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So, the start of the proof is fine. Base case, easy enough. But when I start using the inductive hypothesis to try and prove P(k+1), I run into the following wall:

$\begin{align*} 2\sum_{j=1}^{k+1} \sin x\cos^{2j-1}x&=\sin(2(k+1)x)\\ 2\sum_{j=1}^k \sin x\cos^{2j-1}x +2\sin x\cos^{2(k+1)-1}x &=\sin(2(k+1)x)\\ \sin(2kn)+2\sin x\cos^{2k+1}x &=\sin(2(k+1)x)\space\space\space\text{by our inductive hypothesis} \end{align*}$

Not sure what identity works for that second term. Do I have to switch to $e^{i\theta}$ form? Any insight for the trick would be appreciated. I am like 99% sure I copied the question down correctly as well.

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    $\begingroup$ Try $\,\cos((2j-1)x)\,$ instead of $\,\cos^{2j-1}x.$ $\endgroup$
    – Somos
    Commented Jan 15, 2021 at 12:43
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    $\begingroup$ @Somos Welp, must be a typo in the book then. That works out nicely! Thank you! I am literally looking at it and a couple of lines down is a problem similar to this one but with that value in the angle instead of the exponent haha $\endgroup$
    – Shane Carey
    Commented Jan 15, 2021 at 12:50

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