Prove by induction that $$\sum _{r=1}^n \cos((2r-1)\theta) = \frac{\sin(2n\theta)}{2\sin\theta}$$ is true $\forall \ n \in \mathbb{Z^+}$
So my attempt as this is as follows, I've started the inductive step, but I don't know where to continue from now, any help would be great.
If $n=1$
LHS = $\cos\theta$, RHS = $\frac{\sin(2\theta)}{2\sin\theta} = \cos\theta$ so $\therefore$ true when $n=1$.
Assume true for $n=k$,
$$\sum _{r=1}^k \cos((2r-1)\theta) = \frac{\sin(2k\theta)}{2\sin\theta}$$
If $n=k+1$
$$\sum _{r=1}^{k+1} \cos((2r-1)\theta) = \sum _{r=1}^k \cos((2r-1)\theta) + \cos((2k+1)\theta)$$
$$ = \frac{\sin(2k\theta)}{2\sin\theta} + cos((2k+1)\theta)$$
$$= \frac{\sin(k\theta)\cos(k\theta)}{\sin\theta} + \cos(2k\theta)\cos\theta – \sin(2k\theta)\sin\theta$$
I'm not really sure my last step has lead me anywhere, but i wasn't sure on what else to do than apply the compound angle formulae.
Best Answer
Starting from where you left off:
$$\frac{\sin (2k\theta)}{2\sin \theta} + \cos ((2k+1)\theta)$$
Let $x=2k\theta$. Then the expression is
$$\frac{\sin x}{2\sin\theta} + \cos (x+\theta)$$
Angle sum identity gives
$$\frac{\sin x}{2\sin\theta} + \cos x\cos\theta - \sin x\sin\theta$$
$$=\frac{\sin x + 2\cos x\cos\theta\sin\theta - 2\sin x\sin^2 \theta}{2\sin\theta}.$$
Factoring, we get $$=\frac{(1-2\sin^2\theta)\sin x + (2\cos\theta\sin\theta)\cos x}{2\sin\theta}.$$
We use the double angle formula and angle sum formula in reverse: $$=\frac{\cos 2\theta \sin x + \sin 2\theta \cos x}{2\sin \theta}$$
$$=\frac{\sin(x+2\theta)}{2\sin\theta}$$
$$=\frac{\sin(2k\theta + 2\theta)}{2\sin \theta}$$
$$=\frac{\sin(2(k+1)\theta)}{2\sin \theta}.$$
This completes the inductive step.