Question

If f(x) = cos [π²] x + cos [–π²] x, then $f\left(\frac{\mathrm{\pi }}{2}\right)=$______________.

Answer 1

Given: f(x) = cos [π²] x + cos [–π²] x


$$\begin{gathered} f\left(x\right)=\cos \left[{\mathrm{\pi }}^2\right]x+\cos \left[-{\mathrm{\pi }}^2\right]x \\ =\cos 9x+\cos \left(-10x\right)\left(\because {\mathrm{\pi }}^2=9.85\mathrm{approx}\right) \\ =\cos 9x+\cos 10x \\ \Rightarrow f\left(\frac{\mathrm{\pi }}{2}\right)=\cos 9\left(\frac{\mathrm{\pi }}{2}\right)+\cos 10\left(\frac{\mathrm{\pi }}{2}\right) \\ =\cos \frac{9\mathrm{\pi }}{2}+\cos 5\mathrm{\pi } \\ =0-1 \\ =-1 \end{gathered}$$


Hence, $f\left(\frac{\mathrm{\pi }}{2}\right)=$ ​–1.