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

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## 1 Answer

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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.

answered Feb 20, 2013
edited Mar 19, 2013

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