p2 chapter 7 :: trigonometry and modelling
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P2 Chapter 7 :: Trigonometry And Modelling
jfrost@tiffin.kingston.sch.ukwww.drfrostmaths.com
@DrFrostMaths
Last modified: 23rd November 2017
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Chapter Overview
sin ð + ð = sin ð cos ð + cosð sin ð
1a:: Addition Formulae
âSolve, for 0 †ð¥ < 2ð, the equation2 tan 2ðŠ tan ðŠ = 3
giving your solutions to 3sf.â
1b:: Double Angle Formulae
âFind the maximum value of 2 sin ð¥ + cos ð¥ and the value of ð¥for which this maximum occurs.â
3:: Simplifying ð cos 𥠱 ð sin ð¥
This chapter is very similar to the trigonometry chapters in Year 1. The only difference is that new trig functions: ð ðð, ððð ðð and ððð¡, are introduced.
âThe sea depth of the tide at a beach can be
modelled by ð¥ = ð ð ðð2ðð¡
5+ ðŒ , where ð¡ is
the hours after midnight...â
4:: Modelling
Teacher Note: There is again no change in this chapter relative to the old pre-2017 syllabus.
Addition Formulae
Addition Formulae allow us to deal with a sum or difference of angles.
ð¬ð¢ð§ ðš + ð© = ð¬ð¢ð§ðš ððšð¬ð© + ððšð¬ðš ð¬ð¢ð§ð©ð¬ð¢ð§ ðš â ð© = ððððšðððð© â ððððšðððð©
ððšð¬ ðš + ð© = ððšð¬ðš ððšð¬ð© â ð¬ð¢ð§ðš ð¬ð¢ð§ð©ððšð¬ ðš â ð© = ððšð¬ðš ððšð¬ð© + ð¬ð¢ð§ðš ð¬ð¢ð§ð©
ððð§ ðš + ð© =ððð§ðš + ððð§ð©
ð â ððð§ðš ððð§ð©
ððð ðš â ð© =ððððš â ðððð©
ð + ððððš ðððð©
Do I need to memorise these?Theyâre all technically in the formula booklet, but you REALLY want to eventually memorise these (particularly the ð ðð and ððð ones).
How to memorise:
First notice that for all of these the first thing on the RHS is the same as the first thing on the LHS!
⢠For sin, the operator in the middle is the same as on the LHS.
⢠For cos, itâs the opposite.⢠For tan, itâs the same in
the numerator, opposite in the denominator.
⢠For sin, we mix sin and cos.
⢠For cos, we keep the cosâs and sinâs together.
Common Schoolboy/Schoolgirl Error
Why is sin(ðŽ + ðµ) not just sin ðŽ + sin(ðµ)?Because ððð is a function, not a quantity that can be expanded out like this. Itâs a bit like how ð + ð ð ⢠ðð + ðð.We can easily disprove it with a counterexample.
?
Addition Formulae
Now can you reproduce them without peeking at your notes?
ð¬ð¢ð§ ðš + ð© â¡ ð¬ð¢ð§ðš ððšð¬ð© + ððšð¬ðš ð¬ð¢ð§ð©ð¬ð¢ð§ ðš â ð© â¡ððððšðððð© â ððððšðððð©
ððšð¬ ðš + ð© â¡ððšð¬ðš ððšð¬ð© â ð¬ð¢ð§ðš ð¬ð¢ð§ð©ððšð¬ ðš â ð© â¡ ððšð¬ðš ððšð¬ð© + ð¬ð¢ð§ðš ð¬ð¢ð§ð©
ððð§ ðš + ð© â¡ððð§ðš + ððð§ð©
ð â ððð§ðš ððð§ð©
ððð ðš â ð© â¡ððððš â ðððð©
ð + ððððš ðððð©
How to memorise:
First notice that for all of these the first thing on the RHS is the same as the first thing on the LHS!
⢠For sin, the operator in the middle is the same as on the LHS.
