Question

The value of $co{s}^{-1}\left(cos\frac{3\pi }{2}\right)$ is

(a) $\frac{\pi }{2}$ (b) $\frac{3\pi }{2}$ (c) $\frac{5\pi }{2}$ (d) $\frac{7\pi }{2}$

Answer 1

(a) We have, $co{s}^{-1}\left(cos\frac{3\pi }{2}\right)=co{s}^{-1}\left[cos(2\pi -\frac{\pi }{2})\right][\because cos(2\pi -\frac{\pi }{2})=cos\frac{\pi }{2}]=co{s}^{-1}\left[cos\left(\frac{\pi }{2}\right)\right]=\frac{\pi }{2}\{\because co{s}^{-1}(cosx)=x,x\in [0,\pi \left]\right\}$

Note Remember that, $co{s}^{-1}\left(cos\frac{3\pi }{2}\right)\ne \frac{3\pi }{2}$

$\therefore \frac{3\pi }{2}\notin (0,\pi )$