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Math 20 – 1: Trigonometry

*** Your calculator must be in DEGREE mode during this unit!

DAY 1 – Angles in Standard Position

Standard Position

· Angles are placed in standard position so they can be compared.

· Angles in standard position all have the same perspective and relate to both the x and y axis.

Examples:

Angles not in Standard PositionAngle in Standard Position

(terminal arminitial armvertex)

Putting an Angle in Standard Position

· There are 2 “arms” to any angle

· The initial arm is outstretched from the origin along the x-axis

· The terminal arm is rotated from the initial arm

· Angles can be classified on their quadrant. There are 4 quadrants.

We start the first quadrant from the initial arm and rotate counter-clockwise.

II I

III IV

Examples: Sketch 45°, 120°, 200°, 340° and determine the quadrant of each angle.

Quadrant I Quadrant II Quadrant III Quadrant IV

Co–Terminal Angles

· angles that have the same terminal arm

counter-clockwise (+)Positive angles rotate counter-clockwise

clockwise ()Negative angles rotate clockwise

Examples: Sketch 405°, 540°, 690°, - 210°, 920°, - 370°

Principal Angle

smallest positive co–terminal angle (between 0° and 360°)

Reference Angle

positive acute angle formed by terminal arm & the x–axis (between 0° and 90°)

Example:Find the principal angle and the reference angle. ANS: P = 220, R = 40

(580)Principal = 580 – 360 = 220o

Reference = 220 – 180 = 40o

Example: Find the principal angle and the reference angle for 395° and -470° respectively.

Angle: 395°Angle: -470°

Principal Angle: 35°Principal Angle: 250°

Reference Angle: 35°Reference Angle: 70°

Example: Using a reference angle of 25°, find the related angles in all four quadrants. You can flip and reflect the reference angle diagram to locate related angles in each quadrant.

155° 25°

205° 335°

Example: Find one positive and one negative angle co–terminal with each. Also find the principal angle and reference angle for each.

a.108 ANS: 468, 252, P = 108, R = 72

b.70 ANS: 290, 430, P = 290, R = 70

c.587 ANS: 227, 133, P = 227, R = 47

ASSIGNMENT: Text pg 83 #2, 3, 5, 6, 7; Trig Wkst A #I, II, III, IV, V

DAY 2 – Terminal Arm Length and Special Case Triangles

Using Coordinates to Determine Length of the Terminal Arm

· There are two methods which can be used:

· Pythagorean Theorem

· Distance Formula

· Tip: “Always Sketch First!”

Using the Theorem of Pythagoras

· Given the point (3, 4), draw the terminal arm. Complete the right triangle by joining the terminal point to the x-axis.

· Determine the sides of the triangle. Use the Theorem of Pythagoras.

c2 = a2 + b2

c2 = 32 + 42

c2 = 25

c = 5

· Since we are using angles rotated from the origin, we label the sides as being x, y and r for the radius of the circle that the terminal arm would make.

r

y

x

· Example: Draw the following angle in standard position given any point (x, y) and determine the value of r.

Point: (-2, 3)Point: (-5, -1)Point: (4, -2)

r2 = x2 + y2r2 = x2 + y2r2 = x2 + y2

r2 = (-2)2 + 32r2 = (-5)2 + (-1)2r2 = 42 + (-2)2

r2 = 25r2 = 26r2 = 20

r = 5r = r = 2

Using the Distance Formula

· The distance formula: d = √[(x2 – x1)2 + (y2 – y1)2]

Example: Given point P (-2, -6), determine the length of the terminal arm.

d = √[(x2 – x1)2 + (y2 – y1)2]

d = √[(-2 – 0)2 + (-6 – 0)2]

d =

d = 2

Review of SOH CAH TOA

Example: Solve for x

a) (x) b)

(35) (x) (7)

(200) (74°)

sin 74° = tan x =

x = 7.282 x = 80.074

Example: Determine the ratios for the following:

a) sin 28° =

(17)

b) (28°)cos 28°=

c) tan 62°=

Special Case Triangles – Exact Trigonometric Ratios

· We can use squares or equilateral triangles to calculate exact trigonometric ratios for 30°, 45° and 60°.

· (114545)Draw a square with a diagonal.

· A square with a diagonal will have angles of 45°.

· All sides are equal.

