or curveis “sinus”, or “sine” in English, and beginning with Gherardo of Cremona (about 1114–1187) that becamethe standard term.

Edmund Gunter (1581–1626) seems to have been the First to publish the abbreviations sin and tanfor sin and tangent.

My source for this history is Eli Maor’s Trigonometric Delights (1998, Princeton University Press),pages 35–36. I urge you to consult the book for a fuller account.

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Summary:

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Functions of Special Angles

You need to know the function values of certain special angles, namely 30° (π/6), 45°

(π/4), and 60° (π/3). You also need to be able to go backward and know what angle hasa sine of ½ or a tangent of −√3. While it’s easy to work them out as you go (using easyright triangles), you really need to memorize them because you’ll use them so often thatderiving them or looking them up every time would really slow you down.

Functions of 45°

Look at this 45‐45‐90° triangle, which means sides a and b are equal. Bythe Pythagorean theorem,

a² + b² = c²But a = b and c = 1; therefore

2a² = 1a² = 1/2a = 1/√2 = (√2)/2

Since a = sin 45°,sin 45° = (√2)/2

Also, b = cos 45° and b = a; thereforecos 45° = (√2)/2

Use the deFinition of tan A, equation 3 or equation 4:tan 45° = a/b = 1

sin 45° = cos 45° = (√2)/2

tan 45° = 1

Functions of 30° and 60°

Now look at this diagram. I’ve drawn two 30‐60‐90° triangles backto back, so that the two 30° angles are next to each other. Since2×30° = 60°, the big triangle is a 60‐60‐60° equilateral triangle.Each of the small triangles has hypotenuse 1, so the length 2b isalso 1, which means that

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b = ½2sBut b also equals cos 60°, and therefore

cos 60° = ½You can Find a, which is sin 60°, by using the Pythagorean theorem:

(½)² + a² = c² = 11/4 + a² = 1a² = 3/4 ⇒ a = (√3)/2

Since a = sin 60°, sin 60° = (√3)/2.

Since you know the sine and cosine of 60°, you can easily use the cofunction identities (equation 2) to getthe cosine and sine of 30°:

cos 30° = sin(90°−30°) = sin 60° = (√3)/2sin 30° = cos(90°−30°) = cos 60° = 1/2

As before, use the deFinition of the tangent to Find the tangents of 30° and 60° from the sines and cosines:tan 30° = sin 30° / cos 30°tan 30° = (1/2) / ((√3)/2)tan 30° = 1 / √3 = (√3)/3

andtan 60° = sin 60° / cos 60°tan 60° = ((√3)/2) / (1/2)tan 60° = √3

The values of the trig functions of 30° and 60° can be summarized like this:

sin 30° = ½, sin 60° = (√3)/2

cos 30° = (√3)/2, cos 60° = ½

tan 30° = (√3)/3, tan 60° = √3

Mnemonic for All Special Angles

Incidentally, the sines and cosines of 0, 30°, 45°, 60° and 90° display a pleasing pattern:

for angle A = 0, 30° (π/6), 45° (π/4), 60° (π/3), 90° (π/2):

sin A = (√0)/2, (√1)/2, (√2)/2, (√3)/2, (√4)/2

cos A = (√4)/2, (√3)/2, (√2)/2, (√1)/2, (√0)/2

tan A = 0, (√3)/3, 1, √3, unde�ined

It’s not surprising that the cosine pattern is a mirror image of the sine pattern, since sin(90°−A) = cos A.

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Summary:


Functions of Any Angle

The six trig functions were originally deFined for acute angles in triangles, but now wedeFine them for any angle (or any number). If you want any of the six function values foran angle that’s not between 0 and 90° (π/2), you just Find the function value for thereference angle that is within that interval, and then possibly apply a minus sign.

So far we have deFined the six trig functions as ratios of sides of aright triangle. In a right triangle, the other two angles must be lessthan 90°, as suggested by the picture at left.

Suppose you draw the triangle in a circle this way, with angleA at the origin and the circle’s radius equal to the hypotenuse ofthe triangle. The hypotenuse ends at the point on the circle withcoordinates (x,y), where x and y are the lengths of the two legs ofthe triangle. Then using the standard deFinitions of the trigfunctions, you have

sin A = opposite/hypotenuse = y/rcos A = adjacent/hypotenuse = x/r

This is the key to extending the trig functions to any angle.

Not Just Triangles Any More

The trig functions had their roots in measuring sides of triangles,and chords of a circle (which is practically the same thing). If wethink about an angle in a circle, we can extend the trig functions

to work for any angle.In the diagram, the general angle A is drawn in standard

position, just as we did above for an acute angle. Just as before, itsvertex is at the origin and its initial side lies along the positive xaxis. The point where the terminal side of the angle cuts the circleis labeled (x,y).

(This particular angle happens to be between 90° and 180°(π/2 and π), and we say it lies in Quadrant II. But you could draw asimilar diagram for any angle, even a negative angle or one>360°.)

Now let’s deFine sine and cosine of angle A, in terms of the coordinates (x,y) and the radius r of thecircle:

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(21)sin A = y/r, cos A = x/r

This is nothing new. As you saw above when A was in Quadrant I, this is exactly the deFinition youalready know from equation 1: sin A = opposite/hypotenuse, cos A = adjacent/hypotenuse. We’re justextending it to work for any angle.

The other function deFinitions don’t change at all. From equation 3 we still havetan A = sin A / cos A

which means thattan A = y/x

and the other three functions are still deFined as reciprocals (equation 5).Once again, there’s nothing new here: we’ve just extended the original de)initions to a larger

domain.

Why Bother?

So why go through this? Well, for openers, not every triangle is an acute triangle. Some have an anglegreater than 90°. Even in down‐to‐earth physical triangles, you’ll have to be concerned with functions ofangles greater than 90°.

Beyond that, it turns out that all kinds of physical processes vary in terms of sines and cosines asfunctions of time: height of the tide; length of the day over the course of a year; vibrations of a spring, orof atoms, or of electrons in atoms; voltage and current in an AC circuit; pressure of sound waves, Nearlyevery periodic process can be described in terms of sines and cosines.

And that leads to a subtle shift of emphasis. You started out thinking of trig functions of angles, butreally the domain of trig functions is all real numbers, just like most other functions. How can this be?Well, when you think of an “angle” of so‐and‐so many radians, actually that’s just a pure number. Forinstance, 30°=π/6. We customarily say “radians” just to distinguish from degrees, but really π/6 is a purenumber. When you take sin(π/6), you’re actually evaluating the function sin(x) at x = π/6 (about 0.52),even though traditionally you’re taught to think of π/6 as an angle.

We won’t get too far into that in these pages, but here’s an example. If the average water depth is 8 ftin a certain harbor, and the tide varies by ±3 ft, then the height at time t is given by a function thatresembles y = 8 + 3 cos(0.52t). (It’s actually more complicated, because high tides don’t come at the sametime every day, but that’s the idea.)

Reference Angles

Coming back from philosophy to the nitty‐gritty of computation, how do we Find the value of a functionwhen the angle (or number) is outside the range [0;90°] (which is 0 to π/2)? The key is to deFine areference angle.

Here’s the same picture of angle A again, but with its

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reference angle added. With angle A in standard position, thereference angle is the acute angle between the terminal side of Aand the positive or negative x axis. In this case, angle A is in Q II,the reference angle is 180°−A (π−A). Why? Because the two anglestogether equal 180° (π).

What good does the reference angle do you? Simply this: the six

function values for any angle equal the function values for its

reference angle, give or take a minus sign.That’s an incredibly powerful statement, if you think about it.

In the drawing, A is about 150° and the reference angle istherefore about 30°. Let’s say they’re exactly 150° and 30°, just for discussion. Then sine, cosine, tangent,cotangent, secant, and cosecant of 150° are equal to those same functions of 30°, give or take a minussign.

What’s this “give or take” business? That’s what the next section is about.

Signs of Function Values

Remember the extended deFinitions from equation 21:sin A = y/r, cos A = x/r

The radius r is always taken as positive, and therefore the signs of sine and cosine are the same as thesigns of y and x. But you know which quadrants have positive or negative y and x, so you know for whichangles (or numbers) the sine and cosine are positive or negative. And since the other functions aredeFined in terms of the sine and cosine, you also know where they are positive or negative.

Spend a few minutes thinking about it, and draw some sketches. For instance, is cos 300° positive ornegative? Answer: 300° is in Q IV, which is in the right‐hand half of the circle. Therefore x is positive, andthe cosine must be positive as well. The reference angle is 60° (draw it!), so cos 300° equals cos 60° andnot −cos 60°.

You can check your thinking against the chart that follows. Whatever you do, don’t memorize the

chart! Its purpose is to show you how to reason out the signs of the function values whenever you needthem, not to make you waste storage space in your brain.


Q I

0 to 90°

0 to π/2

Q II

90 to 180°

π/2 to π

Q III

180 to 270°

π to 3π/2

Q IV

270 to 360°

3π/2 to 2π

x positive negative negative positive

y positive positive negative negative

sin A

(= y/r)positive positive negative negative

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Q I

0 to 90°

0 to π/2

Q II

90 to 180°

π/2 to π

Q III

180 to 270°

π to 3π/2

Q IV

270 to 360°

3π/2 to 2π

cos A

(= x/r)positive negative negative positive

tan A

(= y/x)positive negative positive negative

What about other angles? Well, 420° = 360°+60°, and therefore 420° ends in the same position in thecircle as 60°—it’s just going once around the circle and then an additional 60°. So 420° is in Q I, just like60°.

You can analyze negative angles the same way. Take −45°. That occupies the same place on the circleas +315° (360°−45°). −45° is in Q IV.

Examples: Function Values

As you’ve seen, for any function you get the numeric value by considering the reference angle and thepositive or negative sign by looking where the angle is.

Example: What’s cos 240°? Solution: Draw the angle and see that the reference angle is 60°;remember that the reference angle always goes to the x axis, even if the y axis is closer. cos 60° = ½, andtherefore cos 240° will be ½, give or take a minus sign. The angle is in Q III, where x is negative, andtherefore cos 240° is negative. Answer: cos 240° = −½.

Example: What’s tan(−225°)? Solution: Draw the angle and Find the reference angle of 45°.tan 45° = 1. But −225° is in Q II, where x is negative and y is positive; therefore y/x is negative. Answer:

tan(−225°) = −1.

