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Question: If \({\sin ^{ - 1}}\left( x \right) + {\sin ^{ - 1}}\left( y \right) + {\sin ^{ - 1}}\left( z \right...

If sin⁡−1(x)+sin⁡−1(y)+sin⁡−1(z)=π{\sin ^{ - 1}}\left( x \right) + {\sin ^{ - 1}}\left( y \right) + {\sin ^{ - 1}}\left( z \right) = \pi , then prove that x1−x2+y1−y2+z1−z2=2xyzx\sqrt {1 - {x^2}} + y\sqrt {1 - {y^2}} + z\sqrt {1 - {z^2}} = 2xyz

Explanation

Solution

Hint: In this question, we use the concept of inverse trigonometry and also use some basic trigonometric identities. First we assume sin⁡−1x,sin⁡−1y,sin⁡−1z{\sin ^{ - 1}}x,{\sin ^{ - 1}}y,{\sin ^{ - 1}}z are some angle A, B, C and make a relation A+B+C=πA + B + C = \pi and then we use identities sin⁡2x=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x , sin⁡x+sin⁡y=2sin⁡(x+y2)cos⁡(x−y2)\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right) and cos⁡x−cos⁡y=2sin⁡(x+y2)sin⁡(y−x2)\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(x)+sin⁡−1(y)+sin⁡−1(z)=π{\sin ^{ - 1}}\left( x \right) + {\sin ^{ - 1}}\left( y \right) + {\sin ^{ - 1}}\left( z \right) = \pi
Let, sin⁡−1(x)=A⇒x=sin⁡A{\sin ^{ - 1}}\left( x \right) = A \Rightarrow x = \sin A
sin⁡−1(y)=B⇒y=sin⁡B{\sin ^{ - 1}}\left( y \right) = B \Rightarrow y = \sin B
sin⁡−1(z)=C⇒z=sin⁡C{\sin ^{ - 1}}\left( z \right) = C \Rightarrow z = \sin C
Now, A+B+C=π..............(1)A + B + C = \pi ..............\left( 1 \right) where A,B,C∈[−π2,π2]A,B,C \in \left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]
To prove x1−x2+y1−y2+z1−z2=2xyzx\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=x1−x2+y1−y2+z1−z2LHS = 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⁡A1−sin⁡2A+sin⁡B1−sin⁡2B+sin⁡C1−sin⁡2CLHS = \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=1−sin⁡2x\cos x = \sqrt {1 - {{\sin }^2}x}
LHS=sin⁡Acos⁡A+sin⁡Bcos⁡B+sin⁡Ccos⁡C LHS=(2sin⁡Acos⁡A+2sin⁡Bcos⁡B)2+sin⁡Ccos⁡C  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=2sin⁡xcos⁡x\sin 2x = 2\sin x\cos x
LHS=sin⁡2A+sin⁡2B2+sin⁡Ccos⁡CLHS = \dfrac{{\sin 2A + \sin 2B}}{2} + \sin C\cos C
Now use identity sin⁡x+sin⁡y=2sin⁡(x+y2)cos⁡(x−y2)\sin x + \sin y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)

LHS=2sin⁡(2A+2B2)cos⁡(2A−2B2)2+sin⁡Ccos⁡C LHS=sin⁡(A+B)cos⁡(A−B)+sin⁡Ccos⁡C  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=π−CA + B = \pi - C
LHS=sin⁡(π−C)cos⁡(A−B)+sin⁡Ccos⁡CLHS = \sin \left( {\pi - C} \right)\cos \left( {A - B} \right) + \sin C\cos C
We Know, sin⁡(π−C)=sin⁡C\sin \left( {\pi - C} \right) = \sin C
LHS=sin⁡Ccos⁡(A−B)+sin⁡Ccos⁡CLHS = \sin C\cos \left( {A - B} \right) + \sin C\cos C
Take a common sin⁡C\sin C as common,
LHS=sin⁡C(cos⁡(A−B)+cos⁡C)LHS = \sin C\left( {\cos \left( {A - B} \right) + \cos C} \right)
From (1) equation, C=π−A−BC = \pi - A - B
LHS=sin⁡C(cos⁡(A−B)+cos⁡(π−A−B))LHS = \sin C\left( {\cos \left( {A - B} \right) + \cos \left( {\pi - A - B} \right)} \right)
We Know, cos⁡(π−C)=−cos⁡C\cos \left( {\pi - C} \right) = - \cos C
LHS=sin⁡C(cos⁡(A−B)−cos⁡(A+B))LHS = \sin C\left( {\cos \left( {A - B} \right) - \cos \left( {A + B} \right)} \right)
Now use identity cos⁡x−cos⁡y=2sin⁡(x+y2)sin⁡(y−x2)\cos x - \cos y = 2\sin \left( {\dfrac{{x + y}}{2}} \right)\sin \left( {\dfrac{{y - x}}{2}} \right)

LHS=sin⁡C(2sin⁡(A−B+A+B2)sin⁡(A+B−A+B2)) LHS=sin⁡C(2sin⁡(A)sin⁡(B)) LHS=2sin⁡Asin⁡Bsin⁡C  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=RHSLHS = 2xyz = RHS
So, its proved x1−x2+y1−y2+z1−z2=2xyzx\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⁡Cx = \sin A,y = \sin B,z = \sin C (mentioned in above) and also use trigonometric identities. So, we will get the required answer.