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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6First, carefully, do $n=1$ and $n=2.$ – Will Jagy Aug 07 '12 at 05:29
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$\cos^nx-1=\sin^nx$ $=>(\cos x-1)(...)=sinx(sin^{n-1}x)$, $=>-2sin^2\frac{x}{2}(\cos x-1)(...)=2sin\frac{x}{2}cos\frac{x}{2}(sin^{n-1}x)$, Clearly, $sin\frac{x}{2}=0$ is a solution =>$x=2m\pi$ where m is any integer. – lab bhattacharjee Aug 07 '12 at 06:00
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RE lab bhattacharjee: How do you get $-2sin^{2}{x/2}$? – JSCB Aug 07 '12 at 06:11
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@jasoncube, $cos2A=1-2sin^2A$ – lab bhattacharjee Aug 07 '12 at 06:45
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RE lab bhattacharjee: So you have typed the extra $cosx-1$ right? – JSCB Aug 07 '12 at 06:56
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As I've taken it as factor of $cos^nx-1$ as (a-b) is always a factor of $(a^n-b^n)$ – lab bhattacharjee Aug 07 '12 at 19:33
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But the $\cos x-1$ should have been changed to $-2\sin^{2}\frac{x}{2}$. – JSCB Aug 08 '12 at 06:34
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Is the '-' sign discernible? – lab bhattacharjee Aug 11 '12 at 15:31
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What do you mean? – JSCB Aug 21 '12 at 15:40
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@http://www.cut-the-knot.org/wiki-math/index.php?n=MathematicalOlympiads.IMO1961Problem3 – lab bhattacharjee Oct 26 '13 at 06:35
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This problem is IMO $1961$ Problem $3$. Here's an AoPS thread about it. – user236182 Oct 05 '17 at 20:13
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https://math.stackexchange.com/questions/1874442/solve-cosn-x-sinn-x-1 https://math.stackexchange.com/questions/1874477/solve-cosn-x-sinnx-1 https://math.stackexchange.com/questions/1023673/general-solution-of-the-equation-sin2015-phi-cos2015-phi-1 – user236182 Oct 05 '17 at 20:17
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https://socratic.org/questions/what-is-the-solution-for-cos-nx-sin-nx-1-witn-n-in-nn – user236182 Oct 05 '17 at 20:21
3 Answers
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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.
Christian Blatter
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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 $\sin(x)=-1$, $\cos(x)=0$.
robjohn
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8I 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 07 '12 at 06:01
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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
$$1 =y+z \geq y^2+z^2 =1$$
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
$$1 =y^2+z^2 \geq y^n+z^n =1$$
with equality if and only if $y^2=y^n, z^2=z^n$.
N. S.
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Sorry for this late comment, but I was checking this question today, why in case 1 you have $ y+z \leq y^2 +z^2$ ? Knowing that $y,z \leq 1$ ?? – Nizar Jan 30 '18 at 13:50
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