If f(x) = cos([pi^2]x) + cos([- pi^2]x),then | vedclass

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(D) Given $f(x) = \cos([\pi^2]x) + \cos([- \pi^2]x)$.Since $\pi^2 \approx 9.86$,we have $[\pi^2] = 9$.Also,$[-\pi^2] = [-9.86] = -10$.Thus,$f(x) = \co

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$f\left( \frac{\pi}{2} \right) = -1$

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(D) Given $f(x) = \cos([\pi^2]x) + \cos([- \pi^2]x)$.Since $\pi^2 \approx 9.86$,we have $[\pi^2] = 9$.Also,$[-\pi^2] = [-9.86] = -10$.Thus,$f(x) = \cos(9x) + \cos(-10x) = \cos(9x) + \cos(10x)$.Using the sum-to-product formula $\cos A + \cos B = 2\cos\left(\frac{A+B}{2}ight)\cos\left(\frac{A-B}{2}ight)$:$f(x) = 2\cos\left(\frac{19x}{2}ight)\cos\left(\frac{x}{2}ight)$.Now,evaluate $f\left(\frac{\pi}{2}ight)$:$f\left(\frac{\pi}{2}ight) = 2\cos\left(\frac{19\pi}{4}ight)\cos\left(\frac{\pi}{4}ight)$.Since $\cos\left(\frac{19\pi}{4}ight) = \cos\left(4\pi + \frac{3\pi}{4}ight) = \cos\left(\frac{3\pi}{4}ight) = -\frac{1}{\sqrt{2}}$ and $\cos\left(\frac{\pi}{4}ight) = \frac{1}{\sqrt{2}}$:$f\left(\frac{\pi}{2}ight) = 2 	imes \left(-\frac{1}{\sqrt{2}}ight) 	imes \left(\frac{1}{\sqrt{2}}ight) = 2 	imes \left(-\frac{1}{2}ight) = -1$.

If $f(x) = \cos([\pi^2]x) + \cos([- \pi^2]x)$,then which of the following is true?

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