Question

Inverse circular functions,Principal values of ${sin}^{-1}x,{cos}^{-1}x,{tan}^{-1}x$.
${tan}^{-1}x+{tan}^{-1}y={tan}^{-1}\frac{x+y}{1-xy}$, $xy<1$
$\pi +{tan}^{-1}\frac{x+y}{1-xy}$, $xy>1$.
(a) ${sin}^{-1}(1-x)-2{sin}^{-1}x=\pi /2$.
(b) If ${sin}^{-1}x+{sin}^{-1}(1-x)={cos}^{-1}x$, then prove that x is equal to $0,1/2$.

Answer 1

(a) The given equation can be written as
${sin}^{-1}(1-x)=\pi /2+2{sin}^{-1}x$
Taking sine, we get
$1-x=cos\left(2{sin}^{-1}x\right)=1-2{\left(sin{sin}^{-1}x\right)}^2$
or $1-x=1-22^x$ or $x=2x^2$
$\therefore$ $x=0$ or $1/2$,
A check shows that $x=1/2$ is not a root.
Hence $x=0$ is the only root.
(b) ${sin}^{-1}[x\sqrt{1-{(1-x)}^2}+(1-x\left)\sqrt{(1-x^2)}\right]={sin}^{-1}\sqrt{1-x^2}$
$\therefore$ $x\sqrt{2x-x^2}=\sqrt{1-x^2}[1-(1-x\left)\right]=x\sqrt{1-x^2}$
$\therefore$ $x=0$ or $2x-x^2=1-x^2$ or $x=1/2$