QUESTION

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

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.

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}