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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: $F_y: \frac{F_1}{F_2}=\frac{\pi}{\pi}=\frac{1}{1}$ $F_y=\pi\times1=\pi$

4 Answers 4

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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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    I looked at it from a wrong perspective. I understand. Thank you.2012-10-17
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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.2012-10-16