C3 – Chapter 7 Trigonometry – Exam Questions
Section A – Provey/Solvey Questions
[Jan 2014 (I) Q7] 1. (i)
(a)
Prove that
cos 3  4 cos3   3cos 
(You may use the double angle formulae and the identity
cos( A  B)  cos A cos B  sin A sin B )
(4)
(b) Hence solve the equation
2cos3  cos 2 1  0
giving answers in the interval 0     .
Solutions based entirely on graphical or numerical methods are not acceptable.(6)

(ii) Given that   arcsin x and that 0    , show that
2
1  x 
2
cot  
x
,
0<x<1
(3)
[June 2013 Q3] 2. Given that
2 cos (x + 50)° = sin (x + 40)°.
(a) Show, without using a calculator, that
1
tan x° = tan 40°.
3
(4)
(b) Hence solve, for 0 ≤ θ < 360,
2 cos (2θ + 50)° = sin (2θ + 40)°,
giving your answers to 1 decimal place.
[June 2013 (R) Q6] 3.
Use an appropriate double angle formula to show that
cosec 2x = λ cosec x sec x,
and state the value of the constant λ.
(i)
(ii) Solve, for 0 ≤ θ < 2π, the equation
3sec2 θ + 3 sec θ = 2 tan2 θ
You must show all your working. Give your answers in terms of π.
[Jan 2013 Q6] 4.
(4)
Without using a calculator, find the exact value of
(sin 22.5° + cos 22.5°)2.
You must show each stage of your working.
(3)
(6)
(i)
(5)
(ii) (a) Show that cos 2 + sin  = 1 may be written in the form
k sin2  – sin  = 0, stating the value of k.
(2)
(b) Hence solve, for 0   < 360°, the equation
cos 2 + sin  = 1.
(4)
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[June 2012 Q5] 5. (a)
Express 4 cosec2 2θ − cosec2 θ in terms of sin θ and cos θ.
(b) Hence show that
(2)
4 cosec2 2θ − cosec2 θ = sec2 θ .
(4)
(c) Hence or otherwise solve, for 0 < θ < ,
4 cosec2 2θ − cosec2 θ = 4
giving your answers in terms of .
[Jan 2012 Q8] 6.
(a)
(3)
Starting from the formulae for sin (A + B) and cos (A + B), prove that
tan (A + B) =
tan A  tan B
.
1  tan A tan B
(4)
(b) Deduce that
1   3 tan 


tan    =
.
6
 3  tan 

(3)
(c) Hence, or otherwise, solve, for 0  θ  π,
1 + √3 tan θ = (√3 − tan θ) tan (π − θ).
Give your answers as multiples of π.
[June 2011 Q6] 7. (a)
Prove that
1
cos 2

= tan ,
sin 2
sin 2
(6)
  90n, n  ℤ.
(4)
(b) Hence, or otherwise,
(i) show that tan 15 = 2 – 3,
(ii) solve, for 0 < x < 360°,
(3)
cosec 4x – cot 4x = 1.
(5)
[Jan 2011 Q3] 8.
Find all the solutions of
2 cos 2 = 1 – 2 sin 
in the interval 0   < 360°.
[June 2010 Q1] 9. (a)
(6)
Show that
sin 2θ
= tan θ.
1  cos 2θ
(2)
(b) Hence find, for –180° ≤ θ < 180°, all the solutions of
2 sin 2θ
= 1.
1  cos 2θ
Give your answers to 1 decimal place.
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(3)
[June 2009 Q8] 10. (a)
Write down sin 2x in terms of sin x and cos x.
(b) Find, for 0 < x < π, all the solutions of the equation
cosec x − 8 cos x = 0.
giving your answers to 2 decimal places.
[Jan 2009 Q6] 11. (a)
(i)
(1)
(5)
By writing 3θ = (2θ + θ), show that
sin 3θ = 3 sin θ – 4 sin3 θ.
(4)
(ii) Hence, or otherwise, for 0 < θ <

, solve
3
8 sin3 θ – 6 sin θ + 1 = 0.
Give your answers in terms of π.
(b) Using sin (θ – ) = sin θ cos  – cos θ sin , or otherwise, show that
1
sin 15 = (6 – 2).
4
(5)
(4)
[Jan 2010 Q8] 12. Solve
cosec2 2x – cot 2x = 1
for 0  x  180.
Use the double angle formulae and the identity
cos(A + B) ≡ cosA cosB − sinA sinB
to obtain an expression for cos 3x in terms of powers of cos x only.
(7)
[Jan 2008 Q6a] 13. (a)
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(4)
Section B – Rsin(x +) and Min/Max Value Questions
[Jan 2011 Q1] 1. (a)
Express 7 cos x − 24 sin x in the form R cos (x + ) where R > 0 and 0

<< .
2
Give the value of  to 3 decimal places.
(3)
(b) Hence write down the minimum value of 7 cos x – 24 sin x.
(1)
(c) Solve, for 0  x < 2, the equation
7 cos x − 24 sin x = 10,
giving your answers to 2 decimal places.
(5)
[June 2010 Q7] 2. (a)
Express 2 sin θ – 1.5 cos θ in the form R sin (θ – α), where R > 0 and

