# Solve $\cos^{n}x-\sin^{n}x=1$ with $n\in \mathbb{N}$.

Solve $\cos^{n}x-\sin^{n}x=1$ with $n\in \mathbb{N}$

I have no idea how to deal with this crazy question. One idea came into my mine is factorization, but I can't go on... Can anyone help me please? Thank you.

-
First, carefully, do $n=1$ and $n=2.$ – Will Jagy Aug 7 '12 at 5:29
$\cos^nx-1=\sin^nx$ $=>(\cos x-1)(...)=sinx(sin^{n-1}x)$, $=>-2sin^2\frac{x}{2}(\cos x-1)(...)=2sin\frac{x}{2}cos\frac{x}{2}(sin^{n-1}x)$, Clearly, $sin\frac{x}{2}=0$ is a solution =>$x=2m\pi$ where m is any integer. – lab bhattacharjee Aug 7 '12 at 6:00
RE lab bhattacharjee: How do you get $-2sin^{2}{x/2}$? – ᴊ ᴀ s ᴏ ɴ Aug 7 '12 at 6:11
@jasoncube, $cos2A=1-2sin^2A$ – lab bhattacharjee Aug 7 '12 at 6:45
RE lab bhattacharjee: So you have typed the extra $cosx-1$ right? – ᴊ ᴀ s ᴏ ɴ Aug 7 '12 at 6:56

If $n$ is even, then

$$1= \cos^{n}x-\sin^{n}x \leq 1-0=1$$

with equality if and only if $\cos^{n}x=1, \sin^n(x)=0$.

If $n$ is odd,

$$1= \cos^{n}x-\sin^{n}x \,,$$

implies $\cos(x) \geq 0$ and $\sin(x) <0$. Let $\cos(x)=y, \sin(x)=-z$, with $y,z \geq 0$.

$$y^n+z^n=1$$ $$y^2+z^2=1$$

Case 1: $n=1$:

Then , since $0 \leq y,z \leq 1$ we have

$$1 =y+z \leq y^2+z^2 =1$$

with equality if and only if $y=y^2, z=z^2$.

Case 2: $n \geq 3$:

Then , since $0 \leq y,z \leq 1$ we have

$$1 =y^2+z^2 \leq y^n+z^n =1$$

with equality if and only if $y^2=y^n, z^2=z^n$.

-

We consider the $2\pi$-periodic function $$f(x):=\cos^n x-\sin^n x$$ and determine its stationary points in $[0,2\pi[\$. One gets $$f'(x)=-n\cos x\sin x\bigl(\cos^{n-2}x+\sin^{n-2}x\bigr)\ ;$$ therefore the stationary points are the multiples of ${\pi\over2}$, and for odd $n>2$ the points where $\cos x=-\sin x$, i.e., the points ${3\pi\over4}$ and ${7\pi\over4}$. In these points one has the values $$f(0)=1, \quad f({\pi\over2})=-1,\quad f(\pi)=(-1)^n,\quad f({3\pi\over2})=(-1)^{n-1}\ ,$$ furthermore for $n=2m+1$ the values $$f({3\pi\over4})=-{\sqrt{2}\over 2^m}, \quad f({7\pi\over4})={\sqrt{2}\over 2^m}<1\ .$$ It follows that the global maximal value of $f$ is $1$. This value is taken at $0$ and $\pi$ if $n$ is even, and at $0$ and ${3\pi\over2}$ if $n$ is odd.

-

Hint:

For all $n$, when $\cos(x)=1$, $\sin(x)=0$.

For even $n$, when $\cos(x)=-1$, $\sin(x)=0$.

For odd $n$, when $\sin(x)=-1$, $\cos(x)=0$.

-
I guess that, it is implicitly used in robjohn's solution that $x^{n/2} + (1 - x)^{n/2} < 1$ whenever $0 < x < 1$ and $n > 2$. Thus for $n \geq 3$, $$|\cos^n x - \sin^n x| \leq |\cos^n x| + |\sin^n x| < 1$$ unless either $\cos x = 0$ or $\sin x = 0$. – Sangchul Lee Aug 7 '12 at 6:01