⢠For cos, itâs the opposite.⢠For tan, itâs the same in
the numerator, opposite in the denominator.
⢠For sin, we mix sin and cos.
⢠For cos, we keep the cosâs and sinâs together.
??
??
?
?
Proof of sin ðŽ + ðµ â¡ sin ðŽ cosðµ + cosðŽ sin ðµ
ðµ
ðŽ
1
(Not needed for exam)
1: Suppose we had a line of length 1 projected an angle of ðŽ + ðµ above the horizontal.Then the length of ðð =sin ðŽ + ðµIt would seem sensible to try and find this same length in terms of ðŽ and ðµ individually.
ð ð
ð
sin(ðŽ
+ðµ)
2: We can achieve this by forming two right-angled triangles.
ðŽ
3: Then weâre looking for the combined length of these two lines.
4: We can get the lengths of the top triangleâŠ
sinðŽcosðµ
cosðŽsinðµ
5: Which in turn allows us to find the green and blue lengths.
6: Hence sin ðŽ + ðµ =sinðŽ cosðµ + cosðŽ sinðµ
â¡
Proof of other identities
Can you think how to use our geometrically proven result sin ðŽ + ðµ = sinðŽ cosðµ + cos ðŽ sinðµ to prove the identity for sin(ðŽ â ðµ)?
ð¬ð¢ð§ ðš â ð© = ð¬ð¢ð§ ðš + âð©
= ð¬ð¢ð§ ðš ððšð¬ âð© + ððšð¬ ðš ð¬ð¢ð§ âð©= ð¬ð¢ð§ðš ððšð¬ð© â ððšð¬ðš ð¬ð¢ð§ð©
What about cos(ðŽ + ðµ)? (Hint: what links ð ðð and ððð ?)
ððšð¬ ðš + ð© = ð¬ð¢ð§ð
ðâ ðš + ð© = ð¬ð¢ð§
ð
ðâ ðš + âð©
= ð¬ð¢ð§ð
ðâ ðš ððšð¬ âð© + ððšð¬
ð
ðâ ðš ð¬ð¢ð§ âð©
= ððšð¬ðš ððšð¬ð© â ð¬ð¢ð§ðš ð¬ð¢ð§ð©
?
?
?
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Proof of other identities
Edexcel C3 Jan 2012 Q8
The important thing is that you explicit show the division of each term by ððð ðŽ cosðµ.
?
Examples
Given that 2 sin(ð¥ + ðŠ) = 3 cos ð¥ â ðŠ express tan ð¥ in terms of tan ðŠ.
Using your formulae:ð ð¬ð¢ð§ð ððšð¬ ð + ðððšð¬ ð ð¬ð¢ð§ð = ðððšð¬ ð ððšð¬ ð + ð ð¬ð¢ð§ð ð¬ð¢ð§ð
We need to get ððð§ ð and ððð§ð in there. Dividing by ððšð¬ ð ððšð¬ ð would seem like a sensible step:
ð ððð§ ð + ð ððð§ ð = ð + ð ððð§ ð ððð§ðRearranging:
ððð§ ð =ð â ð ððð§ð
ð â ð ððð§ð
Q
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Exercises 7A
Pearson Pure Mathematics Year 2/ASPage 144
(Classes in a rush may wish to skip this exercise)
Uses of Addition Formulae
[Textbook] Using a suitable angle formulae, show that sin 15° =6â 2
4.Q
ð¬ð¢ð§ðð = ð¬ð¢ð§ ðð â ðð = ð¬ð¢ð§ðð ððšð¬ðð â ððšð¬ ðð ð¬ð¢ð§ðð
=ð
ðÃ
ð
ðâ
ð
ðÃð
ð
=ð â ð
ð ð
=ð â ð
ð
[Textbook] Given that sin ðŽ = â3
5and 180° < ðŽ < 270°, and that cosðµ = â
12
13,
and B is obtuse, find the value of: (a) cos ðŽ â ðµ (b) tan ðŽ + ðµ
ððð ðš â ð© â¡ ððð ðš ððð ð© + ððð ðš ððð ð©
ððð ðš â¡ ð â ðððððš = ±ð
ð. But we know from the graph of ððð that it is negative
between ððð° < ðš < ððð°, so ððð ðš = âð
ð.
