Question

If ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\pi$, then prove that $x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$

Answer 1

Let: ${\sin }^{-1}\left(x\right)=A,{\sin }^{-1}\left(y\right)=B,{\sin }^{-1}\left(z\right)=C$

On taking $\sin$ both sides in each of them we get:

$\Rightarrow x=\sin A,y=\sin B,z=\sin C$ ........ $\left[As\sin \right({\sin }^{-1}\left(x\right))=x$]

Given that: $\sin x+\sin y+\sin z=\pi$

$\Rightarrow A+B+C=\pi$

$A+B=\pi -C$ ..... $\left(1\right)$

To prove: $x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$

LHS:

$=\sin A\sqrt{1-{\sin }^{2}A}+\sin B\sqrt{1-{\sin }^{2}B}+\sin C\sqrt{1-{\sin }^{2}C}$

As we know, {$\sqrt{1-{\sin }^{2}A}=\cos A$} , LHS simplifies to:

$=\sin A\cos A+\sin B\cos B+\sin C\cos C$

$=\frac{2\sin A\sin A+2\sin B\cos B+2\sin C\cos C}{2}$ ... Multiply and Divide by $2$

As we know, $2\sin A\cos A=\sin \left(2A\right)$ so LHS:

$=\frac{\sin 2A+\sin 2B+2\sin C\cos C}{2}$

$=\frac{2\sin (A+B)\cos (A-B)+2\sin C\cos C}{2}$ $[Identity:\sin x+\sin y=2\sin (\frac{x+y}{2}\left)\cos \right(\frac{x-y}{2}\left)\right]$

$=\frac{2\sin (\pi -C)\cos (A-B)+2\sin C\cos C}{2}$ ..... From $\left(1\right)$

$=\frac{2\sin C\left(\cos \right(A-B)+\cos (\pi -(A+B)\left)\right)}{2}$ $[Identity:\sin (\pi -C)=\sin C]$

$\frac{2\sin C\left(\cos \right(A-B)-\cos ((A+B)\left)\right)}{2}$ $[Identity:\cos (\pi -x)=-\cos x]$

$\frac{2\sin C\left(2\sin \right(A\left)\sin \right(B\left)\right)}{2}$ $[Identity:\cos y-\cos x=2\sin (\frac{x+y}{2}\left)\sin \right(\frac{x-y}{2}\left)\right]$

$=2\sin A\sin B\sin C$

$=2xyz$

LHS $=$ RHS

Hence proved