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

Besides Mathematical Induction,more solutions are desired.

SBF
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yibotg
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    you cannot get a simple closed form for that. – Beni Bogosel Feb 20 '12 at 12:05
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    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 ... – yibotg Feb 20 '12 at 14:17
  • @tan9p You don't simply get a closed form of $${\sin ^{2k}}\theta + {\cos ^{2k}}\theta $$ – Pedro Feb 20 '12 at 14:38
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    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. –  Feb 20 '12 at 15:30
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    @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. – yibotg Feb 21 '12 at 02:15
  • @Ben Crowell Plot[Sin[x]^(2{1, 2, 3, 4, 5, 6, 7}) + Cos[x]^(2{1, 2, 3, 4, 5, 6, 7}) ,{x,0,2Pi} – yibotg Feb 21 '12 at 02:26
  • Any reason to look at $\sin^{2k}\theta + \cos^{2k}\theta$ instead of $\sin^k\theta + \cos^k\theta$? – asmeurer Jun 17 '12 at 08:58
  • What exactly are you looking for? Using the power reduction formulas you can always reduce this to a linear combination of $\sin(nx)$ and $\cos(nx)$ for $n$ from 1 to $2k$. Or if you limit $n$ to be smaller, it will be a polynomial. Is this what you want? No doubt you could get some "closed form" in terms of some combanitorial coefficients, but unless $k$ equals 0 or 1 it's not going to drastically simplify. – asmeurer Jun 17 '12 at 09:12

2 Answers2

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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:

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

Álvaro Lozano-Robledo
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1

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.

Andrea Mori
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