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Date November 2016 Marks available 6 Reference code 16N.1.AHL.TZ0.H_13
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Hence or otherwise Question number H_13 Adapted from N/A

Question

Find the value of sin π 4 + sin 3 π 4 + sin 5 π 4 + sin 7 π 4 + sin 9 π 4 .

[2]
a.

Show that 1 cos 2 x 2 sin x sin x ,   x k π  where k Z .

[2]
b.

Use the principle of mathematical induction to prove that

sin x + sin 3 x + + sin ( 2 n 1 ) x = 1 cos 2 n x 2 sin x ,   n Z + ,   x k π where k Z .

[9]
c.

Hence or otherwise solve the equation sin x + sin 3 x = cos x  in the interval 0 < x < π .

[6]
d.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

sin π 4 + sin 3 π 4 + sin 5 π 4 + sin 7 π 4 + sin 9 π 4 = 2 2 + 2 2 2 2 2 2 + 2 2 = 2 2    (M1)A1

 

Note: Award M1 for 5 equal terms with \) + \) or signs.

 

[2 marks]

a.

1 cos 2 x 2 sin x 1 ( 1 2 sin 2 x ) 2 sin x    M1

2 sin 2 x 2 sin x    A1

sin x    AG

[2 marks]

b.

let  P ( n ) : sin x + sin 3 x + + sin ( 2 n 1 ) x 1 cos 2 n x 2 sin x

if  n = 1

P ( 1 ) : 1 cos 2 x 2 sin x sin x which is true (as proved in part (b))     R1

assume P ( k )  true, sin x + sin 3 x + + sin ( 2 k 1 ) x 1 cos 2 k x 2 sin x      M1

 

Notes: Only award M1 if the words “assume” and “true” appear. Do not award M1 for “let n = k only. Subsequent marks are independent of this M1.

 

consider P ( k + 1 ) :

P ( k + 1 ) : sin x + sin 3 x + + sin ( 2 k 1 ) x + sin ( 2 k + 1 ) x 1 cos 2 ( k + 1 ) x 2 sin x

L H S = sin x + sin 3 x + + sin ( 2 k 1 ) x + sin ( 2 k + 1 ) x    M1

1 cos 2 k x 2 sin x + sin ( 2 k + 1 ) x    A1

1 cos 2 k x + 2 sin x sin ( 2 k + 1 ) x 2 sin x

1 cos 2 k x + 2 sin x cos x sin 2 k x + 2 sin 2 x cos 2 k x 2 sin x    M1

1 ( ( 1 2 sin 2 x ) cos 2 k x sin 2 x sin 2 k x ) 2 sin x    M1

1 ( cos 2 x cos 2 k x sin 2 x sin 2 k x ) 2 sin x    A1

1 cos ( 2 k x + 2 x ) 2 sin x    A1

1 cos 2 ( k + 1 ) x 2 sin x

so if true for n = k , then also true for  n = k + 1

as true for n = 1 then true for all n Z +      R1

 

Note: Accept answers using transformation formula for product of sines if steps are shown clearly.

 

Note: Award R1 only if candidate is awarded at least 5 marks in the previous steps.

 

[9 marks]

c.

EITHER

sin x + sin 3 x = cos x 1 cos 4 x 2 sin x = cos x    M1

1 cos 4 x = 2 sin x cos x ,   ( sin x 0 )    A1

1 ( 1 2 sin 2 2 x ) = sin 2 x    M1

sin 2 x ( 2 sin 2 x 1 ) = 0    M1

sin 2 x = 0  or sin 2 x = 1 2      A1

2 x = π ,   2 x = π 6 and  2 x = 5 π 6

OR

sin x + sin 3 x = cos x 2 sin 2 x cos x = cos x    M1A1

( 2 sin 2 x 1 ) cos x = 0 ,   ( sin x 0 )    M1A1

sin 2 x = 1 2 of cos x = 0     A1

2 x = π 6 ,   2 x = 5 π 6 and  x = π 2

THEN

x = π 2 ,   x = π 12  and x = 5 π 12      A1

 

Note: Do not award the final A1 if extra solutions are seen.

 

[6 marks]

d.

Examiners report

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d.

Syllabus sections

Topic 3—Geometry and trigonometry » AHL 3.8—Unit circle, Pythag identity, solving trig equations graphically
Topic 3—Geometry and trigonometry

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