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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 3sinx+4cosx can be written in the form Acos(Bx+C)+D, where A,BR+ and C,DR and π<Cπ.

The expression 5sinx+12cosx can be written in the form Acos(Bx+C)+D, where A,BR+ and C,DR and π<Cπ.

In general, the expression asinx+bcosx can be written in the form Acos(Bx+C)+D, where a,b,A,BR+ and C,DR and π<Cπ.

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

The expression asinx+bcosx can also be written in the form a2+b2(aa2+b2sinx+ba2+b2cosx).

Let aa2+b2=sinθ

Sketch the graph y=3sinx+4cosx, for 2πx2π

[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 arctan34, 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=5sinx+12cosx, find the value of ABC and D.

[5]
d.

A.

[1]
e.i.

B.

[1]
e.ii.

C.

[1]
e.iii.

D.

[1]
e.iv.

Show that ba2+b2=cosθ.

[2]
f.i.

Show that ab=tanθ.

[1]
f.ii.

Hence prove your conjectures in part (e).

[6]
g.

Markscheme

      A1

[1 mark]

a.i.

5       A1

[1 mark]

a.ii.

2π      A1

[1 mark]

a.iii.

A=5B=1D=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=arctan34      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  (=arctan512)      A1

[5 marks]

d.

A=a2+b2       A1

[1 mark]

e.i.

B=1       A1

[1 mark]

e.ii.

C=arctanab      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 sin2θ+cos2θ=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 tanθ=sinθcosθ, leading to the required result.       M1

[1 mark]

f.ii.

asinx+bcosx=a2+b2(sinθsinx+cosθcosx)        M1

asinx+bcosx=a2+b2(cos(xθ))        M1A1

So A=a2+b2B=1 and D=0       A1

And C=θ        M1

So C=arctanab       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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