Similarly ððð ð© = ±ð
ðð, and similarly considering the graph for ððð, ððð ð© = +
ð
ðð.
⎠ððð ðš â ð© â¡ âð
ðà â
ðð
ðð+ â
ð
ðÃ
ð
ðð=ðð
ðð
ContinuedâŠ
Q
?
?
Fro Tip: You can get ððð in terms of ð ðð and vice versa by using a rearrangement of sin2 ð¥ + cos2 ð¥ â¡ 1.
So cosðŽ = 1 â sin2 ðŽ
ContinuedâŠ
Given that sinðŽ = â3
5and 180° < ðŽ < 270°, and that cosðµ = â
12
13, find the
value of: (b) tan ðŽ + ðµ
Q
ððð§ ðš + ð© =ððð§ ðš + ððð§(ð©)
ð â ððð§ðš ððð§ð©
We earlier found ð¬ð¢ð§ðš = âð
ð, ð¬ð¢ð§ð© =
ð
ðð, ððšð¬ðš = â
ð
ð, ððšð¬ð© = â
ðð
ðð
⎠ððð§ðš = âð
ð÷ â
ð
ð=
ð
ðand ððð§ð© =
ð
ðð÷ â
ðð
ðð= â
ð
ðð
By substitution:
ððð§ ðš + ð© =ðð
ðð
?
Test Your Understanding
Without using a calculator, determine the exact value of:a) cos 75°b) tan 75°
ððšð¬ ðð° = ððšð¬ ðð° + ðð° = ððšð¬ ðð° ððšð¬ ðð° â ð¬ð¢ð§ ðð° ð¬ð¢ð§ ðð°
=ð
ðÃ
ð
ðâ
ð
ðÃð
ð
=ð â ð
ð ð=
ð â ð
ð
ððð§ ðð° =ð¬ð¢ð§ ðð°
ððšð¬ ðð°ð¬ð¢ð§ ðð° = ð¬ð¢ð§ ðð° + ðð° = ð¬ð¢ð§ ðð° ððšð¬ ðð° + ððšð¬ ðð° ð¬ð¢ð§ ðð°
=ð
ðÃ
ð
ð+
ð
ðÃð
ð=
ð + ð
ð
⎠ððð§ ðð° =ð + ð
ð÷
ð â ð
ð=
ð + ð
ð â ð= ð + ð
?
?
Challenging question
Edexcel June 2013 Q3
âExpandingâ both sides:2 cos ð¥ cos 50 â 2 sin ð¥ sin 50= sin ð¥ cos 40 + cos ð¥ sin 40
Since thing to prove only has 40 in it, use cos 50 = sin 40 and sin 50 =cos 40 .
2 cos ð¥ sin 40 â 2 sin ð¥ cos 40= sin ð¥ cos 40 + cos ð¥ sin 40
As per usual, when we want tans, divide by cos ð¥ cos 40:
2 tan 40 â 2 tan ð¥ = tan ð¥ + tan 40tan 40 = 3 tan ð¥1
3tan 40 = tan ð¥
a
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Exercise 7B
Pearson Pure Mathematics Year 2/ASPages 172-173
[STEP I 2010 Q3] Show thatsin ð¥ + ðŠ â sin ð¥ â ðŠ = 2 cos ð¥ sin ðŠ
and deduce that
sin ðŽ â sin ðµ = 2 cos1
2ðŽ + ðµ sin
1
2ðŽ â ðµ
Show also that
cosðŽ â cosðµ = â2 sin1
2ðŽ + ðµ sin
1
2ðŽ â ðµ
The points ð, ð, ð and ð have coordinates ð cos ð , ð sin ð and ð cos ð , ð sin ð respectively,
where 0 †ð < ð < ð < ð < 2ð, and ð and ð are positive.Given that neither of the lines ðð and ðð is vertical, show that these lines are parallel if and only if
ð + ð â ð â ð = 2ð
Extension
Double Angle Formulae
sin 2ðŽ â¡ 2 sinðŽ cosðŽcos 2ðŽ â¡ cos2 ðŽ â sin2 ðŽ
â¡ 2 cos2 ðŽ â 1â¡ 1 â 2 sin2 ðŽ
tan 2ðŽ =2 tanðŽ
1 â tan2 ðŽ
Tip: The way I remember what way round these go is that the cos on the RHS is âattractedâ to the cos on the LHS, whereas the sin is pushed away.
