Solved: Consider the following trigonometric expression (x+y)-(x-y)... [IB Mathematics Analysis and Approaches (AA)]

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 $\mathrm{P}(1)$ : $\text{LHS}=\cos((2(1)-1)\alpha)=\cos \alpha \text{ and RHS}=\frac{\sin(2(1)\alpha)}{2 \sin \alpha}=\frac{2 \sin \alpha \cos \alpha}{2 \sin \alpha}=\cos \alpha$ So $\text{LHS}=\text{RHS}$ A1 M1
So $\mathrm{P}(1)$ is true
Assume $\mathrm{P}(k)$ is true, i.e. $\sum_{r=1}^k \cos (2 r-1) \alpha=\frac{\sin 2 k \alpha}{2 \sin \alpha}$ for $k \in \mathbb{Z}^{+}$ M1
Considering $\mathrm{P}(k+1)$ : $\sum_{r=1}^{k+1} \cos (2 r-1) \alpha=\sum_{r=1}^k \cos (2 r-1) \alpha+\cos ((2(k+1)-1) \alpha)$ M1
Algebraic manipulation using the identity: $\begin{aligned} &= \frac{\sin 2 k \alpha}{2 \sin \alpha}+\cos ((2(k+1)-1) \alpha) \\ &= \frac{\sin 2 k \alpha+2 \cos ((2 k+1) \alpha) \sin \alpha}{2 \sin \alpha} \\ &= \frac{\sin 2k\alpha +\sin((2k+1)\alpha+\alpha)-\sin((2k+1)\alpha-\alpha)}{2\sin\alpha} \\ &= \frac{\sin 2k\alpha+\sin(2k\alpha+2\alpha)-\sin 2k\alpha}{2\sin \alpha}\\ &= \frac{\sin(2(k+1)\alpha)}{2\sin\alpha} \end{aligned}$ A1 A1 A1
$\mathrm{P}(k+1)$ is true whenever $\mathrm{P}(k)$ is true, and since $\mathrm{P}(1)$ is true, $\mathrm{P}(n)$ is true for $n \in \mathbb{Z}^{+}$ R1

8 marks total

NoteAward M0 for statements such as "let $n=k$ ". Subsequent marks after this M1 are independent of this mark and can be awarded. NoteAward the final R1 mark provided at least five of the previous marks have been awarded.

Mathematics Analysis and Approaches (AA) International Baccalaureate (IB) Practice Question: Consider the following trigonometric expression (x+y)-(x-y) 1. Show that the expression is equal to 2(x)(y) 2. Hence,...