Question

$f\left(x\right)=$ $\left|\begin{array}{ccc}\cos x& \sin x& \cos x\\ \cos 2x& \sin 2x& 2\cos 2x\\ \cos 3x& \sin 3x& 3\cos 3x\end{array}\right|$

Then find $f\left(\pi /2\right)$

Answer 1

$f\left(x\right)=\left|\begin{array}{ccc}\cos x& \sin x& \cos x\\ \cos 2x& \sin 2x& 2\cos 2x\\ \cos 3x& \sin 3x& 3\cos 3x\end{array}\right|$

$f\left(\frac{\pi }{2}\right)\left|\begin{array}{ccc}\cos \frac{\pi }{2}& \sin \frac{\pi }{2}& \cos \frac{\pi }{2}\\ \cos \pi & \sin \pi & 2\cos \pi \\ \cos \frac{3\pi }{2}& \sin \frac{3\pi }{2}& 3\cos \frac{3\pi }{2}\end{array}\right|$

$f\left(\frac{\pi }{2}\right)=\left|\begin{array}{ccc}0& 1& 0\\ -1& 0& -2\\ -\frac{1}{2}& 0& -\frac{3}{2}\end{array}\right|$

$f\left(\frac{\pi }{2}\right)=1\left[\right(-1\times \frac{-3}{2})-(-2\times \frac{-1}{2}\left)\right]$

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