Identities for Related Angles

The techniques we worked out above can be generalized into a set of identities. For instance, if twoangles are supplements then you can write one as A and the other as 180°−A. You know that one will bein Q I and the other in Q II, and you also know that one will be the reference angle of the other. Thereforeyou know at once that the sines of the two angles will be equal, and the cosines of the two will benumerically equal but have opposite signs.

This diagram may help:

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Here you see a unit circle (r = 1) with four identical triangles. Their angles A are at the origin, arrangedso that they’re mirror images of each other, and their hypotenuses form radii of the unit circle. Look atthe triangle in Quadrant I. Since its hypotenuse is 1, its other two sides are cos A and sin A.

The other three triangles are the same size as the First so their sides must be the same length as thesides of the First triangle. But you can also look at the other three radii as belonging to angles 180°−A inQuadrant II, 180°+A in Quadrant III, and −A or 360°−A in Quadrant IV. All the others have a referenceangle equal to A. From the symmetry, you can immediately see things like sin(180°+A) = −sin A andcos(−A) = cos A.

The relations are summarized below. Don’t memorize them! Just draw a diagram whenever you needthem—it’s easiest if you use a hypotenuse of 1. Soon you’ll Find that you can quickly visualize thetriangles in your mind and you won’t even need to draw a diagram. The identities for tangent are easy toderive: just divide sine by cosine as usual.

sin(180°−A) = sin Asin(π−A) = sin A

cos(180°−A) = −cos Acos(π−A) = −cos A

tan(180°−A) = −tan Atan(π−A) = −tan A

(22)sin(180°+A) = −sin Asin(π+A) = −sin A

cos(180°+A) = −cos Acos(π+A) = −cos A

tan(180°+A) = tan Atan(π+A) = tan A

sin(−A) = −sin A cos(−A) = cos A tan(−A) = −tan A

Periodic Functions

You should be able to see that 360° brings you all the way around the circle. That means that an angle of360°+A or 2π+A is the same as angle A. Therefore the function values are unchanged when you add 360°

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or a multiple of 360° (or 2π or a multiple) to the angle. Also, if you move in the opposite direction forangle A, that’s the same angle as 360°−A or 2π−A, so the function values of −A and 360°−A (or 2π−A) arethe same.

For this reason we say that sine and cosine are periodic functions with a period of 360° or 2π.Their values repeat over and over again. Of course secant and cosecant, being reciprocals of cosine andsine, must have the same period.

What about tangent and cotangent? They are periodic too, but their period is 180° or π: they repeattwice as fast as the others. You can see this from equation 22: tan(180°+A) = tan A says that the functionvalues repeat every 180°.

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Summary:


Solving Triangles

A triangle has six parts, three sides and three angles. Given almost any three ofthem—three sides, two sides and an angle, or one side and two angles—you can )ind the

other three values. This is called solving the triangle, and you can do it with any

triangle, not just a right triangle.

Let’s look at a speciFic example to start with. Suppose you have a triangle where one side has a length of180, an adjacent angle is 42°, and the opposite angle is 31°. You’re asked to Find the other angle and theother two sides.

It’s always a good idea to draw a rough sketch, like thisone. Not only does it help you organize your solutionprocess better, but it can help you check your work. Forinstance, since the 31° angle is the smallest, you know thatthe opposite side must also be the shortest. If you were tocome up with an answer of, say, 110 for one of the othersides, you’d know at once that you had made a mistakesomewhere because 110 is < 180 and the other two sidesmust both be > 180.

How would you go about solving this problem? It’s not immediately obvious, I agree. But maybe we canget some help from some useful general techniques in problem solving:

Can you draw a diagram?Can you use what you already know to solve a piece of this problem, or a related problem?If you have a speciFic case, can you solve a more general problem? (Sometimes it works theother way, too, where taking a speciFic example points out a good technique for solving ageneral problem.)

We’ve already got the diagram, but let’s see if those other techniques will be helpful. (By the way, they’renot original with me, but are from a terriFic book on problem‐solving techniques that I think you shouldknow about.)

“Can you use what you already know to solve a piece of this problem?” For example, if this were a righttriangle you’d know right away how to write down the lengths of sides in terms of sines or cosines.

But it’s not a right triangle, alas. Is there any way to turn it into a right triangle? Not exactly, but ifyou construct a line at right angles to one side and passing through the opposite vertex, you’ll have tworight triangles. Maybe solving those right triangles will show how to solve the original triangle.

This diagram shows the same triangle after I drew thatperpendicular. I’ve also used another principle (“Can yousolve a more general problem?”) and replaced the speciFicnumbers with the usual letters for sides and angles.Dropping perpendicular CD in the diagram divides the bigtriangle (which you don’t know how to solve) into two right

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triangles ACD and BCD, with a common side CD. And youcan solve those right triangles.

We’re going to use this simple diagram to develop two important tools for solving triangles: the Lawof Sines and the Law of Cosines. Just drawing this one perpendicular line will show you how to solve notjust the triangle we started with, but any triangle. (Some trig courses teach other laws like the Law ofTangents and the Law of Segments. I’m ignoring them because you can solve triangles just Fine withoutthem.)

Law of Sines

The Law of Sines is simple and beautiful and easy to derive. It’s useful when you know two angles andany side of a triangle, or sometimes when you know two sides and one angle.

Let’s start by writing down things we know that relate the sides and angles of the two right trianglesin the diagram above. You remember how to write down the lengths of the legs of a right triangle? Theleg is always equal to the hypotenuse times either the cosine of the adjacent angle or the sine of theopposite angle. (If that looks like just empty words to you, or even if you’re not 100% conFident about it,please go back and review that section until you feel conFident.)

In the diagram above, look at triangle ADC at the left: the right angle is at D and the hypotenuse is b.We don’t know how much of original angle C is in this triangle, so we can’t use C to Find the lengths of anysides. What can we write down using angle A? By using its cosine and sine we can write the lengths ofboth legs of the triangle:

AD = b×cos A and CD = b×sin ABy the same reasoning, in the other triangle you have

DB = a×cos B and CD = a×sin BThis is striking: you see two different expressions for the length CD. But things that are equal to the samething are equal to each other. That means that

b×sin A = a×sin BDivide through by sin A and you have the solution:

b = a×sin B / sin Ab = 180 ×sin 42° / sin 31° = about 234

What about the third angle, C, and the third side, c? Well, when you have two angles of a triangle you canFind the third one easily:

A+B+C = 180°C = 180°−A−B

In this case, C = 180−31−42° = 107°.

For the third side, there are a couple of ways to go. You wrote expressions above for AD and DB, and youknow that c = AD+DB, so you could compute c = b×cos A + a×cos B.

But that’s two multiplies and an add, a bit more complicated than the one multiply and one divide toFind side b. I’m lazy, and I like to reduce the amount of tapping I do on my calculator. Is there an easierway, even if just slightly easier? Yes, there is. Go back a step, to

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a×sin B = b×sin ADivide left and right by (sin A)(sin B) to get

a / sin A = b / sin BBut there’s nothing special about the two angles A and B.You could just as well have dropped a perpendicular from Ato BC or from B to AC. Shown at right is the result ofdropping a perpendicular from B to line CD.

Because C > 90°, this perpendicular happens to beoutside the triangle and the two right triangles ABD andCBD overlap. But this won’t affect the algebra. By the way,the angle in triangle CBD is not C but 180°−C, thesupplement of C. Angle C belongs to the original triangle ABC.

You can write the length of the common side BD asBD = c×sin A (in triangle ABD)

andBD = a×sin(180−C) (in triangle CBD)

But sin(180−C) = sin C, so you haveBD = a×sin C (in triangle CBD)

Set the two computed lengths of BD equal to each other, and divide by (sin A)(sin C):a×sin C = c×sin Aa / sin A = c / sin C

But we already Figured out earlier thata / sin A = b / sin B

Combining these two equations you have the Law of Sines:a / sin A = b / sin B = c / sin C

It’s nice that the derivation doesn’t make you worry about obtuse versus acute triangles. As you see,when an obtuse angle is involved some dropped perpendicular will lie outside the original triangle, andin that case the derivation uses 180° minus an angle of the original triangle. But since sin x = sin(180°−x),you end up with the same form of the Law of Sines whether the perpendicular is inside or outside thetriangle, whether all three angles are acute or one is obtuse.

Law of Sines—First Form:

a / sin A = b / sin B = c / sin C

This is very simple and beautiful: for any triangle, if you divide any of the three sides by the sine of theopposite angle, you’ll get the same result. This law is valid for any triangle.

The Law of Sines is sometimes given upside down:

Law of Sines—Second Form:

sin A / a = sin B / b = sin C / c

Of course that’s the same law, just as 2/3 = 6/9 and 3/2 = 9/6 are the same statement. Work with it eitherway and you’ll come up with the same answers.

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You can derive the Law of Sines at need, so I don’t speciFically recommend memorizing it. But it’s sosimple and beautiful that it’s pretty hard not to memorize if you use it at all. It’s also pretty hard toremember it wrong: there are no alternating plus and minus signs or combinations of different functions.

In most cases where you use the Law of Sines, you get a unique solution. But sometimes you get twosolutions (or none) in the side‐side‐angle case, where you know two sides and an angle that’s notbetween them. Please see the Special Note below, after the table.

Law of Cosines

The Law of Sines is Fine when you can relate sides and angles. But suppose you know three sides of thetriangle—for instance a = 180, b = 238, c = 340—and you have to Find the three angles. The Law of Sinesis no good for that because it relates two sides and their opposite angles. If you don’t know any angles,you have an equation with two unknowns and you can’t solve it.

But a triangle can be solved when you know all threesides; you just need a different tool. And knowing me, youcan be sure I’m going to help you develop one! It’s calledthe Law of Cosines.

Let’s look back at that generic triangle with aperpendicular dropped from vertex C. You may rememberthat when we First looked at this picture, we pulled outinformation using both the sine and the cosine of the two angles. We used the sine information todevelop the Law of Sines, but we never went anywhere with the cosine information, which was

AD = b cos A and DB or BD = a cos BLet’s see where that can lead us. You remember that the way we came up with the Law of Sines was towrite two equations that featured the length of the construction line CD, and then combine the equationsto eliminate CD. Can we do anything like that here?

