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.

• It seems to be false if $n=2$, because $sin^2x+cos^2x=1$ not $\frac{1}{2}$ for any $x$ – Atul Mishra Jun 28 '17 at 16:13
• are you sure that is $$\frac{1}{2^{n/2}}$$? – Dr. Sonnhard Graubner Jun 28 '17 at 16:14
• @AbhigyanChattopadhyay There may be a $n/2-1$ instead. – Clement C. Jun 28 '17 at 16:16
• Sorry... My mistake – Abhigyan Chattopadhyay Jun 28 '17 at 16:17
• @AbhigyanChattopadhyay most of the answers generalize to arbitrary exponent (mine, for $k$ non integer; Paolo's answer as well) – Clement C. Jun 28 '17 at 16:38

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 magic identity, nice! – Paolo Leonetti Jun 28 '17 at 16:30
• @PaoloLeonetti Thanks! It's kind of exasperating that everybody's trying to do this with calculus when it can be done without it. :P – Franklin Pezzuti Dyer Jun 28 '17 at 16:31
• I agree, this is clearly the best answer. – Paolo Leonetti Jun 28 '17 at 16:34
• I wouldn't say that finding the minimum of $(1-u)^k + u^k$ is really "hard" (and that is definitely something you can do without calculus -- think symmetry)-- trigonometric identities are nice, but something simple is good as well. How long does it take to find this identity, compared to dealing with the above? (After the fact, once the identity is given -- sure, this is a "proof from the Book"; but I'd argue it's not the fastest way if you have never seen it). +1 for it, though. – Clement C. Jun 28 '17 at 16:36
• Not long at all. I used a graphing calculator to form a conjecture, and the proof was quick and painless. :) – Franklin Pezzuti Dyer Jun 28 '17 at 16:37

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 $f$.

• Doing so, you will see that $f$ achieves its minimum for $u=\frac{1}{2}$, leading to a minimum of $\frac{2}{2^k}$. – Clement C. Jun 28 '17 at 16:18
• (and, of course, it does generalize to $k\in\mathbb{R}$, not necessarily integer.) – Clement C. Jun 28 '17 at 16:39
• $k \ge 1$, otherwise you get the oppositive inequality :P – Paolo Leonetti Jun 28 '17 at 16:50
• @PaoloLeonetti Fair point. I forgot there existed reals smaller than 1. – Clement C. Jun 28 '17 at 16:52

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 $\ge 0$.)

• Gosh, thanks for the support! It looks like I'm getting upvoted a bunch now. – Franklin Pezzuti Dyer Jun 28 '17 at 16:40

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.