$\cos^n x-\sin^n x=1$

Question

For $0 < x < 2\pi$ and positive even $n$, the only solution for $\cos^n x-\sin^n x=1$ is $\pi$.
The argument is simple as $0\le\cos^n x, \sin^n x\le1$ and hence $\cos^n x-\sin^n x=1$ iff $\cos^n x=1$ and $\sin^n x=0$.

My question is that any nice argument to show the following statement?

'For $0 < x < 2\pi$ and positive odd $n$, the only solution for $\cos^n x-\sin^n x=1$ is $\frac{3\pi}{2}$.'

http://math.stackexchange.com/questions/179778/solve-cosnx-sinnx-1-with-n-in-mathbbn — lab bhattacharjee, Oct 24 '12 at 04:36
@lab bhattacharjee Thanks for informing. Nevertheless, I think the solution provided here is elegant. — pipi, Oct 24 '12 at 05:03
This problem is IMO $1961$ Problem $3$. Here's an AoPS thread about it. — user236182, Oct 05 '17 at 20:13
Here are some relevant Math.SE question links: (1), (2), (3), (4). — user236182, Oct 05 '17 at 20:29
https://socratic.org/questions/what-is-the-solution-for-cos-nx-sin-nx-1-witn-n-in-nn — user236182, Oct 05 '17 at 20:29

score 9 · Accepted Answer · answered Oct 24 '12 at 00:42

9

We leave the case $n = 1$ and $n = 2$ separately, and assume $n \geq 3$ from now on.

Observe that if $|r| \leq 1$, then $|r^n| \leq r^2$ with equality if and only if $r = 0$ or $|r| = 1$. Then it follows that

This forces every intermediate inequality to be equality. In particular, we must have

$$ \cos x , \sin x \in \{0, \pm 1\}.$$

Thus $x \in \{ \frac{\pi}{2}, \pi, \frac{3\pi}{2} \}$. Now the rest is clear.

answered Oct 24 '12 at 00:42

Sangchul Lee

167,468

This is a nice solution, using the same technique to solve all the cases. Thanks! – pipi Oct 24 '12 at 01:43
Using this technique, $\cos^n x \pm \sin^n x=1$ can be easily solve for all positive integers $n$. – pipi Oct 24 '12 at 04:48

$\cos^n x-\sin^n x=1$

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