(a)     Show that \(\sin 2nx = \sin \left( {(2n + 1)x} \right)\cos x - \cos \left( {(2n + 1)x} \right)\sin x\).

(b)     Hence prove, by induction, that

\[\cos x + \cos 3x + \cos 5x +  \ldots  + \cos \left( {(2n - 1)x} \right) = \frac{{\sin 2nx}}{{2\sin x}},\]

for all \(n \in {\mathbb{Z}^ + }{\text{, }}\sin x \ne 0\).

(c)     Solve the equation \(\cos x + \cos 3x = \frac{1}{2},{\text{ }}0 < x < \pi \).

(a)     \(\sin (2n + 1)x\cos x - \cos (2n + 1)x\sin x = \sin (2n + 1)x - x\)     M1A1

\( = \sin 2nx\)     AG

[2 marks]

 

(b)     if n = 1     M1

\({\text{LHS}} = \cos x\)

\({\text{RHS}} = \frac{{\sin 2x}}{{2\sin x}} = \frac{{2\sin x\cos x}}{{2\sin x}} = \cos x\)     M1

so LHS = RHS and the statement is true for n = 1     R1

assume true for n = k     M1

Note: Only award M1 if the word true appears.

   Do not award M1 for ‘let n = k’ only.

   Subsequent marks are independent of this M1.

 

so \(\cos x + \cos 3x + \cos 5x +  \ldots  + \cos (2k - 1)x = \frac{{\sin 2kx}}{{2\sin x}}\)

if n = k + 1 then

\(\cos x + \cos 3x + \cos 5x +  \ldots  + \cos (2k - 1)x + \cos (2k + 1)x\)     M1

\( = \frac{{\sin 2kx}}{{2\sin x}} + \cos (2k + 1)x\)     A1

\( = \frac{{\sin 2kx + 2\cos (2k + 1)x\sin x}}{{2\sin x}}\)     M1

\( = \frac{{\sin (2k + 1)x\cos x - \cos (2k + 1)x\sin x + 2\cos (2k + 1)x\sin x}}{{2\sin x}}\)     M1

\( = \frac{{\sin (2k + 1)x\cos x + \cos (2k + 1)x\sin x}}{{2\sin x}}\)     A1

\( = \frac{{\sin (2k + 2)x}}{{2\sin x}}\)     M1

\( = \frac{{\sin 2(k + 1)x}}{{2\sin x}}\)     A1

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

as true for n = 1 then true for all \(n \in {\mathbb{Z}^ + }\)     R1

Note: Final R1 is independent of previous work.

[12 marks]

 

(c)     \(\frac{{\sin 4x}}{{2\sin x}} = \frac{1}{2}\)     M1A1

\(\sin 4x = \sin x\)

\(4x = x \Rightarrow x = 0\) but this is impossible

\(4x = \pi - x \Rightarrow x = \frac{\pi }{5}\)     A1

\(4x = 2\pi + x \Rightarrow x = \frac{{2\pi }}{3}\)     A1

\(4x = 3\pi - x \Rightarrow x = \frac{{3\pi }}{5}\)     A1

for not including any answers outside the domain     R1

Note: Award the first M1A1 for correctly obtaining \(8{\cos ^3}x - 4\cos x - 1 = 0\) or equivalent and subsequent marks as appropriate including the answers \(\left( { - \frac{1}{2},\frac{{1 \pm \sqrt 5 }}{4}} \right)\).

[6 marks]

 

Total [20 marks]

This question showed the weaknesses of many candidates in dealing with formal proofs and showing their reasoning in a logical manner. In part (a) just a few candidates clearly showed the result and part (b) showed that most candidates struggle with the formality of a proof by induction. The logic of many solutions was poor, though sometimes contained correct trigonometric work. Very few candidates were successful in answering part (c) using the unit circle. Most candidates attempted to manipulate the equation to obtain a cubic equation but made little progress. A few candidates guessed \(\frac{{2\pi }}{3}\) as a solution but were not able to determine the other solutions.