These are all easily derivable by just setting ðŽ = ðµ in the compound angle formulae. e.g.
sin 2ðŽ = sin ðŽ + ðŽ= sinðŽ cosðŽ + cosðŽ sin ðŽ= 2 sinðŽ cosðŽ
!This first form is relatively rare.
Double-angle formula allow you to halve the angle within a trig function.
Examples
[Textbook] Use the double-angle formulae to write each of the following as a single trigonometric ratio.a) cos2 50° â sin2 50°
b)2 tan
ð
6
1âtan2ð
6
c)4 sin 70°
sec 70°
This matches ððšð¬ ððš = ððšð¬ð ðš â ð¬ð¢ð§ððš:ðððð ðð° â ðððð ðð° = ððšð¬ ððð°
This matches ððð§ ððš =ð ððð§ ðš
ðâððð§ð ðš
ð ðððð ð
ð â ððððð ð
= ððð§ð
ð
ð ð¬ð¢ð§ðð°
ð¬ðð ðð°=
ðð¬ð¢ð§ðð°
ð ÷ ððšð¬ðð°= ðð¬ð¢ð§ðð° ððšð¬ ðð°
This matches ð¬ð¢ð§ ððš = ðð¬ð¢ð§ðš ððšð¬ðšð ð¬ð¢ð§ðð° ððððð° = ð ð¬ð¢ð§ððð°
sin 2ð¥ = 2 sin ð¥ cos ð¥cos 2ð¥ = cos2 ð¥ â sin2 ð¥
= 2 cos2 ð¥ â 1= 1 â 2 sin2 ð¥
tan 2ð¥ =2 tan ð¥
1 â tan2 ð¥
a
b
c
?
?
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Examples
[Textbook] Given that ð¥ = 3 sin ð and ðŠ = 3 â 4ððð 2ð, eliminate ð and express ðŠ in terms of ð¥.
sin 2ð¥ = 2 sin ð¥ cos ð¥cos 2ð¥ = cos2 ð¥ â sin2 ð¥
= 2 cos2 ð¥ â 1= 1 â 2 sin2 ð¥
tan 2ð¥ =2 tan ð¥
1 â tan2 ð¥
Fro Note: This question is an example of turning a set of parametric equations into a single Cartesian one. You will cover this in the next chapter.
ð = ð â ð ð â ðð¬ð¢ð§ð ðœ
= ðð¬ð¢ð§ð ðœ â ð
ð¬ð¢ð§ðœ =ð
ð
⎠ð = ðð
ð
ð
â ð
Given that cos ð¥ =3
4and ð¥ is acute, find the exact value of
(a) sin 2ð¥ (b) tan 2ð¥
Since ððððð + ððšð¬ðð = ð, ððð ð = ± ð â ððððð
ððð ð = ð âð
ð
ð=
ð
ð(and is positive given ð < ð± < ðð°)
⎠ððð ðð = ðððð ð ððð ð = ð Ãð
ðÃð
ð=ð ð
ð
ððð ð =ððð ð
ððð=ð ð/ð
ð/ð=
ð
ð⎠ððððð =
ð Ãðð
ð âðð
ð = ð ð
?
?
