The extreme values of 4 cos left(x^2right) co | vedclass

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(A) Let $f(x) = 4 \cos \left(x^2\right) \cos \left(\frac{\pi}{3}+x^2\right) \cos \left(\frac{\pi}{3}-x^2\right)$.Using the identity $2 \cos A \cos B =

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$-1, 1$

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(A) Let $f(x) = 4 \cos \left(x^2ight) \cos \left(\frac{\pi}{3}+x^2ight) \cos \left(\frac{\pi}{3}-x^2ight)$.Using the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have:$f(x) = 2 \cos \left(x^2ight) \left[ 2 \cos \left(\frac{\pi}{3}+x^2ight) \cos \left(\frac{\pi}{3}-x^2ight) ight]$$f(x) = 2 \cos \left(x^2ight) \left[ \cos \left(\frac{2\pi}{3}ight) + \cos \left(2x^2ight) ight]$Since $\cos \left(\frac{2\pi}{3}ight) = -\frac{1}{2}$,we get:$f(x) = 2 \cos \left(x^2ight) \left[ -\frac{1}{2} + \cos \left(2x^2ight) ight]$$f(x) = -\cos \left(x^2ight) + 2 \cos \left(x^2ight) \cos \left(2x^2ight)$Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ again:$f(x) = -\cos \left(x^2ight) + \cos \left(3x^2ight) + \cos \left(x^2ight)$$f(x) = \cos \left(3x^2ight) \quad \dots (i)$Since the range of $\cos(	heta)$ is $[-1, 1]$,the extreme values of $f(x) = \cos \left(3x^2ight)$ are $-1$ and $1$.

The extreme values of $4 \cos \left(x^2\right) \cos \left(\frac{\pi}{3}+x^2\right) \cos \left(\frac{\pi}{3}-x^2\right)$ over $\mathbb{R}$ are

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