Prove that $\cos{2\alpha}+\cos{4\alpha}+\cdots+\cos{2n\alpha} = \frac{\sin(n+1)\alpha \cos{n \alpha}}{\sin{\alpha}}-1$ I was doing a trigonometry problem and it claimed that $$\cos{2\alpha}+\cos{4\alpha}+\cdots+\cos{2n\alpha} = \dfrac{\sin(n+1)\alpha \cos{n \alpha}}{\sin{\alpha}}-1$$
without proof. Is this supposed to be an obvious result or does it require work to prove it?

Hint: $2\sin \alpha \cdot \cos(2k\alpha)= \sin((2k+1)\alpha)-\sin((2k-1)\alpha)\Rightarrow$
$$ \begin{align*} \sum_{k=1}^n \cos(2k\alpha) &=\frac{ \sum_{k=1}^n 2\sin\alpha\cdot \cos (2k\alpha)}{2\sin\alpha} \\
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&=\frac{ \sum_{k=1}^n \left(\sin((2k+1)\alpha)-\sin((2k-1)\alpha)\right)}{2\sin \alpha} \\
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&=\dfrac{\sin((2n+1)\alpha)-\sin\alpha}{2\sin \alpha} \\
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&=... \end{align*} $$
