Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Besides Mathematical Induction,more solutions are desired.

share|improve this question
3  
you cannot get a simple closed form for that. –  Beni Bogosel Feb 20 '12 at 12:05
2  
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
1  
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
1  
@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

2 Answers 2

up vote 2 down vote accepted

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:

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

share|improve this answer
    
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.