What does  $\sin^{2k}\theta+\cos^{2k}\theta=$? 
What is the sum $\sin^{2k}\theta+\cos^{2k}\theta$ equal to? 

Besides Mathematical Induction,more solutions are desired.
 A: I do not think there is a closed form for all values of $k$, but one can play around with trigonometric identities to simplify the expression for certain values of $k$. For instance:


*

*If $k=2$, then:


$$\sin^4 x + \cos^4 x = (1-\cos^2 x)^2 + \cos^4 x\\
= 1-2\cos^2x + 2\cos^4 x \\
= 1-2\cos^2x(1-\cos^2x)\\
= 1-2\sin^2x\cos^2x\\
= 1 - \frac{\sin^2(2x)}{2}.$$


*

*If $k=3$, then:


$$\sin^6 x + \cos^6 x = (1-\cos^2 x)^3 + \cos^6 x\\
= 1-3\cos^2x + 3\cos^4 x - \cos^6 x + \cos^6 x \\
= 1-3\cos^2x + 3\cos^4x\\
= 1-3\cos^2x(1-\cos^2x)\\
= 1-3\sin^2x\cos^2x\\
= 1 - \frac{3\sin^2(2x)}{4}.$$
A: If you let $z_k=\cos^k(\theta)+i\sin^k(\theta)\in\Bbb C$, it is clear that
$$
\cos^{2k}(\theta)+\sin^{2k}(\theta)=||z_k||^2.
$$
When $k=1$ the complex point $z_1$ describes (under the usual Argand-Gauss identification $\Bbb C=\Bbb R^2$) the circumference of radius $1$ centered in the origin, and your expression gives $1$.
For any other value $k>1$, the point $z_k$ describes a closed curve $\cal C_k\subset\Bbb R^2$ and your expression simply computes the square distance of the generic point from the origin. There's no reason to expect that this expression may take a simpler form than it already has. 
