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Question: If we have an inverse trigonometric expression \({{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi...

If we have an inverse trigonometric expression cos⁡−1x+cos⁡−1y+cos⁡−1z=π{{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi , then prove that x2+y2+z2+2xyz=1{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1

Explanation

Solution

From the question we have cos⁡−1x+cos⁡−1y+cos⁡−1z=π{{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi and we have to prove x2+y2+z2+2xyz=1{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1 using the given condition so we will assume cos⁡−1x=A{{\cos }^{-1}}x=A and cos⁡−1y=B{{\cos }^{-1}}y=B and simplify it further using the expressions π−cos⁡−1z=cos⁡−1(−z)\pi -{{\cos }^{-1}}z={{\cos }^{-1}}\left( -z \right) and cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B and sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.

Complete step-by-step solution:
Now from considering the question we have cos⁡−1x+cos⁡−1y+cos⁡−1z=π{{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi . Let us cos⁡−1x=A{{\cos }^{-1}}x=A and cos⁡−1y=B{{\cos }^{-1}}y=B and write this simply asA+B=π−cos⁡−1zA+B=\pi -{{\cos }^{-1}}z .
As we know that the value of π−cos⁡−1z=cos⁡−1(−z)\pi -{{\cos }^{-1}}z={{\cos }^{-1}}\left( -z \right) we will use it and simplify it and write it as A+B=cos⁡−1(−z)A+B={{\cos }^{-1}}\left( -z \right).
By applying the inverse of cosine function on both sides we will have cos⁡(A+B)=(−z)\cos \left( A+B \right)=\left( -z \right) .
As we know that cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B. We can use this by deriving the values of respective sine functions using the identity sin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1.
Now we can say that sin⁡A=1−x2\sin A=\sqrt{1-{{x}^{2}}} and sin⁡B=1−y2\sin B=\sqrt{1-{{y}^{2}}} because cos⁡A=x\cos A=x and cos⁡B=y\cos B=y and they are connected with the relationsin⁡2θ+cos⁡2θ=1{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 .
By using these values in cos⁡(A+B)=(−z)\cos \left( A+B \right)=\left( -z \right) we will have xy−1−x21−y2=−zxy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}=-z .
This expression can be simplified as 1−x21−y2=xy+z\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}=xy+z .
By squaring on both sides we will have (1−x21−y2)2=(xy+z)2⇒(1−x2)(1−y2)=(xy+z)2{{\left( \sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right)}^{2}}={{\left( xy+z \right)}^{2}}\Rightarrow \left( 1-{{x}^{2}} \right)\left( 1-{{y}^{2}} \right)={{\left( xy+z \right)}^{2}} .
By expanding this we will have (1+x2y2−y2−x2)=(x2y2+z2+2xyz)\left( 1+{{x}^{2}}{{y}^{2}}-{{y}^{2}}-{{x}^{2}} \right)=\left( {{x}^{2}}{{y}^{2}}+{{z}^{2}}+2xyz \right) .
By further simplifying this we will have (1−y2−x2)=(z2+2xyz)\left( 1-{{y}^{2}}-{{x}^{2}} \right)=\left( {{z}^{2}}+2xyz \right) .
By transforming some terms from right to left and some from right to left we will have x2+y2+z2+2xyz=1{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1.
Hence it is proved that when cos⁡−1x+cos⁡−1y+cos⁡−1z=π{{\cos }^{-1}}x+{{\cos }^{-1}}y+{{\cos }^{-1}}z=\pi , the expression x2+y2+z2+2xyz=1{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xyz=1 is valid.

Note: While answering this type of questions we should keep 3 main points in our mind what we have, what is to be proved, and which concepts and formulae can be used if we make a mistake and write π−cos⁡−1z=cos⁡−1z\pi -{{\cos }^{-1}}z={{\cos }^{-1}}z then we will have xy−1−x21−y2=zxy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}}=z which by simplifying we will have x2+y2+z2−2xyz=1{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-2xyz=1 which is completely wrong.