What is the sum $\sin^{2k}\theta+\cos^{2k}\theta$ equal to?
Besides Mathematical Induction,more solutions are desired.
What is the sum $\sin^{2k}\theta+\cos^{2k}\theta$ equal to?
Besides Mathematical Induction,more solutions are desired.
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:
$\sin^4 x + \cos^4 x = (1-\cos^2 x)^2 + \cos^4 x\\ = 1-2\cos^2x + 2\cos^4 x \\ = 1-2\cos^2x(1-\cos^2x)\\ = 1-2\sin^2x\cos^2x\\ = 1 - \frac{\sin^2(2x)}{2}.$
$\sin^6 x + \cos^6 x = (1-\cos^2 x)^3 + \cos^6 x\\ = 1-3\cos^2x + 3\cos^4 x - \cos^6 x + \cos^6 x \\ = 1-3\cos^2x + 3\cos^4x\\ = 1-3\cos^2x(1-\cos^2x)\\ = 1-3\sin^2x\cos^2x\\ = 1 - \frac{3\sin^2(2x)}{4}.$
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.