Question

$f\left(x\right)=\cos \left[{\pi }^2\right]x+\cos \left[-{\pi }^2\right]x$

then $f\left(\frac{-\pi }{4}\right)$ = ___________

• A. $\frac{1}{2}$
• B. 1
• C. $\frac{\sqrt{3}}{2}$
• D. $\frac{-1}{\sqrt{2}}$

Answer 1

Answer: (D)

$\frac{-1}{\sqrt{2}}$

$f\left(x\right)=\cos \left[{\pi }^2\right]x+\cos \left[{-\pi }^2\right]x$

greatest integer of $(\pi {)}^2=9$

greatest integer of $-(\pi {)}^2=-10$

$f\left(\frac{-\pi }{4}\right)=\cos 9\left(\frac{-\pi }{4}\right)+\cos (-10)\left(\frac{-\pi }{4}\right)$

$=\cos \left(\frac{-9\pi }{4}\right)+\cos \left(\frac{10\pi }{4}\right)$

$=\cos \left(\frac{5\pi }{2}\right)+\cos (-3\pi +\frac{3\pi }{4})$

$=0+\cos \left(\frac{3\pi }{4}\right)$

$=\cos (\pi -\frac{\pi }{4})$

$=-\cos \left(\frac{\pi }{4}\right)$

$f\left(\frac{-\pi }{4}\right)=\frac{-1}{\sqrt{2}}$