Well, we know the other two sides of those right triangles, so we can write an expression for theheight CD using the Pythagorean theorem—actually, two expressions, one for each triangle.

a² = (CD)² + (BD)² ⇒ (CD)² = a² − (BD)²b² = (CD)² + (AD)² ⇒ (CD)² = b² − (AD)²

and thereforea² − (BD)² = b² − (AD)²

Substitute the known values of BD and AD in terms of angles and sides of the original triangle, and youhave

a² − a²cos²B = b² − b²cos²ABzzt! No good! That uses two sides and two angles, but we need an equation in three sides and one angle,so that we can solve for that angle. Let’s back up a step, to a² − (BD)² = b² − (AD)², and see if we can go ina different direction.

Maybe the problem is in treating BD and AD as separate entities when actually they’re parts of the same

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line. Since BD+AD = c, we can writeBD = c−AD = c − b×cos A.

Notice that this brings in the third side, c, and angle B drops out. Substituting, we now havea² − (BD)² = b² − (AD)²a² − (c − b×cos A)² = b² − (b×cos A)²

This looks worse than the other one, but actually it’s better because it’s what we’re looking for: anequation for the three sides and one angle. We can solve it with a little algebra:

a² − c² + 2bc cos A − b²cos²A = b² − b²cos²Aa² − c² + 2bc cos A = b²2bc cos A = b² + c² − a²cos A = (b² + c² − a²) / 2bc

We were a long time getting there, but Finally we made it.Now we can plug in the lengths of the sides and come upwith a value for cos A, which in turn will tell us angle A:

cos A = (238² + 340² − 180²) / (2×238×340)cos A = 0.864088A = about 30.2°

Do the same thing to Find the second angle (or use the Lawof Sines, since it’s less work), then subtract the two knownangles from 180° to Find the third angle.

You can Find the Law of Cosines for the other angles by following the same process using the other twoperpendiculars.

Law of Cosines—First Form:

cos A = (b² + c² − a²) / 2bc

cos B = (a² + c² − b²) / 2ac

cos C = (a² + b² − c²) / 2ab

Just for fun, let’s Find angle C for that triangle:cos C = (a² + b² − c²) / 2abcos C = (180² + 238² − 340²) / 2×180×238cos C = −0.309944

Don’t be surprised at the negative number. Remember from the diagram in Functions of Any Angle thatcos A < 0 when A is between 90° and 180°. Because the cosine has unique values all the way from 0° to180°, you never have to worry about multiple solutions of a triangle when you use the Law of Cosines. Inthis case, C is about 108°.

There’s another well‐known form of the Law of Cosines, which may be a bit easier to remember. Startwith the above form, multiply through by 2ab, and isolate c on one side:

cos C = (a² + b² − c²) / 2ab2ab×cos C = a² + b² − c²c² = a² + b² − 2ab×cos C

Notice the arrangement: side c is opposite angle C in the triangle, and they’re at opposite ends of thisequation. Sides a and b are adjacent to angle C both in the triangle and in the equation. And you can play

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the same game to solve for the other two sides:

Law of Cosines—Second Form:

a² = b² + c² − 2bc cos A

b² = a² + c² − 2ac cos B

c² = a² + b² − 2ab cos C

Depending on how you’re using it, you may need the Law of Cosines in either of the two forms that we’veobtained, the First form for Finding an angle and the second form for Finding a side.

Detective Work: Solving All Types of Triangles

With just the deFinitions of sine, cosine, and tangent, you can solve any right triangle. If you’ve got theLaw of Sines and the Law of Cosines under your belt, you can solve any triangle that exists. (Some setsof givens lead to an impossible situation, like a “triangle” with sides 3‐4‐9.)

In this section I’ll run down the various possibilities and give you some pointers. But really it’spretty straightforward: whenever you have to solve a triangle, think about what you have and then thinkabout which formula you can use to get what you need. (When you have two angles, you can always Findthe third by A+B+C = 180°.)

The Cases

Many people Find it easier to think about the known elements of a triangle as a “case”. For instance, if youknow two angles and the side between them, that’s case ASA; if you know two angles and a side that’s notbetween them, that’s case AAS, and so on.

I’m not presenting the following table for you to memorize. Instead, what I hope to do is show youthat between the Law of Sines and the Law of Cosines you can solve any triangle, and that you simplypick which law to use based on which one has just one unknown and otherwise uses information youalready have.

Most cases can be solved with the Law of Sines. But if you have three sides (SSS), or two sides andthe angle between them (SAS), you must begin with the Law of Cosines.

The table is just an exhaustive elaboration of those two principles, so you probably don’t even needto read it! <grin>

If you know this... You can solve the triangle this way...

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three angles

There’s not enough information. Without at least one side you have theshape of the triangle, but no way to scale it correctly. For example, thesame angles could give you a triangle with sides 7‐12‐13, 35‐60‐65, or anyother multiple.

two angles and a side,AAS or ASA

Find the third angle by subtracting from 180°. then use the Law of Sines*(equation 28) twice to Find the second and third sides.

twosidesand ...

the includedangle, SAS

Use the Law of Cosines* (equation 31) to Find the third side. Then useeither the Law of Sines* (equation 29) or the Law of Cosines* (equation30) to Find the second angle.

a non‐includedangle, SSA

Use the Law of Sines* twice, equation 29 to get the second angle andequation 28 to get the third side.

But...This case may have no solutions, one solutions, or two solutions. See moredetails after the table.

three sides, SSS

With the First form equation 30 of the Law of Cosines you use all the sidesto compute one angle. Use that angle and its opposite side in the Law ofSines equation 29 to Find the second angle.

* If a 90° angle is given, the Law of Sines and the Law of Cosines are overkill. Just apply the deFinitionsof the sine and cosine (equation 1) and the tangent (equation 4) to Find the other sides and angles.

Special Note: Side-Side-Angle

For most sets of facts, either there’s a unique solution or they’re obviously absurd. (If you don’t see why a“triangle” with sides 50‐60‐200 is absurd, try to sketch it.) But the SSA case can be tricky.

Suppose you know acute angleB and sides a and b. Given thosefacts, there are two different waysyou could draw the triangle, asshown in the picture. How can thisbe? Well, you use the Law of Sines toFind the sines of angles A and C. Let’ssay you Find sin C = 0.5. That meansC could be either 30° or the supplement, 150°. Remember that the sine of any angle and the sine of itssupplement are the same.

This is the infamous ambiguous case. You can see the problem from the picture: the knownopposite side b can take either of two positions that satisfy the given the lengths of a and b. Those twopositions give rise to two different values for angle A, two different values for angle C, and two differentvalues for side c. Think about it for a while, and you’ll see that this ambiguity can arise only when theknown angle is acute, and the adjacent side is longer than the opposite side, and the opposite side is

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greater than the height.Here’s a complete rundown of all the possibilities with the SSA case:

Possibilities within the SSA Case

known angle < 90° known angle ≥ 90°

adjacent side <

opposite sideone solution one solution

adjacent side =

opposite sideone solution

no solution(Angles that are opposite equal sides must beequal, but you can’t have two angles both ≥ 90°in a triangle.)

adjacent side >

opposite side

Compute the triangle height h(adjacent side times sine of knownangle).

Opposite side < h? no solutionOpposite side = h? one solution(a right triangle)Opposite side > h? twosolutions

no solution(The conditions violate the theorem that thelongest side is always opposite the largestangle.)

For heaven’s sake, don’t try to memorize that table! Instead, always draw a picture. If you can drawtwo pictures that both Fit all the available facts, you have two legitimate solutions. If only one picture Fitsall the facts, it will show you which angle (if any) is > 90°. And if you can’t make any picture that Fits thefacts, the triangle has no solution.

If you do have two solutions, what do you do? If you have no other information to go on, of course youreport both solutions. But check the situation carefully. Maybe you’re told explicitly which is the largestangle, or it’s implied by other facts you know. In that case your solution is constrained, and you reject thesolution that doesn’t meet the constraints.

Solving Triangles on the TI-83 or TI-84

I’ve written a TI‐83/84 program to solve all types of triangles (and Find their area). If you have TI GraphLink software, you can download the program, unzip it, and install it on your TI‐83/84.

For those who don’t have the Graph Link software, here is the program in ASCII form. Special keysor menu selections are indicated between backslashes; for instance \->\ means the STO key. (This is theTI‐83/84 export format.)

Naturally, I’d be happy to know of any corrections or improvements to this program.

\start83P\comment=Program file dated 09/16/02, 00:23\protected=FALSE\name=TRIANGLE\file=H:\TRIANGLE.TXT

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DegreeMenu("WHICH CASE?","AAS",1,"ASA",2,"SAS",3,"SSA",4,"SSS",5)Lbl 1Disp "NEED <A,<B,SA"Input "<A=",AInput "<B=",BInput "SA=",D180-A-B\->\CDsin(B)/sin(A)\->\EDsin(C)/sin(A)\->\FGoto 0Lbl 2Disp "NEED <A,SB,<C"Input "<A=",AInput "SB=",EInput "<C=",C180-A-C\->\BEsin(A)/sin(B)\->\DEsin(C)/sin(B)\->\FGoto 0Lbl 3Disp "NEED <A,SB,SC"Input "<A=",AInput "SB=",EInput "SC=",F\root\(E\^2\+F\^2\-2EFcos(A))\->\Dsin\^-1\(Esin(A)/D)\->\B180-A-B\->\CGoto 0Lbl 4Disp "NEED SA,SB,<A"Input "SA=",DInput "SB=",EInput "<A=",AEsin(A)\->\HIf (D<H)+(A\>=\180)(D\<=\E):ThenDisp "IMPOSS":StopEndIf (H<D)(D<E):Thensin\^-1\(Esin(A)/D)\->\BMenu("AMBIG--IS B","ACUTE",41,"OBTUSE",42)Lbl 42:180-B\->\BGoto 41Endsin\^-1\(Esin(A)/D)\->\BLbl 41180-A-B\->\CD*sin(C)/sin(A)\->\FGoto 0Lbl 5Input "SA=",DInput "SB=",EInput "SC=",Fcos\^-1\((E\^2\+F\^2\-D\^2\)/(2EF))\->\Acos\^-1\((D\^2\+F\^2\-E\^2\)/(2DF))\->\B180-A-B\->\CLbl 0.5DEsin(C)\->\GDisp "AREA <S SIDES",round(G,2),{round(A,1),round(B,1), round(C,1)},round(D,2),round(E,2),round(F,2)

Caution: that long DISP command at the end should be entered as a single command; I’ve just broken itinto two lines for readability.

