The period of sin ^4 x + cos ^4 x is | vedclass

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(D) Let $f(x) = \sin ^4 x + \cos ^4 x$.We know that $\sin ^4 x + \cos ^4 x = (\sin ^2 x + \cos ^2 x)^2 - 2 \sin ^2 x \cos ^2 x$.Since $\sin ^2 x + \co

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$\frac{\pi}{2}$

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(D) Let $f(x) = \sin ^4 x + \cos ^4 x$.We know that $\sin ^4 x + \cos ^4 x = (\sin ^2 x + \cos ^2 x)^2 - 2 \sin ^2 x \cos ^2 x$.Since $\sin ^2 x + \cos ^2 x = 1$, we have $f(x) = 1 - 2 \sin ^2 x \cos ^2 x$.Using the identity $\sin 2x = 2 \sin x \cos x$, we can write $2 \sin ^2 x \cos ^2 x = \frac{1}{2} (2 \sin x \cos x)^2 = \frac{1}{2} \sin ^2 2x$.Thus, $f(x) = 1 - \frac{1}{2} \sin ^2 2x$.Using the identity $\sin ^2 	heta = \frac{1 - \cos 2	heta}{2}$, we get $f(x) = 1 - \frac{1}{2} \left( \frac{1 - \cos 4x}{2} ight) = 1 - \frac{1}{4} + \frac{\cos 4x}{4} = \frac{3}{4} + \frac{1}{4} \cos 4x$.The period of $\cos(kx)$ is $\frac{2\pi}{|k|}$.Therefore, the period of $f(x) = \frac{3}{4} + \frac{1}{4} \cos 4x$ is $\frac{2\pi}{4} = \frac{\pi}{2}$.

The period of $\sin ^4 x + \cos ^4 x$ is

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