Consider the following trigonmetric expression

• $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)$
• $\sin(x-y)=\sin(x)\cos(y)-\sin(y)\cos(x)$ A1
• Hence $\sin(x+y)-\sin(x-y) = \sin(x)\cos(y)+\sin(y)\cos(x)-(\sin(x)\cos(y)-\sin(y)\cos(x))$ M1
• $=2\cos(x)\sin(y)$ A1

let $\mathrm{P}(n)$ be the proposition that $\sum_{r=1}^n \cos (2 r-1) \alpha=\frac{\sin 2 n \alpha}{2 \sin \alpha}$ for $n \in \mathbb{Z}^{+}$

considering $\mathrm{P}(1)$ :

A1 M1

so $\mathrm{P}(1)$ is true

assume $\mathrm{P}(k)$ is true, i.e. $\sum_{r=1}^k \cos (2 r-1) \alpha=\frac{\sin 2 k \alpha}{2 \sin \alpha}\left(k \in \mathbb{Z}^{+}\right)$ M1

Note: Award M0 for statements such as "let $n=k$ ". Note: Subsequent marks after this M1 are independent of this mark and can be awarded.

considering $\mathrm{P}(k+1)$

M1

$\begin{aligned} & =\frac{\sin 2 k \alpha}{2 \sin \alpha}+\cos ((2(k+1)-1) \alpha) \\ & =\frac{\sin 2 k \alpha+2 \cos ((2 k+1) \alpha) \sin \alpha}{2 \sin \alpha} \\ & = \frac{\sin2k\alpha +\sin((2k+1)\alpha)+\alpha)-\sin((2k+1)\alpha-\alpha)}{2\sin\alpha} \\ & =\frac{\sin2k\alpha+\sin(2k\alpha+2\alpha)-\sin2k\alpha}{2\sin \alpha}\\ & =\frac{\sin(2(k+1)\alpha)}{2\sin\alpha}\end{aligned}$ A1 A1 A1

$\mathrm{P}(k+1)$ is true whenever $\mathrm{P}(k)$ is true, $\mathrm{P}(1)$ is true, so $\mathrm{P}(n)$ is true for $n \in \mathbb{Z}^{+}$ R1

Note: Award the final R1 mark provided at least five of the previous marks have been awarded.