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I need to find the frequency of the following trigonometric function.$$y=\sin^4(x)+\cos^4(x)$$ The "answers" section says the answer is: $$F_y=\frac{\pi}{2}$$

This is what i did:

Finding $\sin(x)^4$ frequency (I'll call it F1): $$\cos(2x)=1-\sin^2(x)$$ $$\sin^2(x)=\frac{1-\cos(2x)}{2}$$ $$\sin^4(x)=\frac{\cos^2(2x)-2\cos(2x)+1}{4}=\frac{cos^2(2x)+4\sin^2(x)-1}{4}$$ Finding $\cos(2x)^2$ frequency: $$\cos(4x)=2\cos^2(2x)-1$$ $$\cos^2(2x)=\frac{\cos(4x)+1}{2}$$ $$f_1=\frac{2\pi}{4}=\frac{\pi}{2}$$ Finding $\sin(x)^2$ frequency: $$\cos(2x)=1-2\sin^2(x)$$ $$\sin^2(x)=\frac{1-\cos(2x)}{2}$$ $$f_2=\frac{2\pi}{2}=\pi$$ $$F_1: \frac{f_1}{f_2}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$ $$F_1=\frac{\pi}{2}\times2=\pi$$ Finding $cos(x)^4$ frequency (I'll call it F2): $$\cos(2x)=2\cos^2(x)-1$$ $$\cos^2(x)=\frac{\cos(2x)+1}{2}$$ $$\cos^4(x)=\frac{\cos^2(2x)+2\cos(2x)+1}{4}$$ Finding $\cos(2x)$ frequency (we already have $\cos(2x)^2$ frequency - f1): $$f_3=\frac{2\pi}{2}=\pi$$ $$F_2: \frac{f_1}{f_3}=\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}$$ $$F_2=\frac{\pi}{2}\times2=\pi$$ Finding $y$'s frequency: $$F_y: \frac{F_1}{F_2}=\frac{\pi}{\pi}=\frac{1}{1}$$ $$F_y=\pi\times1=\pi$$

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4 Answers 4

up vote 3 down vote accepted

You have proved that $\pi$ is a period, but you have not shown that it is the smallest period.

I would tackle the problem in more or less the same way that you did, using double angle identities, but the algebra can be simplified. Note that $$1=(\cos^2 x+\sin^2 x)^2=\cos^4 x+\sin^4 x +2\cos^2 x\sin^2 x.$$ It follows that our function is equal to $$1-2 \cos^2 x\sin^2 x.$$ The only interesting part is $2\cos^2 x\sin^2 x$, which is $\frac{1}{2}\sin^2 2x$. It is clear that this has period $\dfrac{\pi}{2}$.

If you wish, you can use the trigonometric identity $\cos 2u=1-2\sin^2 u$ to express our function in terms of $\cos 4x$. We get $\frac{1}{2}\sin^2 2x=\frac{1}{4}(1-\cos 4x)$, and therefore $$\cos^4 x+\sin^4 x=\frac{3}{4}+\frac{1}{4}\cos 4x.$$

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$$\sin^4\left(x+\frac{\pi}{2}\right)=\left(\sin x\cos\left(\frac{\pi}{2}\right)+\sin\left(\frac{\pi}{2}\right)\cos x\right)^4=\cos^4x$$

$$\cos^4\left(x+\frac{\pi}{2}\right)=\left(\cos x\cos\left(\frac{\pi}{2}\right)-\sin x\sin\left(\frac{\pi}{2}\right)\right)^4=(-\sin x)^4=\sin^4 x$$

Thus, the period of $\,\sin^4 x+\cos^4x\,$ indeed is not more than $\,\pi/2\,$

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I won't pretend that I would have thought of this without seeing the answer but we can look at the problem geometrically. This is more of a nice illustration than a genuine 'proof by pictures'.

We know firstly that the curve $x^4+y^4=a$ is symmetric about the $x$- and $y$-axes:

Suppose that $f(\theta)=\sin^4\theta+\cos^4\theta=a$ for some $a$ in the range of $f$. Now we know that $\sin^2\theta+\cos^2\theta$ is equal to one. Hence we can place the point $(\cos\theta,\sin\theta)$ on the intersection of the unit circle and the curve $x^4+y^4=a$ as shown:

Now as the curve has $\pi/2$-rotation symmetry, the point $(\cos(\theta+\pi/2),\sin(\theta+\pi/2))$ is also on the curve $x^4+y^4=a$ and we are done.

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... and I suppose this yields all of the $\sin^{2n}(x)+\cos^{2n}(x)$ frequencies. –  Jp McCarthy Oct 16 '12 at 23:06
1  
I do not yet understand your answer. Maybe it contains concepts which I haven't learnt yet, such as non-injective functions, if that is what you have done here. Thank you for your effort though. Well appreciated. –  yuvalz Oct 17 '12 at 5:40
    
The red dots are $\theta\in[0,2\pi)$ such that $\sin^4\theta+\cos^4\theta=a$. Now rotate by $\pi/2$ --- morryah add $\pi/2$ to $\theta$ and you have another point that satisfies $x^4+y^4=a$. –  Jp McCarthy Oct 17 '12 at 8:40
    
I looked at it from a wrong perspective. I understand. Thank you. –  yuvalz Oct 17 '12 at 20:10

You might consider using Euler's formula, which can be used to obtain $$ \cos(x) = \frac{e^{ix} + e^{-ix}}{2} $$ and $$ \sin(x) = \frac{e^{ix} - e^{-ix}}{2i}. $$

Put $z = e^{ix}$, so that $$\cos(x) =\frac{z+1/z}{2}$$ and $$\sin(x) = \frac{z-1/z}{2i}$$

This gives $$\cos(x)^4+\sin(x)^4 = \left(\frac{z+1/z}{2}\right)^4 + \left(\frac{z-1/z}{2i}\right)^4 = \frac{z^4}{8} + 3/4 + \frac{1/z^4}{8}$$ which is $$\frac{3}{4} + \frac{1}{4} \cos(4x).$$ So your frequency is $$ \frac{2\pi}{4} = \frac{\pi}{2}.$$

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I don't yet know what Euler's formula is, not what some of the "icons" you used mean. But your effort is well appreciated. Thank You. –  yuvalz Oct 16 '12 at 23:12

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