Solving Trigonometric Equations
This is effectively the same type of question you encountered in Chapter 6 and in Year 1, except you may need to use either the addition formulae or double angle formulae.
[Textbook] Solve 3 cos 2ð¥ â cos ð¥ + 2 = 0 for 0 †ð¥ †360°.
3 2 cos2 ð¥ â 1 â cos ð¥ + 2 = 06 cos2 ð¥ â 3 â cos ð¥ + 2 = 06 cos2 ð¥ â cos ð¥ â 1 = 03 cos ð¥ + 1 2 cos ð¥ â 1 = 0
cos ð¥ = â1
3ðð cos ð¥ =
1
2
ð¥ = 109.5°, 250.5° ð¥ = 60°, 300°
Recall that we have a choice of double angle identities for cos 2ð¥. This is clearly the most sensible choice as we end up with an equation just in terms of ððð .?
Further Examples
[Textbook] By noting that 3ðŽ = 2ðŽ + ðŽ, :a) Show that sin 3ðŽ = 3 sinðŽ â 4 sin3ðŽ.b) Hence or otherwise, solve, for 0 < ð < 2ð,
the equation 16 sin3 ð â 12 sinð â 2 3 = 0
ð¬ð¢ð§ ððš + ðš= ð¬ð¢ð§ ððš ððšð¬ðš + ððšð¬ ððš ð¬ð¢ð§ðš
= ð ð¬ð¢ð§ðš ððšð¬ ðš ððšð¬ðš + ð â ð ð¬ð¢ð§ð ðš ð¬ð¢ð§ðš
= ð ð¬ð¢ð§ðš ð â ð¬ð¢ð§ð ðš + ð¬ð¢ð§ ðš â ðð¬ð¢ð§ð ðš
= ð ð¬ð¢ð§ðš â ð ð¬ð¢ð§ð ðš + ð¬ð¢ð§ðš â ð ð¬ð¢ð§ð ðš= ð ð¬ð¢ð§ðš â ð ð¬ð¢ð§ð ðš
Note that ðððððð ðœ â ðð ððððœ
= âð ðð¬ð¢ð§ ðš â ð ð¬ð¢ð§ð ðš = âð ð¬ð¢ð§ ððœ.
⎠âð ð¬ð¢ð§ ððœ = ð ð
ð¬ð¢ð§ ððœ = âð
ð
ððœ =ðð
ð,ðð
ð,ððð
ð,ððð
ð,ððð
ð,ððð
ð
ðœ =ðð
ð,ðð
ð,ððð
ð,ððð
ð,ððð
ð,ððð
ð
Fro Exam Note: A question pretty much just like this came up in an exam once.
?
?
[Textbook] Solve 4 cos ð â 30° = 8 2sin ð in the range 0 †ð < 360°.
Your instinct should be to use the addition formulae first to break up the ðððð ðœ â ðð° .
ð ððšð¬ðœ ððšð¬ðð° + ð ð¬ð¢ð§ðœ ð¬ð¢ð§ðð° = ð ð ð¬ð¢ð§ðœ
ð ð ððšð¬ðœ + ðð¬ð¢ð§ðœ = ð ðð¬ð¢ð§ðœ
ð ðððšð¬ðœ = ð ðð¬ð¢ð§ðœ â ð ð¬ð¢ð§ðœ
ð ð ððšð¬ðœ = ð ð â ð ð¬ð¢ð§ðœ
ð ð
ð ð â ð=ð¬ð¢ð§ðœ
ððšð¬ðœ= ððð§ðœ
ðœ = ðð. ð°, ððð. ð°
?
Test Your Understanding
Edexcel C3 Jan 2013 Q6
?
If you finish that quicklyâŠ
[Textbook] Solve 2 tan2ðŠ tanðŠ = 3 for 0 †ðŠ < 2ð, giving your answer to 2dp.
