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

Toolbox:
• $$cosx-cosy=2sin\large\frac{x+y}{2}.cos\large\frac{y-x}{2}$$
• $$1-sin^2x=cos^2x$$
• $$sin(2\pi-\theta)=-sin\theta$$
• $$sin2A=2sinAcosA$$
• $$cos(\pi-\theta)=-cos\theta$$
Let $$sin^{-1}x=A, \: sin^{-1}y=B, \: sin^{-1}z=C$$
$$\Rightarrow\:sinA=x,sinB=y\:and\:sinC=z$$
By substituting the value of x,y,z in the given eqn.
$$\Rightarrow\:given\: A+B+C=\pi$$
$$\Rightarrow\:A+B=\pi-C$$
L.H.S.

Substituting the value of x y and z in L.H.S. we get
$$x \sqrt{1-x^2}+y\sqrt{1-y^2}+2\sqrt{1-z^2}=$$
$$sinA \sqrt{1-sin^2A}+SinB\sqrt{1-sin^2B}+$$
$$sinC\sqrt{1-sin^2}C$$
$$=sinAcosA+sinBcosB+sinCcosC$$

Multiply and divide by 2
$$=\large\frac{1}{2}\big(2sinAcosA+2sinBcosB+2sinCcosC)$$
Using the formula of sin2A=2sinA.cosA
$$=\large\frac{1}{2}\big(sin2A+sin2B+2sinC.cosC\big)$$

By taking x=2A,y=2B in the above formula of sinx+siny, we get
$$=\large\frac{1}{2}\bigg(\big(2.sin\large\frac{2A+2B}{2}.cos\large\frac{2A-2B}{2}\big)+2sinCcosC\bigg)$$
$$=sin(A+B).cos(A-B)+sinCcosC$$
$$since,\:A+B=\pi-C$$
$$=sin(\pi-C).cos(A-B)+sinCcosC$$

By taking $$\theta=C$$ in the above formula of $$sin(\pi-\theta)$$
$$=sinC.cos(A-B)+sinCcosC$$
$$=sinC(cos(A-B)+cosC)$$
By writing cosC=cos($$\pi-(A+B))=-cos(A+B)$$
$$=sinC(cos(A-B)-cos(A+B))$$

By taking x=A-B, and y=(A+B) in the above formula of cox+cosy we get
$$=sinC\big(2sin\large\frac{A-B+A+B}{2}.sin\frac{A+B-A+B}{2}\big)$$
$$=2sinCsinAsinB=2xyz$$
=R.H.S.

edited Mar 19, 2013