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Summary:

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The “Squared” Identities

This chapter begins exploring trigonometric identities. Three of them involve onlysquares of functions. These are called Pythagorean identities because they’re just thegood old Theorem of Pythagoras in new clothes. Learn the really basic one, namelysin²A + cos²A = 1, and the others are easy to derive from it in a single step.

Students seem to get bogged down in the huge number of trigonometric identities. As I said earlier, Ithink the problem is that students are expected to memorize all of them. But really you don’t have to,because they’re all just forms of a very few basic identities. The next couple of chapters will explore thatidea.

For example, let’s start with the really basic identity:

sin²A + cos²A = 1

That one’s easy to remember: it involves only the basic sine and cosine, and you can’t get the orderwrong unless you try.

But you don’t have to remember even that one, sinceit’s really just another form of the Pythagorean theorem.(You do remember that, I hope?) Just think about a righttriangle with a hypotenuse of one unit, as shown at right.

First convince yourself that the Figure is right, that thelengths of the two legs are sin A and cos A. (Check back inthe section on lengths of sides, if you need to.) Now writedown the Pythagorean theorem for this triangle. Voilà!You’ve got equation 38.

What’s nice is that you can get the other “squared” or Pythagorean identities from this one, and youdon’t have to memorize any of them. Just start with equation 38 and divide through by either sin²A orcos²A.

For example, what about the riddle we started with, the relation between tan²A and sec²A? It’s easy toanswer by a quick derivation—easier than memorizing, in my opinion.

If you want an identity involving tan²A, remember equation 3: tan A is deFined to be sin A/cos A.Therefore, to create an identity involving tan²A you need sin²A/cos²A. So take equation 38 and dividethrough by cos²A:

sin²A + cos²A = 1sin²A/cos²A + cos²A/cos²A = 1/cos²A(sin A/cos A)² + 1 = (1/cos A)²

which leads immediately to the Final form:

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graphic courtesy of TheMathPage

tan²A + 1 = sec²A

You should be able to work out the third identity (involving cot²A and csc²A) easily enough. You caneither start with equation 39 above and use the co‐function rules equation 6 and equation 7, or start withequation 38 and divide by something appropriate. Either way, check to make sure that you get

cot²A + 1 = csc²A

It may be easier for you to visualize these identitiesgeometrically.

In the diagram at right, remember that we showed thatED = tan A and AE = sec A. AD is a radius of the unit circle andtherefore AD = 1. Since angle D is a right angle, you can usethe Pythagorean theorem:

(ED)² + (AD)² = (AE)²tan²A + 1 = sec²A

You can use triangle AFG in a similar way to prove equation40.

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Summary:

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Sum and Difference Formulas

Continuing with trig identities, this page looks at the sum and difference formulas,namely sin(A± B), cos(A± B), and tan(A± B). Remember one and all the rest Flow from it.There’s also a beautiful way to get them from Euler’s formula.

Sine and Cosine of A±B

Formulas for cos(A+B), sin(A−B), and so on are important but hard to remember. Yes, you can derivethem by strictly trigonometric means. But such proofs are lengthy, too hard to reproduce when you’re inthe middle of an exam or of some long calculation.

This brings us to W.W. Sawyer’s marvelous idea, as expressed in chapter 15 of Mathematician’sDelight (1943; reprinted 1991 by Penguin Books). He shows how you can derive the sum and differenceformulas by ordinary algebra and one simple formula.

The ordinary algebra is simply the rules for combining powers:

x x = x

(x ) = x

(If you’re a bit rusty on the laws of exponents, you may want to review them.)

Euler’s Formula

You may already know the “simple formula” that I mentioned above. It’s

memorize:

cos x + i sin x = e

The formula is not Sawyer’s, by the way; it’s commonly called Euler’s formula. I don’t even knowwhether the idea of using Euler’s formula to get the sine and cosine of sum and difference is original withSawyer. But I’m going to give him credit, since his explanation is simple and clear and I’ve never seen itexplained in this way anywhere else.

a b a+b

a b ab

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You’ll sometimes see cos x + i sin x abbreviated as cis x for brevity.I’ve marked Euler’s formula, equation 47, “memorize”. Although it’s not hard to derive (and Sawyer

does it in a few steps by means of power series), you have to start somewhere. And that formula has somany other applications that it’s well worth committing to memory. For instance, you can use it to getthe roots of a complex number and the logarithm of a negative number.

Sine and Cosine of a Sum

Okay, back to Sawyer’s idea. What happens if you substitute x = A+B in equation 47 above? You getcos(A+B) + i sin(A+B) = e

Hmmm, this looks interesting. It involves exactly what we’re looking for, cos(A+B) and sin(A+B). Can yousimplify the right‐hand side? Use equation 46 and then equation 47 to rewrite it:

e = e e = (cos A + i sin A)(cos B + i sin B)Now multiply that out and set it equal to the original left‐hand side:

cos(A+B) + i sin(A+B) = [cos A cos B − sin A sin B] + i[sin A cos B + cos A sin B]

Now here’s the sneaky part. If I told you a+bi = 7−9i and asked you to solve for a and b, you couldimmediately tell me that a = 7 and b = −9, right? More formally, if two complex numbers are equal, theirreal parts must be equal and their imaginary parts must be equal. So the above equation in sines andcosines is actually two equations, one for the real part and one for the imaginary part:

cos(A+B) = cos A cos B − sin A sin B

sin(A+B) = sin A cos B + cos A sin B

In just a few short steps, the formulas for cos(A+B) and sin(A+B) Flow right from equation 47, Euler’sequation for e . No more need to memorize which one has the minus sign and how all the sines andcosines Fit on the right‐hand side: all you have to do is a couple of substitutions and a multiply.

Example: What’s the exact value of cos 75° or cos(5π/12)? Solution: 75° = 45°+30° (5π/12 = π/4+π/6).Using equation 48,

cos 75° = cos(45°+30°)cos 75° = cos 45° cos 30° − sin 45° sin 30°cos 75° = [(√2)/2]×[(√3)/2] − [(√2)/2]×[1/2]cos 75° = (√6)/4 − (√2)/4

Sine and Cosine of a Difference

What about the formulas for the differences of angles? You can write them down at once from equation48 by substituting −B for B and using equation 22. Or, if you prefer, you can get them by substituting x =A−B in equation 47 above. Either way, you get

(iA+iB)

iA+iB iA iB

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cos(A−B) = cos A cos B + sin A sin B

sin(A−B) = sin A cos B − cos A sin B

Some Geometric Proofs

I personally Find the algebraic reasoning given above very easy to follow, though you do have toremember Euler’s formula.

If you prefer geometric derivations of sin(A±B) and cos(A±B), you’ll Find a beautiful set by Len andDeborah Smiley at <http://www.maa.org/pubs/mm_supplements/smiley/trigproofs.html> or<http://math.uaa.alaska.edu/~smiley/trigproofs.html>, worth the wait for the page to load. (Phil Kenny drewmy attention to this page’s original version and to the link at the University of Alaska.) Eric’s TreasureTrove of Mathematics has smaller versions of the pictures at <http://mathworld.wolfram.com

/TrigonometricAdditionFormulas.html>.The fallback position is the standard proof: draw a diagram and use the distance formula or

Pythagorean Theorem to prove the formula for cos(A−B).

Tangent of A±B

Sometimes (not very often) you have to deal with the tangent of the sum or difference of two angles. Ihave only a vague idea of the formula, but it’s easy enough to work out “on the Fly”:

tan(A+B) = sin(A+B) / cos(A+B)tan(A+B) = (sin A cos B + cos A sin B) / (cos A cos B − sin A sin B)

What a mess! There’s no way to factor that and remove common terms—or is there? Suppose you startwith a vague idea that you’d like to know tan(A+B) in terms of tan A and tan B rather than all those sinesand cosines. The numerator and denominator contain sines and cosines, so if you divide by cosines you’dexpect to end up with sines or perhaps sines over cosines. But sine/cosine is tangent, so this seems like apromising line of attack. Since you’ve got cosines of angles A and B to contend with, try dividing thenumerator and denominator of the fraction by cos A cos B:

tan(A+B) = (sin A cos B + cos A sin B) / (cos A cos B − sin A sin B)tan(A+B) = [sin A/cos A + sin B/cos B] / [1 − sin Asin B/cos Acos B]

Hmmm, looks like this is the right track. Simplify it using the deFinition of tan x (see equation 3), and youhave

tan(A+B) = (tan A + tan B) / (1 − tan A tan B)

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Now if you replace B with −B, you have the formula for tan(A−B). (Take a minute to review whytan(−x) = −tan(x).)

tan(A−B) = (tan A − tan B) / (1 + tan A tan B)

Example: What’s the exact value of tan 15° or tan(π/12)? Solution: 15° = 60°−45° (π/12 = π/3 − π/4).Therefore

tan(π/12) = tan(π/3 − π/4)tan(π/12) = [tan(π/3) − tan(π/4)] / [1 + tan(π/3) tan(π/4)]tan(π/12) = [(√3)− 1] / [1 + (√3)×1]tan(π/12) = [(√3)− 1] / [(√3) + 1]

If you like, you can rationalize the denominator:tan(π/12) = [(√3)− 1]² / [(√3) + 1]×[(√3) − 1]tan(π/12) = [3 − 2(√3) + 1] / [3 − 1]tan(π/12) = [4 − 2(√3)] / 2tan(π/12) = 2 − √3

Product-Sum Formulas

Advice to the reader: The formulas in this section aren’t really very useful in trigonometry itself, but areused in calculus. You may wish to skip them, especially on a First reading.

Product to Sum

Sometimes you need to simplify an expression like cos 3x cos 5x. Of course it’s not equal to cos(15x²), butcan it be simpliFied at all? The answer is yes, and in fact you need this technique for calculus work. Thereare four formulas that can be used to break up a product of sines or cosines.

These product‐to‐sum formulas come from the formulas in equation 48 and equation 49 for sine andcosine of A±B. First let’s develop one of these formulas, and then we’ll look at an application beforedeveloping the others.

