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True or false? $$\cos^5(x)-\sin^5(x)=\cos(5x)$$ for all real x. I have no idea how to prove or disprove this. I tried to expand $\cos(5x)$ using double angle formula but I wasn't sure how to go from that to $$\cos^5(x)-\sin^5(x)$$

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    $\begingroup$ If you are interested in what $\cos(5x)$ is equal to, expand both sides of $e^{(5x)i} = (e^{xi})^5$ using Euler's equation $e^{ti} = \cos(t) + i \sin(t)$. $\endgroup$ – Lee Mosher Dec 11 '14 at 14:22
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$x=\dfrac{\pi}{4} \implies \cos^5 x-\sin^5 x=0 \neq\cos \dfrac{5\pi}{4}$.

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Pick values of $x$ where $\cos 5x=1$. Does $\cos^5x-\sin^5x=1$?

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    $\begingroup$ You beat me by < 1 minute! $\endgroup$ – Ahaan S. Rungta Dec 11 '14 at 13:25
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It is false. Quick counterexample: Note that, at $x=\frac{\pi}{5}$, we have $ \cos (5x) = -1 $, but clearly${}^\dagger$, $ \cos^5 \left( \frac {\pi}{5} \right) - \sin^5 \left( \frac {\pi}{5} \right) \ne 1 $. $\Box$


${}^\dagger$ It is clear that, since $ \frac {\pi}{5} $ is acute, $ 0 < \cos^5 \left( \frac {\pi}{5} \right), \sin^5 \left( \frac {\pi}{5} \right) < 1 $. Hence, $$ \cos^5 \left( \frac {\pi}{5} \right) - \sin^5 \left( \frac {\pi}{5} \right) < 1. $$This is how we get that it is not equal to $1$.

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  • $\begingroup$ That is "clearly" without any calculation? $\endgroup$ – Suzu Hirose Dec 11 '14 at 13:30
  • $\begingroup$ @SuzuHirose - Yes, it is clear without doing calculations. I have clarified in my post. $\endgroup$ – Ahaan S. Rungta Dec 11 '14 at 13:33
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    $\begingroup$ I think you mean $\pi/5$ is acute. $\endgroup$ – ganbustein Dec 11 '14 at 13:46
  • $\begingroup$ @ganbustein - Yes, thanks. $\endgroup$ – Ahaan S. Rungta Dec 11 '14 at 13:47

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