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 | Conjecture | 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,B∈R+ and C,D∈R and −π<C⩽π.
The expression 5sinx+12cosx can be written in the form Acos(Bx+C)+D, where A,B∈R+ and C,D∈R and −π<C⩽π.
In general, the expression asinx+bcosx can be written in the form Acos(Bx+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 asinx+bcosx can also be written in the form √a2+b2(a√a2+b2sinx+b√a2+b2cosx).
Let a√a2+b2=sinθ
Sketch the graph y=3sinx+4cosx, for −2π⩽x⩽2π
Write down the amplitude of this graph
Write down the period of this graph
Use your answers from part (a) to write down the value of A, B and D.
Find the value of C.
Find arctan34, giving the answer to 3 significant figures.
Comment on your answer to part (c)(i).
By considering the graph of y=5sinx+12cosx, find the value of A, B, C and D.
A.
B.
C.
D.
Show that b√a2+b2=cosθ.
Show that ab=tanθ.
Hence prove your conjectures in part (e).
Markscheme
A1
[1 mark]
5 A1
[1 mark]
2π A1
[1 mark]
A=5, B=1, D=0 A1
[1 mark]
maximum at x=0.644 M1
So C=−0.644 A1
[2 marks]
0.644 A1
[1 mark]
it appears that C=−arctan34 A1
[1 mark]
M1
A=13 A1
B=1 and D=0 A1
maximum at M1
So C = −0.395 A1
[5 marks]
A1
[1 mark]
A1
[1 mark]
A1
[1 mark]
A1
[1 mark]
EITHER
use of a right triangle and Pythgoras’ to show the missing side length is M1A1
OR
Use of , leading to the required result M1A1
[2 marks]
EITHER
use of a right triangle, leading to the required result. M1
OR
Use of , leading to the required result. M1
[1 mark]
M1
M1A1
So , and A1
And M1
So A1
[6 marks]