# I am working on proving or disproving $\cos^5(x)-\sin^5(x)=\cos(5x)$

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 $$\cos^5(x)-\sin^5(x)$$

• 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 $e^{ti} = \cos(t) + i \sin(t)$. – Lee Mosher Dec 11 '14 at 14:22

$x=\dfrac{\pi}{4} \implies \cos^5 x-\sin^5 x=0 \neq\cos \dfrac{5\pi}{4}$.

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

• You beat me by < 1 minute! – Ahaan S. Rungta Dec 11 '14 at 13:25

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 $1$.

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