ð ððð§ ðð ððð§ ð = ð
ðð ððð§ ð
ð â ððð§ð ðððð§ ð = ð
âŠð ððð§ð ð = ð
ððð§ ð = ±ð
ð
ð = ð. ðð, ð. ðð, ð. ðð, ð. ðð
Forgetting the ± is an incredibly common source of lost marks.
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Exercise 7D
Pearson Pure Mathematics Year 2/ASPages 180-181
Extension
[AEA 2013 Q2]
1
2 [AEA 2009 Q3]
(Solution on next slide)
?
ð sin ð + ð cos ð
Hereâs a sketch of ðŠ = 3 sin ð¥ + 4 cos ð¥.What do you notice?
Itâs a sin graph that seems to be translated on the ð-axis and stretched on the ð axis. This suggests we can represent it as ð = ð¹ ð¬ð¢ð§(ð + ð¶), where ð¶ is the horizontal translation and ð¹ the stretch on the ð-axis.
?
ð sin ð + ð sin ð
Put 3 sin ð¥ + 4 cos ð¥ in the form ð sin ð¥ + ðŒ giving ðŒ in degrees to 1dp.Q
Fro Tip: I recommend you follow this procedure every time â Iâve tutored students whoâve been taught a âshortcutâ (usually skipping Step 1), and they more often than not make a mistake.
STEP 1: Expanding:ð sin ð¥ + ðŒ = ð sin ð¥ cosðŒ + ð cos ð¥ sin ðŒ
STEP 2: Comparing coefficients:ð cosðŒ = 3 ð sin ðŒ = 4
STEP 3: Using the fact that ð 2 sin2 ðŒ + ð 2 cos2 ðŒ = ð 2:
ð = 32 + 42 = 5
STEP 4: Using the fact that ð sin ðŒ
ð cos ðŒ= tanðŒ:
tanðŒ =4
3ðŒ = 53.1°
STEP 5: Put values back into original expression.3 sin ð¥ + 4 cos ð¥ â¡ 5 sin ð¥ + 53.1°
?
?
?
?
?
If ð cos ðŒ = 3 and ð sinðŒ = 4then ð 2 cos2 ðŒ = 32 and ð 2 sin2 ðŒ = 42.ð 2 sin2 ðŒ + ð 2 cos2 ðŒ = 32 + 42
ð 2 sin2ðŒ + cos2 ðŒ = 32 + 42
ð 2 = 32 + 42
ð = 32 + 42
(You can write just the last line in exams)
Test Your Understanding
Put sinð¥ + cosð¥ in the form ð sin ð¥ + ðŒ giving ðŒ in terms of ð.Q
ð sin ð¥ + ðŒ â¡ ð sin ð¥ cos ðŒ + ð cos ð¥ sin ðŒð cosðŒ = 1 ð sinðŒ = 1
ð = 12 + 12 = 2tanðŒ = 1
ðŒ =ð
4
sin ð¥ + cos ð¥ â¡ 2 sin ð¥ +ð
4
Put sinð¥ â 3 cos ð¥ in the form ð sin ð¥ â ðŒ giving ðŒ in terms of ð.Q
ð sin ð¥ + ðŒ â¡ ð sin ð¥ cos ðŒ â ð cos ð¥ sin ðŒ
ð cosðŒ = 1 ð sin ðŒ = 3ð = 2
tanðŒ = 3 ð ð ðŒ =ð
3
sin ð¥ + cos ð¥ â¡ 2 sin ð¥ âð
3
?
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Further Examples
Put 2 cosð + 5 sin ð in the form ð cos ð â ðŒ where 0 < ðŒ < 90°Hence solve, for 0 < ð < 360, the equation 2 cosð + 5 sin ð = 3
Q
2 cos ð + 5 sin ð â¡ 29 cos ð â 68.2°Therefore:
29 cos ð â 68.2° = 3
cos ð â 68.2° =3
29ð â 68.2° = â56.1⊠°, 56.1⊠°ð = 12.1°, 124.3°
(Without using calculus), find the maximum value of 12 cosð + 5 sinð, and give the smallest positive value of ð at which it arises.
