Question

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

Answer 1

Hint: In this question, we use the concept of inverse trigonometry and also use some basic trigonometric identities. First we assume ${\sin ^{ - 1}}x,{\sin ^{ - 1}}y,{\sin ^{ - 1}}z$ are some angle A, B, C and make a relation $A + B + C = \pi $ and then we use identities $\sin 2x = 2\sin x\cos x$ , $\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$ and $\cos x - \cos y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{y - x}}{2}} \right)$.

Complete step-by-step solution:

Given, ${\sin ^{ - 1}}\left( x \right) + {\sin ^{ - 1}}\left( y \right) + {\sin ^{ - 1}}\left( z \right) = \pi $
Let, ${\sin ^{ - 1}}\left( x \right) = A \Rightarrow x = \sin A$
${\sin ^{ - 1}}\left( y \right) = B \Rightarrow y = \sin B$
${\sin ^{ - 1}}\left( z \right) = C \Rightarrow z = \sin C$
Now, $A + B + C = \pi ..............\left( 1 \right)$ where $A,B,C \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$
To prove $x\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} = 2xyz$ this equation . So, we have to prove the Left hand side (LHS) is equal to the right hand side (RHS).
Now take a $LHS = x\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} $
Put the value of x, y and z in the left hand side (LHS).
$LHS = \sin A\sqrt {1 - {{\sin }^2}A} + \sin B\sqrt {1 - {{\sin }^2}B} + \sin C\sqrt {1 - {{\sin }^2}C} $
We have to use an identity $\cos x = \sqrt {1 - {{\sin }^2}x} $
$
  LHS = \sin A\cos A + \sin B\cos B + \sin C\cos C \\
  LHS = \dfrac{{\left( {2\sin A\cos A + 2\sin B\cos B} \right)}}{2} + \sin C\cos C \\
$
Now use identity $\sin 2x = 2\sin x\cos x$
$LHS = \dfrac{{\sin 2A + \sin 2B}}{2} + \sin C\cos C$
Now use identity $\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$
\[
  LHS = \dfrac{{2\sin \left( {\dfrac{{2A + 2B}}{2}} \right)\cos \left( {\dfrac{{2A - 2B}}{2}} \right)}}{2} + \sin C\cos C \\
  LHS = \sin \left( {A + B} \right)\cos \left( {A - B} \right) + \sin C\cos C \\
\]
From (1) equation, $A + B = \pi - C$
\[LHS = \sin \left( {\pi - C} \right)\cos \left( {A - B} \right) + \sin C\cos C\]
We Know, \[\sin \left( {\pi - C} \right) = \sin C\]
\[LHS = \sin C\cos \left( {A - B} \right) + \sin C\cos C\]
Take a common \[\sin C\] as common,
\[LHS = \sin C\left( {\cos \left( {A - B} \right) + \cos C} \right)\]
From (1) equation, $C = \pi - A - B$
\[LHS = \sin C\left( {\cos \left( {A - B} \right) + \cos \left( {\pi - A - B} \right)} \right)\]
We Know, \[\cos \left( {\pi - C} \right) = - \cos C\]
\[LHS = \sin C\left( {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right)\]
Now use identity $\cos x - \cos y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{y - x}}{2}} \right)$
\[
  LHS = \sin C\left( {2\sin \left( {\dfrac{{A - B + A + B}}{2}} \right)\sin \left( {\dfrac{{A + B - A + B}}{2}} \right)} \right) \\
  LHS = \sin C\left( {2\sin \left( A \right)\sin \left( B \right)} \right) \\
  LHS = 2\sin A\sin B\sin C \\
\]
Now put the value of sinA ,sinB and sinC.
\[LHS = 2xyz = RHS\]
 So, its proved $x\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} = 2xyz$.

Note: In such types of problems we have to prove left hand side (LHS) equal to right hand side (RHS). In LHS we use $x = \sin A,y = \sin B,z = \sin C$ (mentioned in above) and also use trigonometric identities. So, we will get the required answer.