Find $\sin^5 x + \cos^5 x$

Question

Given $$\cos \left[\sqrt{\left(\sin x + \cos x\right)\left(1 - \sin x \cos x \right)}\right] \\{}= \sqrt{\cos \left(\sin x + \cos x \right) \cos \left(1 - \sin x \cos x\right)}.$$

Find $\sin^5 x + \cos^5 x.$

My try:
\begin{align*} \sin^5x+\cos^5x &=(\sin x+\cos x)(\sin^4x-\sin^3x\cos x+\sin^2x\cos^2x-\sin x \cos^3x+\sin^4x)\\ &=(\sin x+\cos x)[(\sin^2x+\cos^2x)^2-(\sin^2x+\cos^2x)\sin x \cos x-\sin^2x\cos^2x]\\ &=(\sin x+\cos x)[1-\sin x\cos x-\sin^2x\cos^2x]\\ &=\frac{1}{4}(\sin x+\cos x)(4-2\sin 2x- \sin^2 2x)\\ \end{align*}
I don't know what next?

Do you really mean $$\cos \left[\sqrt{\left(\sin x + \cos x\right)\left(1 - \sin x \cos x \right)}\right]?$$ Having trig expressions inside $\cos$ seems a bit off to me. — A-Level Student, Jul 06 '21 at 10:04
@TobyMak: Evaluate at what point? It is a nonconstant function. — Moishe Kohan, Jul 06 '21 at 10:06
$\sin(x)=1$ and $\cos(x)=1$ are solutions, are these the only ones? — QC_QAOA, Jul 06 '21 at 10:06
Just to simplify the notations $\cos\sqrt{sq}=\sqrt{\cos(s)\cos(q)}$ and we have to calculate $s(1-p-p^2)=s(3q-q^2-1)$ with $s=\sin(x)+\cos(x)$ and $p=\sin(x)\cos(x)=1-q$. I do not see how to get something for $s,q$ in initial equation though. — zwim, Jul 06 '21 at 10:12
The $\sin^5 x + \cos^5 x$ bit is irrelevant since there are only two values where LHS = RHS. — Toby Mak, Jul 06 '21 at 10:20
Yep, everything is trivialized and all expressions equal to $1$. Is this supposed to be a trap exercise ? — zwim, Jul 06 '21 at 10:21
It is the last problem in here (scroll all the way down): https://artofproblemsolving.com/community/q1h1607640p10028934 — husnucuk, Jul 06 '21 at 10:29
Not for all $x$, but RHS=LHS only for $x=0$ and $\pi/2$ (on one period, due to RHS<LHS elsewhere) and in this case $\sin,\cos$ have values $0,1$ or $1,0$ so overall the proposed expressions evaluate to $1$. — zwim, Jul 06 '21 at 11:49
Notice that $(\sin{x}+\cos{x})(1-\sin{x}\cos{x})$ can be written as $(\sin{x}+\cos{x})(\sin^2x+\cos^2x-\sin{x}\cos{x})$ which is equal to $\sin^3x+\cos^3x$ — basilisk, Jul 07 '21 at 16:03
@piteer As a form of source, can you add the link of husnucuk to your question post? It will be helpful, since there are two close votes on the question. — Sarvesh Ravichandran Iyer, Jul 08 '21 at 07:09

basilisk · Answer 1 · 2021-07-08T08:18:51.553

This is my try.. I'm pretty sure that there are shorter methods to solve this problem. And it's not complete as I have ignored some tedious cases.

Let $$A=\sin x + \cos x=\sqrt{2}\cos(\pi/4-x)\tag{1}$$ $$B=1-\sin x\cos x=1-\frac{\sin 2x}{2}=1-\frac{\cos[2(\pi/4-x)]}{2}=\frac{3}{2}-\cos^2(\pi/4-x) \tag{2}$$ Multiplying $(1)$ and $(2)$ $$AB=\sin^3{x} + \cos^3{x}=\frac{3\sqrt{2}}{2}\cos(\pi/4-x)-\sqrt{2}\cos^3(\pi/4-x)$$

