Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
Solve
(a) $cos\left(2{sin}^{-1}x\right)=1/9$
(b) ${cos}^{-1}\left(3/5\right)-{sin}^{-1}\left(4/5\right)={cos}^{-1}x$
(c) If $sin({sin}^{-1}\frac{1}{5}+{cos}^{-1}x)=1$, then prove that x is equal to $1/5$.

Answer 1

(a) Put ${sin}^{-1}x=\theta$ so that $sin\theta =x$.

Then the equation $cos\left(2{sin}^{-1}x\right)=1/9$ takes the form $cos2\theta =1/9$

or $1-2{sin}^2\theta =1/9$ or ${sin}^2\theta =4/9$

or $x^2=4/9$, giving $x=\pm 2/3$.

(b) $\because$ ${sin}^{-1}\left(4/5\right)={cos}^{-1}\left(3/5\right)$.

the given equation can be written as

${sin}^{-1}\left(4/5\right)-{sin}^{-1}\left(4/5\right)={cos}^{-1}x$

or ${cos}^{-1}x=0\Rightarrow x=cos0=1$

$\therefore$ $x=1$.

(c) ${sin}^{-1}\frac{1}{5}+{cos}^{-1}x=\frac{\pi }{2}$

$\therefore$ ${sin}^{-1}\frac{1}{5}=\frac{\pi }{2}-{cos}^{-1}x={sin}^{-1}x$

$\therefore$ $x=1/5$.