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Date May Example questions Marks available 1 Reference code EXM.3.AHL.TZ0.5
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Sketch Question number 5 Adapted from N/A

Question

This question investigates the sum of sine and cosine functions

The expression  3 sin x + 4 cos x can be written in the form  A cos ( B x + C ) + D , where  A , B R + and  C , D R and  π < C π .

The expression  5 sin x + 12 cos x can be written in the form  A cos ( B x + C ) + D , where  A , B R + and  C , D R and π < C π .

In general, the expression  a sin x + b cos x  can be written in the form  A cos ( B x + C ) + D , where  a , b , A , B R + and  C , D R and π < C π .

Conjecture an expression, in terms of a and b , for

The expression  a sin x + b cos x  can also be written in the form  a 2 + b 2 ( a a 2 + b 2 sin x + b a 2 + b 2 cos x ) .

Let  a a 2 + b 2 = sin θ

Sketch the graph  y = 3 sin x + 4 cos x , for  2 π x 2 π

[1]
a.i.

Write down the amplitude of this graph

[1]
a.ii.

Write down the period of this graph

[1]
a.iii.

Use your answers from part (a) to write down the value of  A , B  and D .

[1]
b.i.

Find the value of C .

[2]
b.ii.

Find arctan 3 4 , giving the answer to 3 significant figures.

[1]
c.i.

Comment on your answer to part (c)(i).

[1]
c.ii.

By considering the graph of  y = 5 sin x + 12 cos x , find the value of A B C and  D .

[5]
d.

A .

[1]
e.i.

B .

[1]
e.ii.

C .

[1]
e.iii.

D .

[1]
e.iv.

Show that b a 2 + b 2 = cos θ .

[2]
f.i.

Show that a b = tan θ .

[1]
f.ii.

Hence prove your conjectures in part (e).

[6]
g.

Markscheme

      A1

[1 mark]

a.i.

5       A1

[1 mark]

a.ii.

2 π       A1

[1 mark]

a.iii.

A = 5 B = 1 D = 0       A1

[1 mark]

b.i.

maximum at x = 0.644       M1

So C = 0.644       A1

[2 marks]

b.ii.

0.644      A1

[1 mark]

c.i.

it appears that  C = arctan 3 4       A1

[1 mark]

c.ii.

        M1

A = 13         A1

B = 1 and  D = 0         A1

maximum at  x = 0.395         M1

So C = −0.395   ( = arctan 5 12 )       A1

[5 marks]

d.

A = a 2 + b 2        A1

[1 mark]

e.i.

B = 1        A1

[1 mark]

e.ii.

C = arctan a b       A1

[1 mark]

e.iii.

D = 0       A1

[1 mark]

e.iv.

EITHER

use of a right triangle and Pythgoras’ to show the missing side length is b          M1A1

OR

Use of  si n 2 θ + co s 2 θ = 1 , leading to the required result         M1A1

[2 marks]

f.i.

EITHER

use of a right triangle, leading to the required result.         M1

OR

Use of tan θ = sin θ cos θ , leading to the required result.       M1

[1 mark]

f.ii.

a sin x + b cos x = a 2 + b 2 ( sin θ sin x + cos θ cos x )         M1

a sin x + b cos x = a 2 + b 2 ( cos ( x θ ) )         M1A1

So  A = a 2 + b 2 B = 1 and  D = 0        A1

And  C = θ         M1

So  C = arctan a b        A1

[6 marks]

g.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
a.iii.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.
[N/A]
e.i.
[N/A]
e.ii.
[N/A]
e.iii.
[N/A]
e.iv.
[N/A]
f.i.
[N/A]
f.ii.
[N/A]
g.

Syllabus sections

Topic 3— Geometry and trigonometry » SL 3.7—Circular functions: graphs, composites, transformations
Show 89 related questions
Topic 3— Geometry and trigonometry » AHL 3.10—Compound angle identities
Topic 3— Geometry and trigonometry

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