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Find all $x\in\mathbb{R}$ such that $\sin^{100}(x)+\cos^{100}(x)=1.$

Here, by $\sin^{100}(x),$ I mean $(\sin(x))^{100}.$ I know that some of you may not like this notation as it may be interpreted as the composition of $\sin$ with itself $100$ times, but it's what I'm used to.

I will be providing my own solution. So, think of this question as an invitation for you to give your own ideas/methods. My method uses some calculus, but I think a purely trigonometric solution is possible.

aqualubix
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    I'm almost certain that this is a duplicate, but can't easily find it through a search. $\quad$ The idea is to think of it as $1 = \sin^2 x + \cos^2x \leq \sin^{100} x + \cos^{100}x = 1$. So equality must hold throughout, which means 1/ $\sin (x) = \pm 1, \cos x = \pm 1$ (no solutions here) or 2/ $\sin (x) = \pm 1, \cos x = 0 $ or 3/$\sin(x) = 0 , \cos x = \pm 1$. – Calvin Lin Jun 25 '23 at 15:17
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    Other duplicate: https://math.stackexchange.com/questions/1103954 – Anne Bauval Jun 25 '23 at 15:32

2 Answers2

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Observe that $$1 = \sin^2 x + \cos^2x \geq \sin^{100} x + \cos^{100}x = 1.$$

So equality must hold throughout. Since $|\sin x|, |\cos x| \leq 1$, we have the following cases:

1. $|\sin x| = 1, |\cos x | = 1$
2. $|\sin x| = 1, \cos x = 0 $
3. $\sin(x) = 0 , |\cos x| = 1$.

Thus, equality holds iff $ x = k \frac{\pi}{2}$ for some integer $k$.

Now try it for $ \sin^{99} x + \cos^{99}x = 1$. Why do we get a different answer? Which aspect of this solution needs to be adapted?

Calvin Lin
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    I think you meant $1=\sin^{100}x+\cos^{100}x\le\sin^2x+\cos^2x=1$ – FShrike Jun 25 '23 at 15:27
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    The first case in the spoiler is impossible. Anyway, better not answer (multi)duplicates. – Anne Bauval Jun 25 '23 at 15:36
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    @AnneBauval Yup, I tried to do a search but couldn't find it. Thanks. – Calvin Lin Jun 25 '23 at 15:59
  • https://approach0.xyz/search/?q=AND%20site%3Amath.stackexchange.com%2C%20OR%20content%3A%24%5Ccos%5Enx%2B%5Csin%5Enx%3D1%24&p=1 – Anne Bauval Jun 25 '23 at 16:09
  • Ah, I did the search for $\sin^{100} x + \cos^{100} x = 1$ instead of $\sin^n x + ....$, thinking that Approach0 would be flexible enough, but it wasn't. Thanks! It only gave me $ \sin^2x + \cos^2x = 1$. – Calvin Lin Jun 25 '23 at 16:10
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I will solve a generalization of this question. Let $S=S_1\cup S_2\cup S_3$ where, $$S_1=\{2p\pi:p\in\mathbb{Z}\},$$ $$S_2=\{\frac{\pi}{2}+2q\pi:q\in\mathbb{Z}\},$$ $$S_3=\{\frac{3\pi}{2}+2r\pi:r\in\mathbb{Z}\}.$$

Let $m,n\in\mathbb{N}.$ Then, $\sin^{4m}(x)+\cos^{4n}(x)=1\iff x\in S.$

The reverse implication is easy; just substitute and check.

For the forward implication, let $y=\sin^{2}(x).$ Then, using the Pythagorean identity, $\cos^{2}(x)=1-y.$ After this substitution, the equation becomes $y^{2m}+(1-y)^{2n}=1.$ Consider the polynomial $p(y)=y^{2m}+(1-y)^{2n}-1.$ We are interested in roots of this polynomial in $[0,1].$ This is because $y=\sin^{2}(x).$ Notice that $p(0)=p(1)=0.$ So, there is some $y_0\in(0,1),$ such that $p'(y_0)=0.$ Further, $p''(y)$ is strictly positive for all $y.$ So, $y_0$ is a point of local minima. Also, since $p''(y)>0$ for all $y, p'$ is strictly increasing. So, for any $y<y_0, p'(y)<p'(y_0)=0.$ Similarly, for any $y>y_0, p(y)'>p'(y_0)=0.$ That means, in the interval $[0,y_0), p$ is strictly decreasing. So, for any $y\in(0,y_0), p(y)<p(0)=0.$ So, $p$ has no root in $(0,y_0).$ Similarly, one can show that $p$ has no root in $(y_0,1).$ In this case, $0=p(1)>p(y)$ for any $y\in(y_0,1).$ Moreover, $p(y_0)$ itself is not zero, since it's a local minima, and points in it's neighborhood are less than $0,$ so it must be less than $0$ too. In conclusion, the only roots of $p$ in $[0,1]$ are $0$ and $1.$ This gives $\sin^{2}(x)=0$ or $1.$ This implies that $x\in S.$

Note: In fact, the only real roots of $p$ are $0$ and $1.$

aqualubix
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