Finding the maximum value of $(\sin x)^{\cos x}$

Question

I found a question to find the maximum value of $(\sin x)^{\cos x}$.

Since $\sin x < 1$, I think the maximum value is $1$. Is it correct?

Anyone please help. Thanks in advance.

Answer 1

Hint:

What is

$$\lim_{x\to\pi^-}\sin x^{\cos x}\ ?$$

Answer 2

Unfortunately, no that is not correct. Remember that $x^{-a}=\frac{1}{x^{a}}$.

$\cos(x)$ takes on negative values. In particular, for $x$ values slightly less than $\pi$, we have $\sin(x)$ slightly greater than $0$, and $\cos(x)$ slightly greater than $-1$.

Ultimately, the function tends to $\infty$ as $x$ approaches $\pi$ from the left, so there is no maximum value.

Answer 3

As @Yves Daoust, let $x=\pi-y$ $$\big[\sin(x)\big]^{\cos(x)}=\big[\sin(y)\big]^{-\cos(y)}$$ Take logarithms and use Taylor expansion to end with $$\big[\sin(y)\big]^{-\cos(y)}=\frac 1y +\frac 16(1+3\log(y))\, y+O\left(y^3\right)$$ Now, consider $y \to 0$.