Take the two formulas for cos(A±B) and add them:cos(A−B) = cos A cos B + sin A sin Bcos(A+B) = cos A cos B − sin A sin Bcos(A−B) + cos(A+B) = 2 cos A cos B½ [cos(A−B) + cos(A+B)] = cos A cos B

Example: Suppose you need to graph the function

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f(x) = cos 2x cos 3x,or perhaps you need to Find its integral. Both of these are rather hard to do with the function as it stands.But you can use the product‐to‐sum formula, with A = 2x and B = 3x, to rewrite the function as a sum:

f(x) = cos 2x cos 3xf(x) = ½ [cos(2x−3x) + cos(2x+3x)]f(x) = ½ [cos(−x) + cos 5x]

Use equation 22, cos(−x) = cos x:f(x) = ½ [cos x + cos 5x]f(x) = ½ cos x + ½ cos 5x

This is quite easy to integrate. And while it’s not exactly trivial to graph, it’s much easier than theoriginal, because cos x and cos 5x are easy to graph.

The other three product‐to‐sum formulas come from the other three ways to add or subtract theformulas in equation 48 and equation 49. If you subtract the two cosine formulas instead of adding:

cos(A−B) = cos A cos B + sin A sin B−cos(A+B) = −cos A cos B + sin A sin B

you getcos(A−B) − cos(A+B) = 2 sin A sin B½ [cos(A−B) − cos(A+B)] = sin A sin B

To get the other two product‐to sum formulas, add the two sine formulas from equation 48 and equation49, or subtract them. Here are all four formulas together:

cos A cos B = ½ cos(A−B) + ½ cos(A+B)

sin A sin B = ½ cos(A−B) − ½ cos(A+B)

sin A cos B = ½ sin(A+B) + ½ sin(A−B)

cos A sin B = ½ sin(A+B) − ½ sin(A−B)

The fourth one of those formulas really isn’t needed, because you can always evaluate cos p sin q assin q cos p. But it’s traditional to present all four formulas.

Sum to Product

There are also formulas that combine a sum or difference into a product. Heon Joon Choi, a physicsstudent from Cornell, has kindly told me of an application: “superposing two waves and trying to Figureout the nodes is much easier if they are multiplied, rather than added.” This makes sense: solving mostequations is easier once you’ve factored them. The sum‐to‐product formulas are also used to prove theLaw of Tangents, though that itself is no longer used in solving triangles.

Here’s how to get the sum‐to‐product formulas. First make these deFinitions:A = ½(u+v), and B = ½(u−v)

Then you can see that

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A+B = u, and A−B = vNow make those substitutions in all four formulas of equation 52, and after simplifying you will have thesum‐to‐product formulas:

cos u + cos v = 2 cos(½(u+v)) cos(½(u−v))

cos u − cos v = −2 sin(½(u+v)) sin(½(u−v))

sin u + sin v = 2 sin(½(u+v)) cos(½(u−v))

sin u − sin v = 2 sin(½(u−v)) cos(½(u+v))

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Double Angle and Half Angle Formulas

Very often you can simplify your work by expanding something like sin(2A) or cos(½A)into functions of plain A. Sometimes it works the other way and a complicated

expression becomes simpler if you see it as a function of half an angle or twice an angle.The formulas seem intimidating, but they’re really just variations on equation 48 andequation 50.

Sine or Cosine of a Double Angle

With equation 48, you can Find sin(A+B). What happens if you set B=A?sin(A+A) = sin A cos A + cos A sin A

But A+A is just 2A, and the two terms on the right‐hand side are equal. Therefore:sin(2A) = 2 sin A cos A

The cosine formula is just as easy:cos(A+A) = cos A cos A − sin A sin Acos(2A) = cos²A − sin²A

Though this is valid, it’s not completely satisfying. It would be nice to have a formula for cos(2A) in termsof just a sine or just a cosine. Fortunately, we can use sin²x + cos²x = 1 to eliminate either the sine or thecosine from that formula:

cos(2A) = cos²A − sin²A = cos²A − (1 − cos²A) = 2 cos²A − 1cos(2A) = cos²A − sin²A = (1 − sin²A) − sin²A = 1 − 2 sin²A

On different occasions you’ll have occasion to use all three forms of the formula for cos(2A). Don’t worrytoo much about where the minus signs and 1s go; just remember that you can always transform any ofthem into the others by using good old sin²x + cos²x = 1.

sin(2A) = 2 sin A cos A

cos(2A) = cos²A − sin²A = 2 cos²A − 1 = 1 − 2 sin²A

There’s a very cool second proof of these formulas, using Sawyer’s marvelous idea. Also, there’s an easyway to Find functions of higher multiples: 3A, 4A, and so on.

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Tangent of a Double Angle

To get the formula for tan(2A), you can either start with equation 50 and put B = A to get tan(A+A), or useequation 59 for sin(2A) / cos(2A) and divide both parts of the fraction by cos²A. Either way, you get

tan(2A) = 2 tan A / (1 − tan²A)

Sine or Cosine of a Half Angle

What about the formulas for sine, cosine, and tangent of half an angle? Since A = (2A)/2, you mightexpect the double‐angle formulas equation 59 and equation 60 to be some use. And indeed they are,though you have to pick carefully.

For instance, sin(2A) isn’t much help. Put A = B/2 and you havesin B = 2 sin(B/2) cos(B/2)

That’s true enough, but there’s no easy way to solve for sin(B/2) or cos(B/2).

There’s much more help in equation 59 for cos(2A). Put 2A = B or A = B/2 and you getcos B = cos²(B/2) − sin²(B/2) = 2 cos²(B/2) − 1 = 1 − 2 sin²(B/2)

Use just the First and last parts of that:cos B = 1 − 2 sin²(B/2)

Rearrange a bit:sin²(B/2) = (1 − cos B) / 2

and take the square rootsin(B/2) = ± sqrt([1 − cos B] / 2)

You need the plus or minus sign because sin(B/2) may be positive or negative, depending on B. For anygiven B (or B/2) there will be only one correct sign, which you already know from the diagram that weexplored back in Functions of Any Angle.

Example: If B = 280°, then B/2 = 140°, and you know that sin 140° is positive because the angle is inQuadrant II (above the axis).

To Find cos(B/2), start with a different piece of the cos(2A) formula from equation 59:cos(2A) = 2 cos²A − 1

As before, put A = B/2:cos B = 2 cos²(B/2) − 1

Rearrange and solve for cos(B/2):cos²(B/2) = [1 + cos B]/2cos(B/2) = ± sqrt([1 + cos B] / 2)

You have to pick the correct sign for cos(B/2) depending on the value of B/2, just as you did with

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sin(B/2). But of course the sign of the sine is not always the sign of the cosine.

sin(B/2) = ± sqrt([1 − cos B] / 2)

cos(B/2) = ± sqrt([1 + cos B] / 2)

Tangent of a Half Angle

You can Find tan(B/2) in the usual way, dividing sine by cosine from equation 61:tan(B/2) = sin(B/2)/cos(B/2) = ± sqrt([1 − cos B]/[1 + cos B])

In the sine and cosine formulas we couldn’t avoid the square roots, but in this tangent formula we can.Multiply top and bottom by sqrt(1+cos B):

tan(B/2) = sqrt([1 − cos B] / [1 + cos B])tan(B/2) = sqrt([1−cos B][1+cos B] / [1+cos B]²)tan(B/2) = [sqrt(1−cos²B)] / [1+cos B]

Then use equation 38, your old friend: sin²x + cos²x = 1;tan(B/2) = sqrt(sin²B) / (1+cos B) = sin B / (1+cos B)

If you had multiplied top and bottom by sqrt(1−cos B) instead of sqrt(1+cos B), you would gettan(B/2) = sin(B/2)/cos(B/2) = ± sqrt([1−cos B]/[1+cos B])tan(B/2) = sqrt([1−cos B] / [1+cos B])tan(B/2) = sqrt([1−cos B]² / [1+cos B][1−cos B])tan(B/2) = (1−cos B) / sqrt(1−cos²B)tan(B/2) = (1−cos B) / sqrt(sin²B) = (1−cos B) / sin B

The half‐angle tangent formulas can be summarized like this:

tan(B/2) = (1 − cos B) / sin B = sin B / (1 + cos B)

You may wonder what happened to the plus or minus sign in tan(B/2). Fortuitously, it drops out. Sincecos B is always between −1 and +1, (1 − cos B) and (1 + cos B) are both positive for any B. And the sine ofany angle always has the same sign as the tangent of the corresponding half‐angle.

Don’t take my word for that last statement, please. There are only four possibilities, and they’re easyenough to work out in a table. (Review interval notation if you need to.)

B/2 Q I, (0°;90°) Q II, (90°;180°) Q III, (180°;270°) Q IV, (270°;360°)

tan(B/2) + − + −

B (0°;180°), Q I or II (180°;360°), Q III or IV (360°;540°), Q I or II (540°;720°), Q III or IV

sin B + − + −

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1−cos B + + + +

Of course, you can ignore the whole matter of the sign of the sine and just assign the proper sign whenyou do the computation.

Another question you may have about equation 62: what happens if cos B = −1, so that (1 + cos B) = 0?Don’t we have division by zero then? Well, take a little closer look at those circumstances. The angles Bfor which cos B = −1 are ±180°, ±540°, and so on. But in this case the half angles B/2 are ±90°, ±270°, andso on: angles for which the tangent is not deFined anyway. So the problem of division by zero neverarises.

And in the other formula, sin B = 0 is not a problem. Excluding the cases where cos B = −1, thiscorresponds to B = 0°, ±360°, ±720°, etc. But the half angles B/2 are 0°, ±180°, ±360°, and so on. For all ofthem, tan(B/2) = 0, as you can verify from the second half of formula equation 62.

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Summary:


Inverse Functions

The inverse trig functions (also called arcfunctions) are similar to any other inversefunctions: they go from the function value back to the angle (or number). Theirranges are restricted, by deFinition, because an inverse function must not give multipleanswers. Once you understand the inverse functions, you can simplify composite

functions like sin(Arctan x).

Sometimes you have a sine or cosine or tangent and need to Find the associated angle. For instance, thishappens whenever you solve a triangle. When you have a sine function value and )ind the

corresponding angle, we say you are Finding the arcsine or inverse sine of that value, and similarly forthe other functions.

Different books use different notation: sin with a superscript −1, or arcsin. I prefer the “arc” formsbecause the superscript −1 looks too much like an exponent.