Q
Use either ð sin(ð + ðŒ) or ð cos(ð â ðŒ) before that way the + sign in the middle matches up.
â¡ 13 cos ð â 22.6°
ððð is at most 1,thus the expression has value at most 13.This occurs when ð â 22.6 = 0 (as cos 0 = 1) thus ð = 22.6
?
?
Fro Tip: This is an exam favourite!
Quickfire Maxima
What is the maximum value of the expression and determine the smallest positive value of ð (in degrees) at which it occurs.
Expression Maximum (Smallest) ðœ at max
20 sin ð 20 90°
5 â 10 sin ð 15 270°
3 cos ð + 20° 3 340°
2
10 + 3 sin ð â 30
2
7
300°
? ?? ?? ?
? ?
Proving Trigonometric Identities
Prove that tan 2ð â¡2
cot ðâtan ð
Prove that 1âcos 2ð
sin 2ðâ¡ tan ð
Just like Chapter 6 had âproveyâ and âsolveyâ questions, we also get the âproveyâ questions in Chapter 7. Just use the appropriate double angle or addition formula.
tan 2ð â¡2 tan ð
1 â tan2 ðâ¡
2
1tan ð
â tanð
â¡2
cot ð â tanð
?
1 â 1 â 2 sin2 ð
2 sin ð cos ðâ¡
2 sin2 ð
2 sin ð cos ðâ¡ tanð
?
Test Your Understanding
[OCR] Prove that cot 2ð¥ + ððð ðð 2ð¥ â¡ cot ð¥
ððšð¬ ðð
ð¬ð¢ð§ðð+
ð
ð¬ð¢ð§ðð
â¡ððšð¬ ðð + ð
ð¬ð¢ð§ððâ¡ðððšð¬ð ð â ð + ð
ðð¬ð¢ð§ð ððšð¬ ð
â¡ðððšð¬ð ð
ð ð¬ð¢ð§ð ððšð¬ ðâ¡ððšð¬ ð
ð¬ð¢ð§ðâ¡ ððšð ð
[OCR] By writing cos ð¥ = cos 2 Ãð¥
2or otherwise,
prove the identity 1âcos ð¥
1+cos ð¥â¡ tan2
ð¥
2
ððšð¬ ð â¡ ððð ð Ãð
ðâ¡ ðððšð¬ð
ð
ðâ ð
âŽð â ððð ð
ð + ððð ðâ¡ð â ðððšð¬ð
ðð
â ð
ð + ððððððð
â ð
â¡ð ð â ððšð¬ð
ðð
ð ððšð¬ððð
â¡ð¬ð¢ð§ð
ðð
ððšð¬ððð
â¡ ððð§ðð
ð
?
?
Exercise 7F
Pearson Pure Mathematics Year 2/ASPages 187-188
Extension:
[STEP I 2005 Q4]
(a) â117
125(b) ð¡ðð ð = 8 + 75? ?
Very Challenging Exam Example
Edexcel C3 June 2015 Q8
? b
We eventually want cos ðŽ â sinðŽso cos2 ðŽ â sin2ðŽ is best choice of double-angle formula because this can be factorise to give cos ðŽ â sinðŽ as a factor.
This is even less obvious. Knowing that weâll have cosðŽ â sin ðŽ cosðŽ + sin ðŽ in the denominator and that the cosðŽ + sin ðŽ will cancel, we might work backwards from the final result and multiply by cosðŽ + sin ðŽ! Working backwards from the thing weâre trying to prove is occasionally a good strategy (provided the steps are reversible!)
? a
Modelling
? a
? b
? c
? d
When trigonometric equations are in the form ð ð ðð ð𥠱 ð or ð ððð ð𥠱 ð , they can be used to model various things which have an oscillating behaviour, e.g. tides, the swing of a pendulum and sound waves.
Test Your Understanding
? a? bi? bii
? c
? d
Tip: Reflect carefully on the substitution you use to allow (bii) to match your identity in (a). ð =?
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