Question & Answer
QUESTION

Write the value of \[\cos \left( {{\sin }^{-1}}x+{{\cos }^{-1}}x \right),\left| x \right|\le 1.\]

ANSWER Verified Verified
Hint: First find the value of \[\left( {{\sin }^{-1}}x+{{\cos }^{-1}}x \right)\] by substituting \[\left( {{\sin }^{-1}}x \right)=\alpha \] and \[\left( {{\cos }^{-1}}x \right)=\beta \]. Thus find its value and substitute it in the expression. Use a trigonometric table and find the values.

Complete step-by-step answer:
We have been given the expression \[\cos \left( {{\sin }^{-1}}x+{{\cos }^{-1}}x \right)......(1)\]
Let us assume that \[\left( {{\sin }^{-1}}x \right)=\alpha \] and \[\left( {{\cos }^{-1}}x \right)=\beta \].

\[\begin{align}

  & {{\sin }^{-1}}x=\alpha \\

 & \therefore x=\sin \alpha \\

\end{align}\] and \[\begin{align}

  & {{\cos }^{-1}}x=\beta \\

 & \therefore x=\cos \beta \\

\end{align}\]

Thus, by equating both the expressions, we get,

\[\sin \alpha =\cos \beta =x\].

Thus we can say that \[\sin \alpha =\sin \left( {}^{\pi }/{}_{2}-\beta \right)\].

From the trigonometric identities, we know that.

\[\begin{align}

  & \sin \left( {{90}^{\circ }}-\theta \right)=\cos \theta \\

 & \therefore \sin \left( {{90}^{\circ }}-\beta \right)=\cos \beta \\

 & \therefore \sin \alpha =\sin \left( {}^{\pi }/{}_{2}-\beta \right) \\

\end{align}\]

Cancelling out sin from both the sides, we get,

\[\begin{align}

  & \alpha ={}^{\pi }/{}_{2}-\beta \\

 & \therefore \alpha +\beta ={}^{\pi }/{}_{2}. \\

\end{align}\]

We said that \[\left( {{\sin }^{-1}}x \right)=\alpha \] and \[\left( {{\cos }^{-1}}x \right)=\beta \].

\[\therefore {{\sin }^{-1}}x+{{\cos }^{-1}}x={}^{\pi }/{}_{2}........(2)\]

Now let us substitute equation (2) in (1).

\[\cos \left( {{\sin }^{-1}}x+{{\cos }^{-1}}x \right)=\cos {}^{\pi }/{}_{2}=0\].

We know from the trigonometric table that \[\cos {}^{\pi }/{}_{2}=0\].


Note:

We can also say that

 \[\begin{align}

  & \left( {{\cos }^{-1}}x \right)=A \\

 & x=\cos A \\

\end{align}\]

We know,

\[\begin{align}

  & \sin A=\sqrt{1-{{\cos }^{2}}A} \\

 & \therefore A={{\sin }^{-1}}\sqrt{1-{{x}^{2}}} \\

 & \therefore {{\sin }^{-1}}x+{{\cos }^{-1}}x={{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{1-{{x}^{2}}} \\

 & \therefore {{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{1-{{x}^{2}}} \\

 & ={{\sin }^{-1}}\left( x\sqrt{1-{{\left( \sqrt{1-{{x}^{2}}} \right)}^{2}}}+\sqrt{1-{{x}^{2}}} \right)
\\

 & ={{\sin }^{-1}}\left( x\sqrt{{{x}^{2}}}+1-{{x}^{2}} \right) \\

 & ={{\sin }^{-1}}\left( {{x}^{2}}+1-{{x}^{2}} \right)={{\sin }^{-1}}(1)={}^{\pi }/{}_{2}. \\

\end{align}\]