# 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.

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you cannot get a simple closed form for that. – Beni Bogosel Feb 20 '12 at 12:05
Use complexe numbers: sin x = (exp xi - exp -xi)/2... – Hassan Feb 20 '12 at 12:45
yep,not a simple form: if k is odd,2^(-2k+2) ( cos2kt + C_2k^2 cos (2k-2)t ... – tan9p Feb 20 '12 at 14:17
To build intuition, try graphing it for k=20. It's a picket fence. Looking at the graph, it's pretty unlikely that it will have a simpler representation than the one already given. – Ben Crowell Feb 20 '12 at 15:30
@Ben Crowell by graphing it for k = 1..7, I thought it would like to converges to a function.but I do not know the exact funtion. – tan9p Feb 21 '12 at 2:15

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:

$$\sin^6 x + \cos^6 x = (1-\cos^2 x)^3 + \cos^6 x\\ = 1-3\cos^2x + 3\cos^4 x - \cos^6 x + \cos^6 x \\ = 1-3\cos^2x + 3\cos^4x\\ = 1-3\cos^2x(1-\cos^2x)\\ = 1-3\sin^2x\cos^2x\\ = 1 - \frac{3\sin^2(2x)}{4}.$$

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thank you,but I want to generalized. – tan9p Feb 21 '12 at 2:16

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.

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The original problem is to discuss the Range of the function f(x) = cos^{2k}\theta + sin^{2k}\theta. – tan9p Feb 21 '12 at 2:17
Wait, you need to know the range of $f(x)$? That is not what you asked in your original post! – Álvaro Lozano-Robledo Feb 21 '12 at 2:30
yeah,but I thought I'd get the range if I know the exact form. – tan9p Feb 21 '12 at 3:44
If you want the range of $f(\theta)=\cos^{2k}(\theta)+\sin^{2k}(\theta)$ a useful intermediate step may be finding the points where $df/d\theta=0$. – Andrea Mori Feb 21 '12 at 9:09
you are right,the calculus is a good method. – tan9p Feb 21 '12 at 11:52