The value of cos ^4 x is | vedclass

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(A) We know that $\cos 2x = 2\cos^2 x - 1$,which implies $\cos^2 x = \frac{1 + \cos 2x}{2}$.Now,$\cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}

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$\frac{3}{8}+\frac{1}{2} \cos 2 x+\frac{1}{8} \cos 4 x$

Vedclass · Answer

(A) We know that $\cos 2x = 2\cos^2 x - 1$,which implies $\cos^2 x = \frac{1 + \cos 2x}{2}$.Now,$\cos^4 x = (\cos^2 x)^2 = \left(\frac{1 + \cos 2x}{2}\right)^2$.Expanding this,we get $\cos^4 x = \frac{1 + \cos^2 2x + 2\cos 2x}{4} = \frac{1}{4} + \frac{1}{2}\cos 2x + \frac{1}{4}\cos^2 2x$.Using the identity $\cos^2 2x = \frac{1 + \cos 4x}{2}$,we substitute:$\cos^4 x = \frac{1}{4} + \frac{1}{2}\cos 2x + \frac{1}{4}\left(\frac{1 + \cos 4x}{2}\right)$.$\cos^4 x = \frac{1}{4} + \frac{1}{2}\cos 2x + \frac{1}{8} + \frac{1}{8}\cos 4x$.$\cos^4 x = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$.

The value of $\cos ^4 x$ is

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