Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.

${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$

$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.

Evaluate

(a) ${cos}^{-1}x+{cos}^{-1}[\frac{x}{2}+\frac{\sqrt{(3-3x^2)}}{2}]$ $(\frac{1}{2}\le x\le 1)$

(b) $cos(2{cos}^{-1}x+{sin}^{-1}x)$ at $x=1/5$,

where $0\le {cos}^{-1}x\le \pi$

and $-\pi /2\le {sin}^{-1}x\le \pi /2$

Answer 1

(a) If ${cos}^{-1}x=y$, then $x=cosy$.

we have $\frac{1}{2}\le x\le 1$

$\Rightarrow$ ${cos}^{-1}1\le {cos}^{-1}x\le {cos}^{-1}\frac{1}{2}$

$\Rightarrow$ $0\le y\le \pi /3$.

$\therefore$ $\frac{x}{2}+\frac{\sqrt{3-3x^2}}{2}+\frac{1}{2}cosy+\frac{\sqrt{3}}{2}siny$

$=cos\left(\pi /3\right)cosy+sin\left(\pi /3\right)siny$

$=cos(\pi /3-y)$

It follows that

${cos}^{-1}[\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2}]=\frac{1}{3}\pi -y$

$[\because {cos}^{-1}cos\theta =\theta$ for $0\le \theta \le \pi$ and here $0\le \frac{\pi }{3}-y\le \frac{\pi }{3}]$

$\therefore$ The given expression $=y+\frac{\pi }{3}-y=\frac{\pi }{3}$.

(b) We have $cos\left[2{cos}^{-1}\right(1/5)+{sin}^{-1}(1/5\left)\right]$

$=cos\left[\right\{{cos}^{-1}\left(1/5\right)+{sin}^{-1}\left(1/5\right)\}+{cos}^{-1}(1/5\left)\right]$

$=cos[\pi /2+{cos}^{-1}(1/5\left)\right]=-sin{cos}^{-1}\left(1/5\right)$

$=-sin{sin}^{-1}\left[\sqrt{24}/5\right]=-\sqrt{24}/5=-2\sqrt{6}/5$