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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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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 by thanvigandhi_1
edited Mar 19, 2013 by thanvigandhi_1
 

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