The graph of the function f(x) = cos x cos(x | vedclass

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(D) Let $y = \cos x \cos(x + 2) - \cos^2(x + 1)$.Using the identity $\cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)]$,we have:$y = \frac{1}{2}[

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$A$ straight line passing through the point $(\frac{\pi}{2}, -\sin^2 1)$ and parallel to the $x$-axis

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(D) Let $y = \cos x \cos(x + 2) - \cos^2(x + 1)$.Using the identity $\cos A \cos B = \frac{1}{2}[\cos(A - B) + \cos(A + B)]$,we have:$y = \frac{1}{2}[\cos(x - (x + 2)) + \cos(x + x + 2)] - \cos^2(x + 1)$$y = \frac{1}{2}[\cos(-2) + \cos(2x + 2)] - \cos^2(x + 1)$Since $\cos(-2) = \cos 2$,we get:$y = \frac{1}{2}\cos 2 + \frac{1}{2}\cos(2(x + 1)) - \cos^2(x + 1)$Using $\cos(2	heta) = 2\cos^2 	heta - 1$:$y = \frac{1}{2}\cos 2 + \frac{1}{2}(2\cos^2(x + 1) - 1) - \cos^2(x + 1)$$y = \frac{1}{2}\cos 2 + \cos^2(x + 1) - \frac{1}{2} - \cos^2(x + 1)$$y = \frac{1}{2}(\cos 2 - 1)$Using $1 - \cos 2 = 2\sin^2 1$,we get $y = -\sin^2 1$.This represents a straight line parallel to the $x$-axis,passing through $(\frac{\pi}{2}, -\sin^2 1)$.

The graph of the function $f(x) = \cos x \cos(x + 2) - \cos^2(x + 1)$ is

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