· Let the sides equal 1

· By the Theorem of Pythagoras, r =

· Draw an equilateral triangle with a perpendicular line from the top straight down

· All angles are equal in an equilateral triangle (60°)

· After drawing the perpendicular line, we know the small angle is 30°

· Let each side equal 2

· (306021)By the Theorem of Pythagoras, y =

Finding Exact Values

· Steps:

· Sketch the special case triangles and label

· Sketch the given angle

· Find the reference angle

(306012) (114545)

· Example: Cos 45°Example: Sin 60°

cos 45° = = sin 60° =

· Example: Tan 30°Example: Cos 30°

tan 30° = = cos 30° =

Solving Equations using Exact Values, Quadrant I ONLY

Examples:

a) sin x = x = 30°

b) cos x = x = 60°

c) tan x = 1x = 45°

d) cos x = x = 45°

e) sin x = x = 60°

f) cos x = x = 30°

g) tan x = x = 60°

h) sin x = x = 45°

i) tan x = x = 30°

ASSIGNMENT: Text pg 83 #8; 84 #10, 11, 12, 13; Trig Wkst A #VI, VII

DAY 3: Trig Ratios of Angles in Standard Position

(rxy(x, y)) Primary Trig FunctionsReciprocal Trig Functions

sin = cosecant (csc) =

cos = secant (sec) =

tan = cotangent (cot) =

Note: r is the reference angle so is always positive and is always between r and the x axis.

(SATC II I III IV)

The CAST rule shows where each trig ratio is positive.

Quadrant I – all trig ratios are positive.

Quadrant II – only Sine is positive.

Quadrant III – only Tangent is positive.

Quadrant IV – only cosine is positive.

Examples:Compare the following ratios:

a.cos 43 ANS: 0.73b.sin 43 ANS: 0.68

cos 137 ANS: 0.73sin 137 ANS: 0.68

cos 223 ANS: 0.73sin 223 ANS: 0.68

cos 317 ANS: 0.73sin 317 ANS: 0.68

cos 43 ANS: 0.73sin 43 ANS: 0.68

· All families of angles (those with the same reference angle), use the same congruent triangle to calculate trig ratios

· Due to the position of the triangle, the x and/or y-values will change from positive to negative as you change quadrants.

Example 1: Given P (3, -4); sketch the triangle and use it to calculate the trig ratios.

· Sketch, Find Length of Terminal Arm, Determine the 6 Trig Ratios

r2 = x2 + y2sin θ = csc θ = =

r2 = (3)2 + (-4)2

(θ)r2 = 25 cos θ = sec θ =

r = 5

tan θ = cot θ = =

Example 2: The following points are on the terminal arm of angle . Find the 6 trig ratios.

a.(5, 12)

ANS: Draw (5, 12)

Solve r using Pythagorean Theorem.

Determine 6 trig ratios. sin θ = = csc θ = =

cos θ = = sec θ = =

tan θ = = cot θ = =

b.(2, 3)

ANS: Draw (2, –3)

Solve r using Pythagorean theorem.

Determine the 6 trig ratios. sin θ = = = csc θ = =

cos θ = = = sec θ = =

tan θ = = cot θ = =

c.(m, m 5) Find sin . ANS: “c” is not required in 20-1

Example 3: Find the following ratios using your calculator accurate to 4 decimal places. Simply type them in your calculator. Watch out with your mode!

a. cos 147 b. tan 280° c. sin 61

ANS: – 0.8387 ANS: – 5.6713ANS: 0.8746

Example 4: Given 140°, sketch and determine the sign of the 3 trig ratios

sin is positive

cos is negative

tan is negative

Example 5: Find other angles with the same ratio for the given function. (Between 0 & 360)

a.sin 38 ANS: 142b.tan 115 ANS: 295 Put on exam/quiz

c.cos 72 ANS: 72, 288d.csc 198 ANS: 342

e. cot 163 ANS: 17, 197

Solving Degree 1 Trigonometric Equations

· Solving a trig equation means finding the angle.

· Since we now know that the positive or negative nature of trig ratios affects the quadrant we are in, we have to consider more than one possible answer.

