Question:
If $f(x)=\cos [e] x+\cos [-e] x$, then $f(\pi)=$ ___________.
Solution:
Given: f(x) = cos[e]x + cos[–e]x
$f(x)=\cos [e] x+\cos [-e] x$
$=\cos 2 x+\cos (-3 x) \quad(\because e=2.718$ approx $)$
$=\cos 2 x+\cos 3 x$
$\Rightarrow f(\pi)=\cos 2 \pi+\cos 3 \pi$
$=1-1$
$=0$
Hence, $f(\pi)=\underline{0}$
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