Prove that the minimum value of $\sin^n(x)+\cos^n(x)$ is $\frac{1}{2^{\frac{n}{2}-1}}$, if n is even A few days back, I discovered a relationship between $\sin^n(x)+\cos^n(x)$, when n is even.
Its minimum value was always $\frac{1}{2^{\frac{n}{2}-1}}$.
I tried to prove this, and to extend it to the case where $n$ was odd, but I failed to do so. Can someone tell me the first step to it? I really want to prove this by myself.
Thanks in advance
 A: HINT: Prove that
$$\sin^{2n}(x)+\cos^{2n}(x)$$
and
$$\frac{2^{n-1}-1}{2^n}\cos(4x)+\frac{2^{n-1}+1}{2^n}$$
have the same minimum points and period.
This can be done using trigonometric identities. Once you have turned the sum of trigonometric functions into a single trigonometric function, you can find its minima without even differentiating, since it is just a cosine wave.
EDIT: It seems that my "magic identity" was flawed, so I have improved my answer.
BETTER HINT: Show, using your knowledge of trig identities, that the period of $\sin^{2n}(x)+\cos^{2n}(x)$ is $\frac{\pi}{2}$, and that the minima occur at odd multiples of $\frac{\pi}{4}$. Then you can just evaluate it at those values, since call of them will be the same.
A: An outline:
Write $n=2k$ for some integer $k$. Then, you want to find the minimum of
$$
\sin^{2k}x+\cos^{2k }x = (1-\cos^2 x)^k + (\cos^2 x)^k
$$
over all $x\in\mathbb{R}$. Because $\cos$ is surjective on $[-1,1]$, it is equivalent to minimize
$$
f(u) = (1-u)^k + u^{k}
$$
over all $u\in[0,1]$. For that, you can differentiate the (smooth) function $f$.
A: Fix $n>1$ real and use
$$
\left(\frac{a^{n}+b^{n}}{2}\right)^{1/n} \ge \frac{a+b}{2},
$$
with $a=\sin^2 x$ and $b=\cos^2 x$ (hence, we are assuming that they are $\ge 0$.)
A: just a hint to improve
The derivative is
$n\sin (x)\cos (x)\Bigl(\sin^{n-2}(x)-\cos^{n-2}(x)\Bigr)$
the minimum is attained for $x $ such that
$$\tan (x)=1$$
or
$$x=\frac {\pi}{4} $$
you can finish.