· Steps:

· Calculate the reference value for theta

· Determine the possible quadrants based on the positive or negative value

· Determine possible values theta (θ)

Example 6: Cosine θ = 0.4315 Cos is positive in Quadrant I and IV

Reference θ = cos-1(0.4315) = 64.437°

Possible Values of θ = 64.437° and 295.563°

Example 7: Sine θ = 0.615Sin is positive in Quadrant I and II

Reference θ = sin-1(0.615) = 37.952°


· To simplify our calculations we use a positive ration to find the reference angle.

· This ensures your calculator gives a solution in quadrant I.

Example 8: Cosine θ = -0.235Cos is negative in Quadrant II and III

POSITIVE Reference θ = cos-1(0.235) = 76.408°


Example 9: Tangent θ = -2.43Cos is negative in Quadrant II and IV

POSITIVE Reference θ = tan-1(2.43) = 67.632°


Example 10: Tangent θ = 1/3Tan is positive in Quadrant I and III

Reference θ = tan-1(1/3) = 18.435°


Example 11: Sine θ = -7/9Sin is negative in Quadrant III and IV

POSITIVE Reference θ = sin-1(7/9) = 51.056°


Example 12: cos = , sin > 0. Find the other 5 ratios.

ANS: Think CAST. Cos is negative and sine is positive. This can only happen in quadrant II.

Thus, x = –3 and r = 4.

To get y, use the Pythagorean Theorem.

(Since we are in quadrant II, y must be positive so y = stays as y = )

ASSIGNMENT: Trig Wkst A #VIII, IX, X, XI, XII, XIII, XIV

DAY 4 – Sine Law

· The Sine Law is used when you have a side-angle pair and any other angle or side

· You may have ASA or SSA

· The Sine Law is:

= =

· (BACbacxyh)Proof:

sin A = sin C =

h = c sin Ah = a sin C

Substitute: c sin A = a sin C

Rearrange: =

(B)

(y)sin A = sin B =

(c)

(a)h = b sin Ah = a sin B

(x) (h)

Substitute: b sin A = a sin B

(b) (C) (A)Rearrange: =

Conclusion:

= =

Example 1: Solve the following triangle.

(55°1612) = =

=

B = 37.9°

C = 180° - 55° - 37.9° = 87.1°

=

c = 19.5

Example 2: In ΔABC, BC = 20, ˂A = 31°, and ˂C = 104°. Find AC to the hundredths.

(31°104°20ABC) = =

=

c = 37.68

Example 3: In ΔXYZ, ˂X = 105°, x = 20.3 and y = 15.1. Find ˂ Y to the nearest degree.

(15.1105°20.3ZYX) = =

=

Y = 46°

ASSIGNMENT: Text: pg. 138 # 3-6, 8, 10; Trig Wkst A #XV; Trig Wkst B

DAY 5 – The Ambiguous Case

· The Ambiguous Case occurs when you have SAS (angle not included)

· The side length opposite the given angle MUST be smaller than the other given side length.

· The ambiguous case is solved with the Sine Law BUT you must also check for alternate solutions because the solution gives a positive Quadrant I value. Sine is also positive in Quadrant II.

(C)Example 1: Solve triangle ABC if A = 29.3°, b = 20.5 cm and a = 12.8 cm.

(20.5) (12.8) =

(29.3°) =

(B) (A)

sin B = 51.6 OR 180 – 51.6 = 128.4°

We have two possibilities for the triangle.

(29.3°ACB12.820.551.6°) (C)

(20.5)

(12.8)

(128.4°) (B) (A) (29.3°)

B = 128.4°B = 51.6°

C = 22.3°C = 99.1°

c = = 9.9c = = 25.8

(30°ABC1626D)Example 2: Calculate the measures of triangle BCD and triangle BDA to the nearest tenth of a centimetre.

C = 26 sin 30 / 16 = 54.3 OR 180 – 54.3 = 125.7

B = 180 – 54.3 – 30 = 95.7 180 – 125.7 – 30 = 24.3

AC = 16 sin 95.7 / sin 30 = 31.8AD = 16 sin 24.3 / sin 30 = 13.2

Example 3: Solve the following triangles: ΔKLM where M = 37.3°, m = 85 km and l = 90 km.

M = 37.3°

m = 85 km

l = 90 km

L = 39.9° OR 140.1°

K = 102.8° OR 2.6°

k = 135.2 km OR 6.3 km

Example 4: Solve the following triangles: ΔXYZ where X = 120°, x = 40 and z = 20.