You may be less intimidated by the arcfunctions if you pronounce them as the angle whose. Forinstance, arccos 0.6 becomes “the angle whose cosine is 0.6”.

Principal Values

What is arcsin(0.5)? You probably recognize that 0.5 is ½, and it must be a sine of one of the specialangles. In fact, equation 15 tells you that sin(30°) = ½. So you can say that arcsin(0.5) = 30° or π/6.

But wait, there’s more! You know from equation 22 thatsin(30°) = sin(150°) = sin(390°) = sin(−210°)

and so on, and therefore they all equal ½. In fact,sin(30° + 360°k) = sin(150° + 360°k) = 0.5 for all integer k

So which of this inFinite number of values is the arcsine?To make an arcsine function, we have to restrict the range (the possible answers given by each

function) so that each number has at most one arcsine. (Why do I say “at most one” and not “one andonly one”? The sine of any angle is between −1 and +1 inclusive; therefore only those numbers havearcsines. There is o angle whose sine is 2, so 2 has no arcsines and arcsin 2 is a meaningless noise.)

The arcsine is deFined so that its range is the interval [−π/2;+π/2], which is the same as [−90°;+90°].Some books use a capital letter for the function Arcsin distinguish it from the multi‐valued relationarcsin. So we could say

arcsin(0.5) = π/6+2kπ or 5π/6+2kπ for all integer kbut

Arcsin(0.5) = π/6Why the particular range −π/2 to +π/2? To start with, it seems tidy that any arcfunction of a positive

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number should fall in Quadrant I, [0;+π/2]. So the only real question is arcfunctions of negative numbers.If we prefer the numerically smallest values for the arcsine function, then Arcsin(−0.5) = −30° = −π/6 Fitsthat rule, and a negative number’s arcsine (and arctangent, too) will be in Quadrant IV, [−π/2;0].

What about the arccosine? The cosine is positive in both Quadrant I and Quadrant IV, so the arccosine ofa negative number must fall in Quadrant II or Quadrant III. Thomas (Calculus and Analytic Geometry, 4thedition) resolves this in a neat way. Remember from equation 2 that

cos A = sin(π/2 − A)It makes a nice symmetry to write

Arccos x = π/2 − Arcsin xAnd that is how Thomas deFines the inverse cosine function. Since the range of Arcsin is the closedinterval [−π/2;+π/2], the range of Arccos is π/2 minus that, or [0;π].

You can remember the range of the arctangent function this way: the only two choices that make senseare 0 to π and −π/2 to +π/2. But tan(π/2) is undeFined, so if you picked [0;π] for the range of Arctan itwould have an ugly hole in it at π/2. Therefore the range is deFined to be (−π/2;+π/2), an open intervalbut with no points missing inside.

Once the range for Arctan is deFined, there’s really only one sensible way to deFine Arccot:cot x = tan(π/2 − x) ⇒ Arccot x = π/2 − Arctan x

which gives the single open interval (0;π) as the range.

Thomas deFines the arcsecant and arccosecant functions using the reciprocal relationships fromequation 5:

sec x = 1/(cos x) ⇒ Arcsec x = Arccos(1/x)csc x = 1/(sin x) ⇒ Arccsc x = Arcsin(1/x)

This means that Arcsec and Arccsc have the same ranges as Arccos and Arcsin, respectively.Here are the domains (inputs) and ranges (outputs) of all six inverse trig functions:

function derived from domain range

Arcsin inverse of sine function [−1; +1] [−π/2; +π/2]

Arccos Arccos x = π/2 − Arcsin x [−1; +1] [0; π]

Arctan inverse of tangent function all reals (−π/2; +π/2)

Arccot Arccot x = π/2 − Arctan x all reals (0; π)

Arcsec Arcsec x = Arccos(1/x) (−∞; −1], [1; ∞) [0; π]

Arccsc Arccsc x = Arcsin(1/x) (−∞; −1], [1; ∞) [−π/2; +π/2]

Remember that the relations are many‐valued, not limited to the above ranges of the functions. If you seethe capital A in the function name, you know you’re talking about the function; otherwise you have todepend on context.

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Functions of Arcfunctions

Advice to the reader: The formulas in this section aren’t really very useful in trigonometry itself, but areused in integral calculus and some physics or engineering courses. You may wish to skip them, especiallyon a First reading.

Sometimes you have to evaluate expressions likecos(Arctan x)

That looks scary, but actually it’s a piece of cake. You can simplify any trig function of any inverse

function in two easy steps, using this method:

Think of the inner arcfunction as an angle. Draw a right triangle and label that angle and thetwo relevant sides.

1.

Use the Pythagorean Theorem to Find the third side of the triangle, then write down the valueof the outer function according to its deFinition.

2.

(If the answer contains variables raised to odd powers, you may need to add some absolute value signs.See Example 3.)

Example 1: cos(Arctan x)

It may be helpful to read the expression out in words: “the cosine of Arctan x.” Doesn’t help much? Well,remember what Arctan x is. It’s the (principal) angle whose tangent is x. So what you have to Find readsas “the cosine of the angle whose tangent is x.” And that suggests your plan of attack: )irst identify that

angle, then Find its cosine.

Let’s give a name to that “angle whose”. Call it A:A = Arctan x

from which you know thattan A = x

Now all you have to do is Find cos A, and that’s easy if you draw a littlepicture.

Start by drawing a right triangle, and mark one acute angle as A.Using the deFinition of A, write down the lengths of two sides of the

triangle. Since tan A = x, and the deFinition of tangent is opposite side overadjacent side, the simplest choice is to label the opposite side x and theadjacent side 1. Then, by deFinition, tan A = x/1 = x, which is how we

needed to set up angle A.

The next step is to Find the third side. Here you know the two legs, so youuse the theorem of Pythagoras to Find the hypotenuse, sqrt(1+x²). (For

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some problems, you’ll know one leg and the hypotenuse, and you’ll use thetheorem to Find the other leg.)

Once you have all three sides’ lengths, you can write down the valueof any function of A. In this case you need cos A, which is adjacent sideover hypotenuse:

cos A = 1/sqrt(1+x²)But cos A = cos(Arctan x). Therefore

cos(Arctan x) = 1/sqrt(1+x²)and there’s your answer.

Example 2: cos(Arcsin x)

Read this as “the cosine of the angle A whose sine is x”. Draw your triangle, and label angle A. (Pleasetake a minute and make the drawing.) You know from equation 1 that

sin A = x = opposite/hypotenuseand therefore you label the opposite side x and the hypotenuse 1.

Next, solve for the third side, which is sqrt(1−x²), and write that down. Now you need cos A, which isthe adjacent side over the hypotenuse, which is sqrt(1−x²)/1. Answer:

cos(Arcsin x) = sqrt(1−x²)There you go: quick and painless.

Example 3: cos(Arctan 1/x))

This looks similar to Example 1, but as you’ll see there’s an additional wrinkle. (Thanks to Brian Scott,who raised the issue, albeit inadvertently, in his article “Re: Expression” on 10 Dec 2000 inalt.algebra.help.)

Begin in the regular way by drawing your triangle. Since A =Arctan(1/x), or tan A = 1/x, you make 1 the length of the opposite side andx the length of the adjacent side. The hypotenuse is then sqrt(1+x²).

Now you can write down cos A, which is adjacent over hypotenuse:cos A = x / sqrt(1+x²)cos(Arctan 1/x) = x / sqrt(1+x²)

But this example has a problem that does not occur in the earlier examples.Suppose x is negative, say −√3. Then Arctan(−1/√3) = −π/6, and cos(−π/6) = +(√3)/2. But the

answer in the previous paragraph, x/sqrt(1+x²), yields −√3/√(1+3) = −(√3)/2, which has the wrong sign.What went wrong? The trouble is that Arctan yields values in (−π/2;+π/2), which is Quadrants IV

and I. But the cosine is always positive on that interval. Therefore cos(Arctan x) always yields a positiveresult. Remember also from equation 22 that cos(−A) = cos A. To ensure this, use the absolute value sign,and the true Final answer is

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cos(Arctan(1/x)) = | x | / sqrt(1+x²)Why doesn’t every example have this problem? The earlier examples involved only the square of avariable, which is naturally nonnegative. Only here, where we have an odd power, does it matter.

Summary: When your answer contains an odd power (1, 3, 5, etc.) of a variable, you must add a third

step to the process: carefully examine the signs and adjust your answer so that it has the correct sign forboth positive and negative values of the variable.

Arcfunctions of Functions

Advice to the reader: The formulas in this section are for the really hard‐core trig fan. They aren’t reallyvery useful in trigonometry itself, but are used in integral calculus and some physics or engineeringcourses. You may wish to skip them, especially on a First reading.

You may be wondering about the inside‐out versions, taking the arcfunction of a function. Some ofthese expressions can be solved algebraically, on a restricted domain at least, but some cannot. (I amgrateful to David Cantrell for help with analysis of these problems in general and Example 6 inparticular.)

We can say at once that there will be no pure algebraic equivalent to an arcfunction of a trigfunction. This means there will be no nice neat procedure as there was for functions of arcfunctions

Why? The six trig functions are all periodic, and therefore any function of any of them must also beperiodic. But no algebraic functions are periodic, except trivial ones like f(x) = 2, and therefore nofunction of a trig function can be represented by purely algebraic operations. As we will see, some can berepresented if we add non‐algebraic functions like mod and Floor.

Example 4: Arccos(sin u)

This is the angle whose cosine is sin u. To come up with a simpler form, set x equal to the desiredexpression, and solve the equation by taking cosine of both sides:

x = Arccos(sin u)cos x = sin u

This could be solved if we could somehow transform it to sin(something) = sin(u) or cos(x) =cos(something else). In fact, we can use equation 2 to do that. It tells us that

sin u = cos(π/2 − u)and combining that with the above we have

cos x = cos(π/2 − u)Now if x is in Quadrant I, the interval [0;π/2], then u will be in Quadrant I also, and we can write

x = π/2−uand therefore

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Arccos(sin u) = π/2−u for u in Quadrant I

But this solution does not work for all quadrants. For instance, try a number from Quadrant II:Arccos(sin(5π/6)) = Arccos(½) = π/6

butπ/2 − 5π/6 = −π/3

Obviously π/2−u isn’t a general solution for Arccos(sin u). Try graphing Arccos(sin x) and π/2−x and you’llsee the problem: one is a sawtooth and the other is a straight line.

