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

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

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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 $f(x) = \cos x \cos(x + 2) - \cos^2(x + 1)$.Using the formula $2 \cos A \cos B = \cos(A + B) + \cos(A - B)$,we have:$f(x) = \frac{1}{2} [\cos(2x + 2) + \cos(-2)] - \cos^2(x + 1)$.Recall that $2 \cos^2 	heta = 1 + \cos(2	heta)$,so $\cos^2(x + 1) = \frac{1 + \cos(2x + 2)}{2}$.Substituting this into the expression for $f(x)$:$f(x) = \frac{1}{2} \cos(2x + 2) + \frac{1}{2} \cos 2 - \frac{1}{2} - \frac{1}{2} \cos(2x + 2)$.$f(x) = \frac{1}{2} \cos 2 - \frac{1}{2} = -\frac{1}{2} (1 - \cos 2)$.Using $1 - \cos 2	heta = 2 \sin^2 	heta$,we get $1 - \cos 2 = 2 \sin^2 1$.Thus,$f(x) = -\frac{1}{2} (2 \sin^2 1) = -\sin^2 1$.Since $f(x) = -\sin^2 1$ is a constant,the graph is a horizontal straight line parallel to the $x$-axis passing through the point $(0, -\sin^2 1)$ and $(\frac{\pi}{2}, -\sin^2 1)$.

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

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