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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 $\frac{3\pi}{2}$.'

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@lab bhattacharjee Thanks for informing. Nevertheless, I think the solution provided here is elegant. – pipi Oct 24 '12 at 5:03
ya, it's exquisite. – lab bhattacharjee Oct 24 '12 at 5:08
up vote 6 down vote accepted

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

$$1 = \left|\cos^n x - \sin^n x\right| \leq \left|\cos^n x\right| + \left|\sin^n x\right| \leq \cos^2 x + \sin^2 x = 1. $$

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.

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This is a nice solution, using the same technique to solve all the cases. Thanks! – pipi Oct 24 '12 at 1:43
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 4:48

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