I need to somehow show that $$\cos(x)+\cos(3x)+…\cos\left((2n-1)\,x\right)=\frac{\sin(2nx)}{2\sin(x)}$$ for some integer $n>0$.
This seems impossible to me since if I consider the left-hand side, I don't know any identity of the form $\cos(ax)+\cos(bx)$. So, I can’t do anything. On the other hand, if I consider the right-hand side, the only identity I know is $\sin(2nx)=2\sin(nx)\cos(nx)$, which doesn’t seem to help me. Finally, Taylor expansion is not helpful either.
How can I show this?
Best Answer
You can prove this by induction. Hints below.
Base case
For $n=1$, choose an appropriate trig identity to expand the numerator $\sin 2nx =\sin 2x$ on the right hand side.
Inductive case
Assume
$$\cos x + \cos 3x + \ldots + \cos ((2n-3)x) = \frac{\sin((2n-2)x)}{2\sin x}$$
This gives
\begin{align} \cos x + \cos 3x + \ldots + \cos ((2n-1)x) &= \frac{\sin((2n-2)x)}{2\sin x} + \cos ((2n-1)x) \\ &=\frac{\sin(2n-2)x + 2\cos ((2n-1)x) \sin x}{2\sin x} \\ &=\frac{\sin(2(n-1)x-x) + 2\cos ((2n-1)x) \sin x}{2\sin x} \end{align}
Then use a trig identity to expand $\sin(2(n-1)x-x)$, simplify and you should see fairly immediately that another trig identity gets you to $2\sin 2nx$ on the numerator.