Prove by induction that $\sum _{r=1}^n \cos((2r-1)\theta) = \frac{\sin(2n\theta)}{2\sin\theta}$ is true $\forall \ n \in \mathbb{Z^+}$ 
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.
 A: 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.
A: $$\sin 2(k+1)\theta=\sin (2k+1)\theta \cos\theta+\cos (2k+1)\theta\sin\theta$$
$$\sin 2(k-1)\theta=\sin (2k+1)\theta \cos\theta-\cos (2k+1)\theta\sin\theta$$
whence
$$\sin 2(k+1)\theta-\sin 2(k-1)\theta=2\cos(2k+1)\theta\sin\theta$$
which with a bit of rearrangement gives you your inductive step. 
A: You may use the fact that:
$$1+\cosθ+\cos(2θ)+\cdots+\cos(nθ)=\frac12+\frac{\sin\left[\left(n+\frac12\right)θ\right]}{2\sin\left(\frac\theta2\right)}$$
Here's a very simple solution that doesn't use complex numbers, just some basic trigonometric identities.  Recall that $$2 \cos \alpha \sin \beta = \sin(\alpha + \beta) - \sin(\alpha - \beta).$$  With the choice $\alpha = k \theta$, $\beta = \theta/2$, we then have $$2 \cos k\theta \sin \frac{\theta}{2} = \sin\bigl((k + {\textstyle \frac{1}{2}})\theta\bigr) - \sin\bigl((k - {\textstyle \frac{1}{2}})\theta\bigr).$$  Summing of both sides over $k = 0, 1, \ldots, n$ and observing that the RHS telescopes, $$\sum_{k=0}^n 2 \cos k\theta \sin \frac{\theta}{2} = \sin\bigl((n + {\textstyle \frac{1}{2}})\theta\bigr) - \sin\bigl(-{\textstyle \frac{1}{2}}\theta\bigr),$$ from which it immediately follows that $$\sum_{k=0}^n \cos k\theta = \frac{1}{2}\left( 1 + \frac{\sin((k+\frac{1}{2})\theta)}{\sin\frac{\theta}{2}}\right).$$