X = 120°

x = 40

z = 20

Z = 25.7° *** No other possibilities. This triangle is not ambiguous.

Y = 34.3°

y = 26.0

ASSIGNMENT: Trig Wkst B

Solve the following triangles:

1. ΔABC where B = 27°, b = 25 and c = 30.

2. ΔTUV where U = 48°, v = 15.6 cm and l = 12.6 cm.

3. ΔPQR where P = 30°, p = 24 m and q = 48 m.

Answer the following:

4. Sam is receiving a signal from a beacon that is directly east of his position. Due to poor terrain, he cannot go directly to the beacon. He travels 50 metres at an angle of 20° South of East. Then he travels 30 metres to the beacon.

a. Sketch and label a possible triangle to represent his path and the beacon.

b. Calculate the angle at the beacon of the triangle.

c. Are there any other possibilities?

d. How far from Sam’s original position is the beacon?

DAY 6 – The Cosine Law

· The Cosine Law is used when you do NOT have a side-angle pair

· You may have SSS or SAS

· The Cosine Law is:

· The Cosine Law can be rearranged as:

· The Cosine Law is an extension of the Theorem of Pythagoras. Cosine 90° is zero so the last term will disappear when dealing with a right-angle triangle.

· Proof:

(BACbacxb - xh)

cos A =

x = c cos A

Rearrange:

Substitute:

Rearrange:

· The Cosine Law can be rearranged to solve for the angle.

Example 1: In triangle DEF, what is the value of side e to the nearest thousandth if d = 5, ˂E = 74° and f = 3.

(3574°)

Example 2: Find ˂C to the nearest degree if a = 12, b = 18 and c = 15 in ΔABC.

(B)

(121815)

(A) (C)

Example 3: Solve the following triangle. Give all answers to the nearest tenth.

(P58°RQ13.816.4)

= =

=

P = 69.8

= =

=

R = 52.2

ASSIGNMENT: Trig Wkst A #XVI, XVII, XVIII; Trig Wkst B

DAY 7

Solving Linear Trig Equations

A.Calculator

Solve (1 revolution)

1.Steps1. Isolate the function

ref =22.2o2. Where are the terminal arms?

Sine is positive in quadrants 1 and 2 (Cast)3. Ref

2.3.

ref =53.1oref =63.4o

Cos is positive in quadrants 2 and 3 (Cast)Tan is negative in quadrants 2 & 4

4.5.

ref =74.1oref =8.2o

Tan is positive in quadrants 1 and 4 (Cast)Sin is positive in quadrants 1&2

Solve (1 revolution) Exact

Examples:

1.2 + 1 = 0

=

ANS: 210, 330

2.4 + 4 = 0

= –1

tan x = –1

ANS: 135, 315

Solving Trig Equations

0 < 360 (0.01)

1.3 1 = 02.5 + 3 = 0

ref70.53 ref36.87

ANS: 70.53, 289.47ANS: 323.13, 216.87

3.5 + 2 = 3 34.9 1 = 0

2=–5

ref68.20ref19.47

ANS: 291.80, 111.80ANS: 19.47, 160.53, 199.47, 340.53

5.3 7 6 = 06.cos2 + 3cos = 0

ref

ref

ANS: 326.31, 146.31ANS: 90, 270

Trig Eqtn 20-1 WC.docx

14M20-1 LP Trigonometry.docx

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Math 20

–

1: Trigonometry

*** Your calculator must be in DEGREE mode during this unit!

DAY

1

–

Angles in Standard Position

Standard Position

·

Angles are placed in standard position so they can be compared.

·

Angles in

standard position all have the same perspective and relate to both the x and y

axis.

Example

s

:

Angles not in Standard Position

Angle in Standard Position

Putting an Angle in Standard Position

·

There are 2 “arms” to any angle

·

The initial arm is outstretched from the origin along the x

-

axis

·

The terminal arm is rotated from the initial arm

·

Angles can be classified on their quadrant. There are 4 quadrants.

We start the first quadrant from the ini

tial arm and rotate counter

-

clockwise

.

II I

III IV

Examples: Sketch 45°, 120°, 200°, 340°

and determine the quadrant of each angle.

Quadrant I Quadrant II Quadrant III Quadrant IV

q

terminal arm

initial

arm

v

ertex

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