Induction proof for sum $\sin(\theta)+\sin(3\theta)+\cdots$

Question

I would like to prove by induction that

$$\sin(\theta)+\sin(3\theta)+\dots+\sin((2n-1)\theta)=\frac{\sin^2(n\theta)}{\sin(\theta)}$$

for $n\in\mathbb{Z}$. (This is a question from the International Baccalaureate Mathematics HL textbook by Fannon et al.; I teach the course.)

I am aware that this answer deals with the category of problems, but the solutions provided there are not induction proofs and do not obviously translate to induction proofs. In the office we managed to 'brute force' it but the solution seemed to require identities beyond the apparent level of the question.

Answer 1

you have to prove that $$\frac{\sin^2(nx)}{\sin(x)}+\sin((2n+1)x)=\frac{\sin^2((n+1)x)}{\sin(x)}$$ and work backwards it must be $$\sin(x)\sin((2n+1)x)=\sin^2((n+1)x)-\sin^2(nx)$$ can you prove this? and use $$\sin(x)-\sin(y)=2 \sin \left(\frac{x}{2}-\frac{y}{2}\right) \cos \left(\frac{x}{2}+\frac{y}{2}\right)$$ and $$\sin(x)+\sin(y)=2 \sin \left(\frac{x}{2}+\frac{y}{2}\right) \cos \left(\frac{x}{2}-\frac{y}{2}\right)$$

Answer 2

Base case: n=1

$$\sin(\theta)=\frac{\sin^2(\theta)}{\sin(\theta)}=\sin(\theta)$$

Inductive step: n -> n+1

Let's assume that the statement is true for n:

$$\sum_{k=1}^{n}\sin[(2k-1)\theta]= \sin(\theta)+\sin(3\theta)+\dots+\sin[(2n-1)\theta]=\frac{\sin^2(n\theta)}{\sin(\theta)}$$

Let's verify that it holds also for $n+1$, that is:

$$\sum_{k=1}^{n+1}\sin[(2k-1)\theta]=\frac{\sin^2[(n+1)\theta]}{\sin(\theta)}$$

Since:

$$\sum_{k=1}^{n+1}\sin[(2k-1)\theta]=\sum_{k=1}^{n}\sin[(2k-1)\theta]+\sin[(2n+1)\theta]= \frac{\sin^2(n\theta)}{\sin(\theta)}+\sin[(2n+1)\theta]$$

Finally we have t prove that:

$$\frac{\sin^2[(n+1)\theta]}{\sin(\theta)}=\frac{\sin^2(n\theta)}{\sin(\theta)}+\sin[(2n+1)\theta]$$