Finding range of values of trigonometric function

Question

$f(n)= (\sin x)^n + (\cos x)^n$

I wanted to find the range of the values of this function in terms of n .

What I tried

I tried to use various inequalities like AM, GM ,HM but failed to derive s definite value due to the restriction on domain of sin x and cos x. I was hoping someone would help me derive an expression for range in terms of n or more specifically help me calculate it for n=4,6.

Answer 1

$f(x)=\sin^m x+\cos^m x$

I found the range is $[-1,1]$ if $n$ is odd

$\left[2^{1-\frac{m}{2}},1\right]$ if $n$ is even

Indeed

$f'(x)=m \cos x \sin ^{m-1} x-m \sin x \cos ^{m-1} x$

$f'(x)=0$ if $m \sin x \cos x \left(\sin ^{m-2} x-\cos ^{m-2} x\right)=0$

that is if $\sin 2x =0$ or $\tan^{m-2} x=1$

if $m$ is odd: $\quad 2x=k\pi\to x=k\dfrac{\pi}{2};\;x=\dfrac{\pi}{4}+k\pi;\;x=\dfrac{3\pi}{4}+k\pi$

if $m$ is even: $\quad 2x=k\pi\to x=k\dfrac{\pi}{2};\;x=\pm\dfrac{\pi}{4}+k\pi;\;x=\pm\dfrac{3\pi}{4}+k\pi$

Anyway when $m$ is even minimum is achieved when $x=\dfrac{\pi}{4}+k\pi$

because $f''(x)=m \left((m-1) \sin ^{m-2} x+(m-1) \cos ^{m-2} x-m \left(\sin ^m x+\cos ^m x\right)\right)$

and $f''\left(\frac{\pi}{4}\right)=2^{1-\frac{m}{2}} (m-2) m>0$ for $m>2$

$f\left(\frac{\pi}{4}\right)=2^{1 - \frac{m}{2}}$

While if $m$ is odd the absolute minimum is

$f\left(\frac{3\pi}{2}\right)=-1$

because $f''\left(\frac{3\pi}{2}\right)=(-1)^{m+1} m>0$ for odd $m$

To understand that local maxima/minima were or not global I helped myself with graphs

Answer 2

For $n\geq2$ we obtain: $$\sin^nx+\cos^nx\leq\sin^2x+\cos^2x=1.$$ The equality occurs for $x=0$, which says that $1$ is a maximal value. $$\sin^4x+\cos^4x=1-2\sin^2x\cos^2x=1-\frac{1}{2}\sin^22x\geq\frac{1}{2}.$$ The equality occurs for $x=\frac{\pi}{4}$, which says that $\frac{1}{2}$ is a minimal value. $$\sin^6x+\cos^6x=1-3\sin^2x\cos^2x=1-\frac{3}{4}\sin^22x=\frac{1}{4}.$$ The equality occurs for $x=\frac{\pi}{4}$ again.

Since $f$ is a continuous function we obtain the answer for $n=4$: $$\left[\frac{1}{2},1\right]$$ and for $n=6$: $$\left[\frac{1}{4},1\right].$$ Done!

Answer 3

For all $n \ge 2$, $$ \sin^n(x) + \cos^n(x) \le \vert \sin(x) \vert^n + \vert \cos(x) \vert^n \le \sin^2(x) + \cos^2(x) = 1 $$ with equality for $x = 0$ (as already stated in another answer).

For odd $n \ge 3$, $$ \sin^n(x) + \cos^n(x) \ge -\vert \sin(x) \vert^n - \vert \cos(x) \vert^n \ge - \sin^2(x) - \cos^2(x) = -1 $$ with equality for $x = \pi$.

For even $n = 2k \ge 2$, substitute $u = \sin^2(x)$, then $$ \sin^n(x) + \cos^n(x) = u^k + (1-u)^k \ge 2 \left( \frac {u + (1-u)}{2} \right)^k = \frac{1}{2^{k-1}} $$ because $t \to t^k$ is a convex function for $t \ge 0$. Equality holds for $x = \frac \pi 4$, corresponding to $u = \frac 12$.

Putting it together, the range is $$ [-1, 1] \quad \text{for odd $n \ge 3$,} $$ and $$ [\frac{1}{2^{n/2-1}}, 1] \quad \text{for even $n \ge 2$.} $$

Finally, for $n=1$ the range can be deduced from the identity $$ \sin(x) + \cos(x) = \sqrt 2 \sin(x + \frac \pi 4) \, . $$

Graph for $n=4$ and $n = 5$ (created with wxMaxima):