Finding range of values of trigonometric function $f(n)= (\sin x)^n + (\cos x)^n$
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.
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):
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!