Given, $$\cos(\sqrt{AB})=\sqrt{\cos A\cos B}$$ $$\implies2\cos^2(\sqrt{AB})=\cos(A+B)+\cos(A-B)$$ $$2\cos^2(\sqrt{AB})-1=\cos(A+B)+\cos(A-B)-1$$ $$\implies \cos(2\sqrt{AB})=\cos(A+B)+\cos(A-B)-1$$ $$\cos(2\sqrt{AB})-\cos(A-B)=\cos(A+B)-\cos(0)$$ $$-2\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)=-2\sin^2\bigg(\frac{A+B}{2}\bigg)$$ $$\implies\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)= \sin\bigg(\frac{A+B}{2}\bigg)\sin\bigg(\frac{A+B}{2}\bigg)\tag{$\star$}$$ $$\implies\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)=\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)$$ $$\implies\sin\bigg(\frac{2\sqrt{AB}+A-B}{2}\bigg)-\sin\bigg(\frac{2\sqrt{AB}-A+B}{2}\bigg)=0$$ $$\implies2\cos\big(2\sqrt{AB}\big)\sin\big(A-B\big)=0$$

Note that, for $(\star)$,the graph looks like this.I have only taken the simplest case.

For the sake of simplicity, we will consider $A=B$ as of now and ignore the other cases.. We have, $$\sqrt{2}\cos(\pi/4-x)=\frac{3}{2}-\cos^2(\pi/4-x)$$ Let $\cos(\pi/4-x)=k$ $$\implies k^2 +\sqrt{2}k-\frac{3}{2}=0$$ $$k=\frac{-3\sqrt{2}}{2},k=\frac{1}{\sqrt{2}}$$ $$\cos(\pi/4-x)=\cos(\pi/4)$$ $$\implies x=2n\pi +\frac{\pi}{2}\text{ or } x=2n\pi\text{ (where }n\in\mathbb{Z})$$

Hence $\sin^5x+\cos^5x=1$ provided that the condition given in the question is true

@Breakingnotsobad $1$ is not of the form $2n\pi$ or $2n\pi + \pi/2$. As $\sin(2n\pi+\pi/2)=\sin(\pi/2)=1$ and $\cos(2n\pi+\pi/2)=cos(\pi/2)=0$. Hence $\sin^5(x)+\cos^5(x)=1$. Similarly $\sin(2n\pi)=0$ and $\cos(2n\pi)=1$ — basilisk, Jul 08 '21 at 07:57
@Breakingnotsobad, I have edited my post. Are you asking whether $\sin^5x+\cos^5x$ is equal to $1$ for all values of $x$? If so it's not. The values of $x$ which satisfies the condition given in the question will have $\sin^5x+\cos^5x=1$ — basilisk, Jul 08 '21 at 08:08

score 0 · Answer 2 · answered Jul 09 '21 at 18:29

I wanted to develop the idea expressed in comments that LHS is bounded below by $\cos(1)$ while RHS is bounded above by $\cos(1)$ and that we get equality for $x=0\text{ or }\frac{\pi}2$ only.

The LHS part and equality case is not too complicated to prove:

$\require{cancel}\begin{align}f(x) &=\Big(\cos(x)+\sin(x)\Big)\Big(\overbrace{1}^{\cos(x)^2+\sin(x)^2}-\sin(x)\cos(x)\Big)\\ &=\cos(x)^3+\cancel{\cos(x)\sin(x)^2}-\cancel{\sin(x)\cos(x)^2}+\cancel{\sin(x)\cos(x)^2}+\sin(x)^3-\cancel{\sin(x)^2\cos(x)}\\ &=\cos(x)^3+\sin(x)^3\end{align}$

Notice also that $|f(x)|$ has period $\pi$ and that $\cos\searrow$ on $[0,\pi]$

Therefore (and since RHS undefined where $f(x)$ is negative):

$$\cos\left(\sqrt{|f(x)|}\right)\ge \cos(1)\quad\text{and}\quad LHS\ge \cos(1)$$

The equality case is for extremal points.

So let search for critical points $f'(x)=3\sin(x)\cos(x)(\sin(x)-\cos(x))=0\iff x=0,\frac{\pi}2,\frac{\pi}4$ over the $[0,\pi]$ interval, among which only $0,\frac{\pi}2$ lead to $f(x)=1$.

However, I faced a wall when I tried to prove the RHS part, so it seems basilik's approach despite looking more complicated at first sight, is eventually more effective.

Does anyone attempted this approach too ?

Sorry my comment is not immediately on topic but I wanted to mention that your mastery of TeX is very impressive to me! — Saegusa, Jul 09 '21 at 18:34
@Saegusa All thanks to this meta https://math.meta.stackexchange.com/q/5020/399263, just try to digest it over the years. — zwim, Jul 09 '21 at 18:39

Find $\sin^5 x + \cos^5 x$

2 Answers2