Sparing you the gory details, π/2−u is right only in Quadrants IV and I. We have to “decorate” itrather a lot to make it match Arccos(sin u) in the other quadrants, and also to account for the repetitionof values every 2π. The First modiFication is not too hard: On the interval [−π/2;+3π/2], theabsolute‐value expression |π/2−u| matches the sawtooth graph of Arccos(sin u).

The repetition every 2π is harder to reFlect, but this manages it:Arccos(sin u) = | π/2 − u + 2π*Floor[(u+π/2)/2π] |

where “Floor” means the greatest integer less than or equal to. Messy, eh? (Note also that “Floor” is not analgebraic function.)

It could be made a bit shorter with mod (which is also not algebraic):Arccos(sin u) = | π − mod(u+π/2,2π) |

where mod(a,b) is the nonnegative remainder when a is divided by b.

Example 5: Arcsec(cos u)

This one, the angle whose secant is cos u, has a very odd solution. Try the solution method from Example4 and you get

x = Arcsec(cos u)sec x = cos u

But sec x = 1/cos x, and therefore1/( cos x ) = cos u

Now think about that equation. The cosine’s values are all between −1 and +1. So the only way one cosinecan be the reciprocal of another is if they’re both equal to 1 or both equal to −1; no other solutions exist.

First case: If cos u = 1 then u is an even multiple of π. But Arcsec 1 = 0, and thereforeArcsec(cos u) = 0 when u = 2kπ

Second case: If cos u = −1 then u is an odd multiple of π. But Arcsec(−1) = π, and thereforeArcsec(cos u) = π when u = (2k+1)π

If u is not a multiple of π, cos u will be less than 1 and greater than −1. The Arcsec function is not deFinedfor such values, and therefore

Arcsec(cos u) does not exist when u is not a multiple of π

The graph of Arcsec(cos u) is rather curious: single points at the ends of an inFinite sawtooth: ..., (−3π,π),(−2π,0), (−π,π), (0,0), (π,π), (2π,0), (3π,π), ...

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Example 6: Arctan(sin u)

Proceeding in the regular way, we havex = Arctan(sin u)tan x = sin u

The most likely approach is the one from Example 4: try to transform the above into tan(x) =tan(something) or sin(something else) = sin(u).

If there is any trig identity or combination that can be used to do that, it is unknown to me. I suspectstrongly that Arctan(sin u) can’t be converted to an algebraic expression, even with the use of mod orFloor, but I can’t prove it.

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Trig without Tears:

Notes and Digressions

The Problem with Memorizing

Dad sighed. “Kip, do you think that table was brought down from on high by an archangel?”Robert A. Heinlein, in Have Space Suit—Will Travel (1958)

It’s not just that there are so many trig identities; they seem so arbitrary. Of course they’re not reallyarbitrary, since all can be proved; but when you try to memorize all of them they seem like a jumble of

symbols where the right ones aren’t more obviously right than the wrong ones. For example, is itsec²A = 1 + tan²A or tan²A = 1 + sec²A ? I doubt you know off hand which is right; I certainly don’tremember. Who can remember a dozen or more like that, and remember all of them accurately?

Too many teachers expect (or allow) students to memorize the trig identities and parrot them ondemand, much like a series of Bible verses. In other words, even if they’re originally taught as a series ofconnected propositions, they’re remembered and used as a set of unrelated facts. And that, I think, is theproblem. The trig identities were not brought down by an archangel; they were developed bymathematicians, and it’s well within your grasp to re‐develop them when you need to. With effort, wecan remember a few key facts about anything. But it’s much easier if we can Fit them into a context, sothat the identities work together as a whole.

Why bother? Well, of course it will make your life easier in trig class. But you’ll also need the trigidentities in later math classes, especially calculus, and in physics and engineering classes. In all of those,you’ll Find the going much easier if you’re thoroughly grounded in trigonometry as a uniFied Field ofknowledge instead of a collection of unrelated facts.

This is why it’s easier to remember almost any song than an equivalent length of prose: the songgives you additional cues in the rhythm, common patterns of emphasis, and usually rhymes at the ends oflines. With prose you have only the general thought to hold it together, so that you must memorize itessentially as a series of words. With the song there are internal structures that help you, even if you’renot aware of them.

If you’re memorizing Lincoln’s Gettysburg Address, you might have trouble remembering whetherhe said “recall” or “remember” at a certain point; in a song, there’s no possible doubt which of thosewords is right because the wrong one won’t Fit in the rhythm.

On the Other Hand ...

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I’m not against all memorization. Some things have to be memorized because they’re a matter ofdeFinition. Others you may choose to memorize because you use them very often, you’re conFident youcan memorize them correctly, and the derivation takes more time than you’re comfortable with. Stillothers you may not set out to memorize, but after using them many times you Find you’ve memorizedthem without trying to—much like a telephone number that you dial often.

I’m not against all memorization; I’m against needless memorization used as a substitute for

thought. If you decide in particular cases that memory works well for you, I won’t argue. But I do hopeyou see the need to be able to re‐derive things on the spot, in case your memory fails. Have you everdialed a friend’s telephone number and found you couldn’t quite remember whether it was 6821 or8621? If you can’t remember a phone number, you have to look it up in the book. My goal is to free you

from having to look up trig identities in the book.Thanks to David Dixon at [email protected] for an illuminating exchange of notes on this topic.

He made me realize that I was sounding more more anti‐memory than I intended to, and in consequenceI’ve added this note. But he may not necessarily agree with what I say here.

I wrote these pages to show you how to make the trig identities “Fit” as a coherent whole, so thatyou’ll have no more doubt about them than you do about the words of a song you know well. Thedifference is that you won’t need to do it from memory. And you’ll gain the sense of power that comesfrom mastering your subject instead of groping tentatively and hoping to strike the right answer by goodluck.

Proof of Euler’s Formula

Euler’s formula (equation 47) is easily proved by means of power series. Start with the formulas

e = SUM [ x / n! ] = 1 + x + x /2! + x /3! + ...

cos x = SUM [ (−1) x / (2n)! ] = 1 − x /2! + x /4! − x /6! + ...

sin x = SUM [ (−1) x / (2n+1)! ] = x − x /3! + x /5! − x /7! + ...

(These are how the function values are actually calculated, by the way. If you want to know the value ofe , you just substitute 2 for x in the formula and compute until the additional terms fall within yourdesired accuracy.)

Now we have to Find the value of e , where i = √(−1). Use the First formula to Find e , by substituting ixfor x:

e = SUM [ (ix) / n! ]e = 1 + (ix) + (ix) /2! + (ix) /3! + (ix) /4! + (ix) /5! + (ix) /6! + (ix) /7! + ...

Simplify the powers of i, using i² = −1:e = 1 + ix − x /2! − ix /3! + x /4! + ix /5! − x /6! − ix /7! + ...

Finally, group the real and imaginary terms separately:e = [1 − x /2! + x /4! − x /6! + ...] + i[x − x /3! + x /5! − x /7! + ...]

x n 2 3

n 2n 2 4 6

n 2n+1 3 5 7

2

ix ix

ix n

ix 2 3 4 5 6 7

ix 2 3 4 5 6 7

ix 2 4 6 3 5 7

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Those should look familiar. If you refer back to the power series at the start of this section you’ll see thatthe First group of terms is just cos x and the second group is just sin x. So you have

e = cos x + i sin xwhich is Euler’s formula, as advertised!

You may wonder where the series for cos x, sin x, and e come from. The answer is that they are theTaylor series expansions of the functions. (You’ll probably study Taylor series in second‐ or third‐semester calculus.)

Polar Form of a Complex Number

As you may remember, complex numbers like 3+4i and 2−7i are oftenplotted on a complex plane. This makes it easier to visualize adding andsubtracting. The illustration at right shows that

(3+4i) + (2−7i) = 5−3i

It turns out that for multiplying, dividing,and Finding powers and roots, complexnumbers are easier to work with in polar

form. This means that instead of thinking about the real and imaginarycomponents (the “−3” and “5” in −3+5i), you think of the length anddirection of the line.

The length is easy, just the good old Pythagorean theorem in fact:r = √(3²+5²) = √34 ≅ 5.83

(The length r is called the absolute value or modulus of the complex number.)

But what about the direction? You can see from the picture that θ about 120° or about 2 radians(measured counterclockwise from the positive real axis, which points east), but what is it exactly? Well,you know that tan θ = −5/3. If you take Arctan(−5/3) on your calculator, you get −1.03, and adding π toget into the proper quadrant gives θ = 2.11 radians. (In degrees, Arctan(−5/3) = −59.04, and adding 180gives θ = 120.96°.) You can write it this way:

−3+5i ≅ 5.83 e or 5.83[cos 2.11 + i sin 2.11] or 5.83 cis 2.11 or 5.83∠120.96°

There’s a nice trick that Finds the angle in the correct quadrant automatically:θ = 2 Arctan(y/(x+r))

This builds in the adjustment to the correct quadrant, so you never have to worry about whether to addor subtract 180° or π. If you test 2 Arctan(5/(−3+√34)) on your calculator, you get 2.11 radians or120.96° as before.

This formula is particularly handy when you’re writing a program or spreadsheet formula to Find θ.It Flows directly from the formulas for tangent of half an angle, where the functions are expressed interms of x, y, and r. A July 2003 article by Rob Johnson, archived here, inspired me to add this section.

ix

x

2.11i

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Powers and Roots of a Complex Number

One of the applications of Euler’s formula (equation 47) is Finding any power or root of any complexnumber. (Sawyer doesn’t do this, or at least not in the same book.) It’s not hard to develop the formula,using just Euler’s formula and the laws of exponents.

Start by writing down Euler’s formula, multiplied left and right by a scale factor r:r (cos x + i sin x) = r e

Next raise both sides to the Nth power:[r (cos x + i sin x)] = [r e ]

Let’s leave the left‐hand side alone for a while, and simplify the right‐hand side. First, by the laws ofexponents

[r e ] = r eApply Euler’s formula to the right‐hand side and you have

[r e ] = r (cos Nx + i sin Nx)Connect that right‐hand side to the original left‐hand side and you have DeMoivre’s theorem:

[r(cos x + i sin x)] = r (cos Nx + i sin Nx)

This tells you that if you put a number into polar form, you can Find any power or root of it. (Rememberthat the nth root of a number is the same as the 1/n power.)

