$\cos(x)+\cos(3x)+...\cos(2n-1)x=\sin(2nx)/2\sin(x)$ 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?
 A: 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.
A: Use Euler's identity
$$
e^{ix}=\cos x + i \sin x
$$
So your case can be written as the real part of
\begin{align*}
\cos x + \cos 3x + \cdots + \cos(2n-1)x +i(\sin x + \sin 3x + \cdots + \sin (2n-1)x )
&= \sum_{k=1}^n e^{i(2k-1)x}\\
&= e^{ix}\frac{e^{2nix}-1}{e^{2ix}-1}\\ 
&= \frac{\sin 2nx}{2\sin x} + i (\cdots)
\end{align*}
A: Using the identity $e^{i\theta}+e^{-i\theta}=2\cos(\theta)$, you can express the sum of the cosine terms as:-
$\cos(x)+\cos(3x)+\cdots+\cos((2n-1)x)=\frac{1}{2}(\sum_{k=1}^ne^{j(2k-1)x}+\sum_{k=1}^ne^{-j(2k-1)x})$
The right hand side of the above equation is the sum of two Geometric progressions for the complex exponential terms, where
$$\sum_{k=1}^ne^{j(2k-1)x}=\frac{e^{jx}(1-(e^{j2x})^n)}{1-e^{j2x}}=\frac{e^{jx}-e^{j(2n+1)x}}{1-e^{j2x}}$$
$$\sum_{k=1}^ne^{-j(2k-1)x}=\frac{e^{-jx}(1-(e^{-j2x})^n)}{1-e^{-j2x}}=\frac{e^{-jx}-e^{-j(2n+1)x}}{1-e^{-j2x}}$$
This leads (after some algebraic manipulation) to
$\large \sum_{k=1}^ne^{j(2k-1)x}+\sum_{k=1}^ne^{-j(2k-1)x}=\frac{e^{j(2n-1)x}+e^{-j(2n-1)x}-(e^{j(2n+1)x}+e^{-j(2n+1)x})}{2-(e^{j2x}+e^{-j2x})}\\\large=\frac{2\cos((2n-1)x)-2\cos((2n+1)x)}{2(1-\cos(2x))}=\frac{\cos((2n-1)x)-\cos((2n+1)x)}{1-\cos(2x)}$ 
Using the sum of angles identities for trigonometry (as stated below)
$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ 
$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$
$\cos(2a)=\cos^2(a)-\sin^2(a)=1-2\sin^2(a)$
we can further simplify the sum of the two Geometric progressions to
$$ \frac{\cos(2nx)\cos(x)+\sin(2nx)\sin(x)-\cos(2nx)\cos(x)+\sin(2nx)\sin(x)}{1-1+2\sin^2(x)}\\=\frac{2\sin(2nx)\sin(x)}{2\sin^2(x)}=\frac{\sin(2nx)}{\sin(x)}$$
Thus we have 
$$ \frac{1}{2}(\sum_{k=1}^ne^{j(2k-1)x}+\sum_{k=1}^ne^{-j(2k-1)x})=\frac{\sin(2nx)}{2\sin(x)}$$
so that the sum of the Cosine terms evaluates as follows:- 
$$\cos(x)+\cos(3x)+\cdots+\cos((2n-1)x)=\frac{\sin(2nx)}{2\sin(x)}$$
