Prove by induction that $\sum _{r=1}^n \cos((2r-1)\theta) = \frac{\sin(2n\theta)}{2\sin\theta}$ is true $\forall \ n \in \mathbb{Z^+}$

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.