Square Root of i

For instance, you know the square roots of −1 are i and −i, but what’s the square root of i? You can useDeMoivre’s theorem to Find it.

First, put i into e form (polar form), using the above technique. i = 0+1i. cos x = 0 and sin x = 1 whenx = 90° or π/2. Therefore

i = cos(π/2) + i sin(π/2)√i = i = [cos(π/2) + i sin(π/2)]

And by equation 82√i = cos(½×π/2) + i sin(½×π/2)√i = cos(π/4) + i sin(π/4)√i = 1/√2 + i×1/√2√i = (1+i)/√2

The other square root is minus that, as usual.

ix

N ix N

ix N N iNx

ix N N

N N

ix

½ ½

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Principal Root of Any Number

You can Find any root of any complex number in a similar way, but usually with one preliminary step.For instance, suppose you want the cube roots of 3+4i. The First step is to put that number into polar

form. The absolute value is sqrt(3²+4²) = 5, and you use the Arctan technique given above to Find theangle θ = 2 Arctan(4/(3+5)) = 2 Arctan(½) ≅ 53.13° or 0.9273 radians. The polar form is

3+4i ≅ 5(cos 53.13° + i sin 53.13°)To take a cube root of that, use equation 82 with N = 1/3:

cube root of (3+4i) = 5 [cos(53.13°/3) + i sin(53.13°/3)]The cube root of 5 is about 1.71. Dividing the angle θ by 3, 53.13°/3 ≅ 17.71°; the cosine and sine of thatare about 0.95 and 0.30. Therefore

cube root of (3+4i) ≅ 1.71 × (cos 17.71° + i sin 17.71°)cube root of (3+4i) ≅ 1.63 + 0.52i

Multiple Roots

You may have noticed that I talked about “the cube roots [plural] of 3+4i” and what we found was “a

cube root”. Even with the square root of i, I waved my hand and said that the “other” square root wasminus the First one, “as usual”.

You already know that every positive real has two square roots. In fact, every complex number has

n nth roots.

How can you Find them? It depends on the periodic nature of the cosine and sine functions. To start, lookback at Euler’s formula,

e = cos x + i sin xWhat happens if you add 2π or 360° to x? You have

e = cos(x+2π) + i sin(x+2π)But taking sine or cosine of 2π plus an angle is exactly the same as taking sine or cosine of the originalangle. So the right‐hand side is equal to cos x + i sin x, which is equal to e . Therefore

e = eIn fact, you can keep adding 2π or 360° to x as long as you like, and never change the value of the result.Symbolically,

e = e for all integer kWhen you take an nth root, you simply use that identity.

Getting back to the cube roots of 3+4i, recall that that number is the same as 5e , where θ is about 53.13°or 0.9273 radians. The three cube roots of e are

e , e , and e for k = 0,1,2or

e , e , and e for k = 0,1,2Compute those as cos x + i sin x in the usual way, and then multiply by the (principal) cube root of 5. I get

1/3

ix

i(x+2π)

ix

i(x+2π) ix

i(x+2πk) ix

iθ

iθ

iθ/3 i(θ+2π)/3 i(θ+4π)/3

iθ/3 i(θ/3+2π/3) i(θ/3+4π/3)

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these three roots:cube roots of 3+4i ≅ 1.63+0.52i, −1.26+1.15i, −0.36−1.67i

More generally, we can extend DeMoivre’s theorem (equation 82) to show the N Nth roots of anycomplex number:

[r (cos x + i sin x] = r [cos(x/N+2πk/N) + i sin(x/N+2πk/N)] for k = 0,1,2,...,N−1

Logarithm of a Negative Number

Another application Flows from a famous special case of Euler’s formula. Substitute x = π or 180° inequation 47. Since sin 180° = 0, the imaginary term drops out. And cos 180° = −1, so you have the famousformula

−1 = e

It’s interesting to take the natural log of both sides:ln(−1) = ln(e )

which givesln(−1) = iπ

It’s easy enough to Find the logarithm of any other negative number. Sinceln(ab) = ln a + ln b

then for all real a you haveln(−a) = ln[a × −1] = ln a + ln(−1) = ln a + iπ

ln(−a) = ln a + iπ

I don’t honestly know whether all of this has any practical application. But if you’ve ever wondered aboutthe logarithm of a negative number, now you know.

Cool Proof of Double-Angle Formulas

I can’t resist pointing out another cool thing about Sawyer’s marvelous idea: you can also use it to provethe double‐angle formulas equation 59 directly. From Euler’s formula for e you can immediately obtainthe formulas for cos(2A) and sin(2A) without going through the formulas for sums of angles. Here’s how.

(1/N) (1/N)

iπ

iπ

ix

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Remember the laws of exponents: x = (x ) . One important special case is that x = (x )². Use thatwith Euler’s formula (equation 47):

cos(2A) + i sin(2A) = e = (e )² = (cos A + i sin A)² = cos²A + 2i sinA cosA + i²sin²A = cos²A + 2i sinA cosA − sin²A = cos²A − sin²A + 2i sinA cosA

Since the real parts on left and right must be equal, you have the formula for cos(2A). Since the imaginaryparts must be equal, you have the formula for sin(2A). That’s all there is to it.

Multiple-Angle Formulas

The above technique is even more powerful for deriving formulas for functions of 3A, 4A, or any multipleof angle A. To derive the formulas for nA, expand the nth power of (cos A + i sin A), then collect real andimaginary terms.

This was inspired by a May 2009 reading of Paul Nahin’s An Imaginary Tale: The Story of √−1(Princeton, 1998).

Just to show the method, I’ll derive the functions of 3A and 4A. You can try the brute‐force approach ofcos(2A+2A) and sin(2A+2A) and see how much effort my method saves.

Functions of 3A

De Moivre’s theorem (equation 82) tells us thatcos 3A + i sin 3A = (cos A + i sin A)³

Expand the right‐hand side via the binomial theorem, remembering that i² = −1 and i³ = −i:cos 3A + i sin 3A = cos³A + 3 i cos²A sin A − 3 cos A sin²A − i sin³A

Collect real and imaginary terms:cos 3A + i sin 3A = cos³A − 3 cos A sin²A + 3 i cos²A sin A − i sin³Acos 3A + i sin 3A = [cos A (cos²A − 3 sin²A)] + i [sin A (3 cos²A − sin²A)]

Set the real part on the left equal to the real part on the right, and similarly for the imaginary parts, andyou have the formulas for the cosine and sine of 3A:

cos 3A = cos A (cos²A − 3 sin²A) and sin 3A = sin A (3 cos²A − sin²A)

The tangent formula is easy to get: just divide.tan 3A = sin 3A / cos 3Atan 3A = [sin A (3 cos²A − sin²A)] / [cos A (cos²A − 3 sin²A)]tan 3A = tan A (3 cos²A − sin²A) / (cos²A − 3 sin²A)

ab a b 2b b

(2A)i

iA

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Divide top and bottom by cos²A to simplify:tan 3A = tan A (3 − tan²A) / (1 − 3 tan²A)

cos 3A = cos A (cos²A − 3 sin²A)

sin 3A = sin A (3 cos²A − sin²A)

tan 3A = tan A (3 − tan²A) / (1 − 3 tan²A)

Functions of 4A

It’s no more work to Find the functions of 4A. I’ll just run through the steps without commentary. First,cosine and sine of 4A:

cos 4A + i sin 4A = (cos A + i sin A)cos 4A + i sin 4A = cos A + 4 i cos³A sin A − 6 cos²A sin²A − 4 i cos A sin³A + sin Acos 4A + i sin 4A = cos A − 6 cos²A sin²A + sin A + 4 i cos³A sin A − 4 i cos A sin³Acos 4A = cos A − 6 cos²A sin²A + sin A and sin 4A = 4 cos³A sin A − 4 cos A sin³Acos 4A = (cos²A − sin²A)² − 4 sin²A cos²A and sin 4A = 4 sin A cos A (cos²A − sin²A)

Now the tangent formula:tan 4A = sin 4A / cos 4Atan 4A = [4 sin A cos A (cos²A − sin²A)] / [(cos²A − sin²A)² − 4 sin²A cos²A]

Divide top and bottom by cos A:tan 4A = 4 tan A (1 − tan²A) / [(1 − tan²A)² − 4 tan²A]

cos 4A = (cos²A − sin²A)² − 4 sin²A cos²A

sin 4A = 4 sin A cos A (cos²A − sin²A)

tan 4A = 4 tan A (1 − tan²A) / [(1 − tan²A)² − 4 tan²A]

Great Book on Problem Solving

I have to recommend a terriFic little book, How To Solve It by G. Polya. Most teachers aren’t very good atteaching you how to solve problems and do proofs. They show you how they do them, and expect you topick up their techniques by a sort of osmosis. But most of them aren’t very good at explaining the thoughtprocess that goes into doing a geometrical proof, or solving a dreaded “story problem”.

Polya’s book does a great job of teaching you how to solve problems. He shows you the kinds ofquestions you should ask yourself when you see a problem. In other words, he teaches you how to getyourself over the hum, past the Floundering that most people do when they see an unfamiliar problem.

4

4 4

4 4

4 4

4

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And he does it with lots of examples, so that you can develop conFidence in your techniques and compareyour methods with his. The techniques I’ve mentioned above are just three out of the many in his book.

There’s even a handy checklist of questions you can ask yourself whenever you’re stuck on aproblem.

How To Solve It was First published in 1945, and it’s periodically in and out of print. If you can’t get itfrom your bookstore, go to the library and borrow a copy. You won’t be sorry.

What’s New

25 Jan 2012: Correct a substitution in the First attempt at a derivation of the Law of Cosines, thanks toMark Schoonover.

5 Jul 2010: Add a disclaimer about complex trigonometry, following a note from Altan Kartaltepe.

19 May 2010: Add link to a Youtube video mnemonic for the basic trig identities.

25 Aug 2009: Correct typo in Reference Angles: 0 to 90° was listed as 0 to π/4, and that should havebeen 0 to π/2.

25 May 2009: Add Multiple‐Angle Formulas.

(intervening changes suppressed)

19 Feb 1997: new document

this page: http://oakroadsystems.com/twt/pftwt.htm

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trig without tears

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