Question:
If $f(x)=\cos \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x$, then $f\left(\frac{\pi}{2}\right)=$
Solution:
Given: $f(x)=\cos \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x$
$f(x)=\cos \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x$
$=\cos 9 x+\cos (-10 x) \quad\left(\because \pi^{2}=9.85\right.$ approx $)$
$=\cos 9 x+\cos 10 x$
$\Rightarrow f\left(\frac{\pi}{2}\right)=\cos 9\left(\frac{\pi}{2}\right)+\cos 10\left(\frac{\pi}{2}\right)$
$=\cos \frac{9 \pi}{2}+\cos 5 \pi$
$=0-1$
$=-1$
Hence, $f\left(\frac{\pi}{2}\right)=\underline{-1}$
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