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.

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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 $y^2=y^n, z^2=z^n$.

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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.

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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 $\sin(x)=-1$, $\cos(x)=0$.

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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 $\sin x = 0$. –  sos440 Aug 7 '12 at 6:01

Let $\sin^nx=y$ so, $\cos^nx=1+y$.

For real $x$, $1+y \le 1$ $\Rightarrow$ $y\le 0$ $\Rightarrow$ $\sin^nx\le0$

For even $n$, $\sin x$ must be $0$ and $\cos x=\pm 1$ $\Rightarrow$ $x=m\pi$ where $m$ is any integer for even $n$.

For odd $n$, $-1\le y\le 1$ and $-1\le 1+y\le 1$ $\Rightarrow$ $-2\le y\le 0$, $\Rightarrow$ $-1\le y\le 0$

$(\sin x)^n=y=-z$(say) so $1\ge z\ge0$

$=>\sin x=(-z)^{\frac{1}{n}}=-z^\frac{1}{n}$ as n is odd and $(\cos x)^n=(1-z)=>\cos x=(1-z)^\frac{1}{n}$

Solving these two, we can get the values of x.

If z=0, $\sin x=0$ and $\cos x=(1-0)^\frac{1}{n}$ which has only one real value 1 as n is odd. So, $x=2r\pi$ where $r$ is any integer.

If z=1, $\sin x=-1^\frac{1}{n}=-1$ for the same reason and $\cos x=(1-1)^\frac{1}{n}=0$ So, $x=s\pi-\frac{\pi}{2}$ where $s$ is any integer.

If $z=\frac{1}{2}, sinx= -\frac{1}{2^{\frac{1}{n}}}$ and $cosx=(1-\frac{1}{2})^{\frac{1}{n}}=\frac{1}{2^{\frac{1}{n}}}$

But, $sin^2x+cos^2x=1=>({-\frac{1}{2^{\frac{1}{n}}}})^2+({\frac{1}{2^{\frac{1}{n}}}})^2=1$

$=>2.2^{-\frac{2}{n}}=1=>n=2$ which is impossible as n is odd.

If $z=\frac{1}{4}, sinx= -\frac{1}{4^{\frac{1}{n}}}$ and $cosx=(1-\frac{1}{4})^{\frac{1}{n}}=(\frac{3}{4})^{\frac{1}{n}}$

But, $sin^2x+cos^2x=1$=>1+3$^{\frac{2}{n}}=4^{\frac{2}{n}}$ =>n=2 which as earlier explained, is impossible as n is odd.

If $z=\frac{p}{q}$ where (p,q)=1, $sin^2x+cos^2x=1$ we get,

$p^{\frac{2}{n}}+(q-p)^{\frac{2}{n}}=q^{\frac{2}{n}}$

If $p^{\frac{2}{n}}=a=>p=a^{\frac{n}{2}}$,

$(q-p)^{\frac{2}{n}}=b=>q-p=b^{\frac{n}{2}}$ and

$q^{\frac{2}{n}}=c=>q=c^{\frac{n}{2}}$

$=>c^{\frac{n}{2}}=a^{\frac{n}{2}}+b^{\frac{n}{2}}$ (observe that n is odd)

If a,b,c are perfect squares, let $a=A^2,b=B^2, c=C^2$=> $C^n=A^n+B^n$ which is unsolvable in integers if n>2 according to FLT.

If some or all of them are not perfect squares, the respective term will be irrational as n is odd.

If one of the term is irrational, there will be no solution as the rest two are rational and (rational ± rational=rational).

I will try to figure out the rest two cases.

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Why "For even n, sinx must be 0" ? –  ᴊ ᴀ s ᴏ ɴ Aug 7 '12 at 7:45
For real x,$(sinx)^{2t}≥0$, but here $sin^nx≤0$ –  lab bhattacharjee Aug 7 '12 at 8:04
What if z ≠ 0, 1 ? –  ᴊ ᴀ s ᴏ ɴ Aug 7 '12 at 8:15
I'm not sure if the last 2 lines of your answer is right... Can you further explain it a little bit? thx~ –  ᴊ ᴀ s ᴏ ɴ Aug 8 '12 at 6:59
Fermat's Last Theorem +_+ –  ᴊ ᴀ s ᴏ ɴ Aug 9 '12 at 14:20