0<α< .
2
Give the value of α to 4 decimal places.
(3)
(b) (i) Find the maximum value of 2 sin θ – 1.5 cos θ.
(ii) Find the value of θ, for 0 ≤ θ < π, at which this maximum occurs.
(3)
Tom models the height of sea water, H metres, on a particular day by the equation
 4 t 
 4 t 
H = 6 + 2 sin 
 – 1.5 cos 
 , 0 ≤ t <12,
 25 
 25 
where t hours is the number of hours after midday.
(c) Calculate the maximum value of H predicted by this model and the value of t, to 2
decimal places, when this maximum occurs.
(3)
(d) Calculate, to the nearest minute, the times when the height of sea water is predicted, by
this model, to be 7 metres.
(6)
[Jan 2010 Q3] 3. (a)
Express 5 cos x – 3 sin x in the form R cos(x + α), where R > 0 and 0 <
1
α< 2.
(4)
(b) Hence, or otherwise, solve the equation
5 cos x – 3 sin x = 4
for 0  x < 2, giving your answers to 2 decimal places.
(5)
Use the identity cos (A + B) = cos A cos B – sin A sin B, to show that
cos 2A = 1 − 2 sin2 A
(2)
The curves C1 and C2 have equations
C1: y = 3 sin 2x
[June 2009 Q6] 4. (a)
C2: y = 4 sin2 x − 2 cos 2x
(b) Show that the x-coordinates of the points where C1 and C2 intersect satisfy the equation
4 cos 2x + 3 sin 2x = 2
(3)
(c) Express 4cos 2x + 3 sin 2x in the form R cos (2x – α), where R > 0 and 0 < α < 90°,
giving the value of α to 2 decimal places.
(3)
(d) Hence find, for 0  x < 180°, all the solutions of
4 cos 2x + 3 sin 2x = 2,
giving your answers to 1 decimal place.
(4)
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[Jan 2009 Q8] 5. (a)
Express 3 cos θ + 4 sin θ in the form R cos (θ – α), where R and α are
constants, R > 0 and 0 < α < 90°.
(4)
(b) Hence find the maximum value of 3 cos θ + 4 sin θ and the smallest positive value of θ
for which this maximum occurs.
(3)
The temperature, f(t), of a warehouse is modelled using the equation
f (t) = 10 + 3 cos (15t)° + 4 sin (15t)°,
where t is the time in hours from midday and 0  t < 24.
(c) Calculate the minimum temperature of the warehouse as given by this model.
(d) Find the value of t when this minimum temperature occurs.
(2)
(3)
[June 2008 Q2] 6.
f(x) = 5 cos x + 12 sin x.

Given that f(x) = R cos (x – α), where R > 0 and 0 < α < ,
2
(a) find the value of R and the value of α to 3 decimal places.
(b) Hence solve the equation
5 cos x + 12 sin x = 6
(4)
for 0  x < 2π.
(c) (i) Write down the maximum value of 5 cos x + 12 sin x.
(ii) Find the smallest positive value of x for which this maximum value occurs.
(5)
(1)
(2)
[June 2013 Q8] 7.
Kate crosses a road, of constant width 7 m, in order to take a photograph of a marathon
runner, John, approaching at 3 m s–1.
Kate is 24 m ahead of John when she starts to cross the road from the fixed point A.
John passes her as she reaches the other side of the road at a variable point B, as shown in
Figure 2.
Kate’s speed is V m s–1 and she moves in a straight line, which makes an angle θ,
0 < θ < 150°, with the edge of the road, as shown in Figure 2.
You may assume that V is given by the formula
V
21
,
24sin   7 cos 
0 < θ < 150°
(a) Express 24sin θ + 7cos θ in the form R cos (θ – α), where R and α are constants and
where R > 0 and 0 < α < 90°, giving the value of α to 2 decimal places.
(3)
Given that θ varies,
(b) find the minimum value of V.
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(2)
Given that Kate’s speed has the value found in part (b),
(c) find the distance AB.
(3)
Given instead that Kate’s speed is 1.68 m s–1,
(d) find the two possible values of the angle θ, given that 0 < θ < 150°.
(6)
[June 2013 (Withdrawn) Q8] 8.
[June 2013 (R) Q3] 9.
f(x) = 7cos x + sin x
Given that f(x) = Rcos(x – a), where R > 0 and 0 < a < 90°,
(a) find the exact value of R and the value of a to one decimal place.
(b) Hence solve the equation
(3)
7cos x + sin x = 5
for 0 ≤ x < 360°, giving your answers to one decimal place.
(c) State the values of k for which the equation
7cos x + sin x = k
has only one solution in the interval 0 ≤ x < 360°.
[Jan 2013 Q4] 10. (a)

0<α< .
2
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(5)
(2)
Express 6 cos  + 8 sin  in the form R cos ( – α), where R > 0 and
Give the value of α to 3 decimal places.
p( ) =
(b)
(4)
4
,
12  6 cos   8 sin 
0    2.
Calculate
(i) the maximum value of p( ),
(ii) the value of  at which the maximum occurs.
(4)
f(x) = 7 cos 2x − 24 sin 2x.
[June 2012 Q8] 11.
Given that f(x) = R cos (2x + α), where R > 0 and 0 < α < 90,
(a) find the value of R and the value of α.
(b) Hence solve the equation
(3)
7 cos 2x − 24 sin 2x = 12.5
for 0  x < 180, giving your answers to 1 decimal place.
(5)
(c) Express 14 cos2 x − 48 sin x cos x in the form a cos 2x + b sin 2x + c, where a, b, and c
are constants to be found.
(2)
(d) Hence, using your answers to parts (a) and (c), deduce the maximum value of
14 cos2 x − 48 sin x cos x.
(2)
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Mark Schemes
Section A
Question 1
Question 2
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Question 3
Question 4
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Question 5
Question 6
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Question 7
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Question 8
Question 9
Question 10
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Question 11
Question 12
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Question 13
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Section B
Question 1
Question 2
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Question 3
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Question 4
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Question 5
Question 6
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Question 7
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Question 8
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Question 9
Question 10
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Question 11
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