Chapter 7.2, Problem 46E

To determine

To write: the expression in terms of x and y only.

Answer to Problem 46E

  $\sin \left({\sin }^{-1}x+{\cos }^{-1}y\right)=xy+\sqrt{1-x^2}\sqrt{1-y^2}$

Explanation of Solution

Given information:

Given expression,

  $\sin \left({\sin }^{-1}x+{\cos }^{-1}y\right)$

Formula used:

Addition formula of sine

  $\sin \left(u-v\right)=\sin u\cos v-\cos u\sin v$

Calculation:

Now, consider the given expression

  $\sin \left({\sin }^{-1}x+{\cos }^{-1}y\right)$

Now, let us take $u={\sin }^{-1}x,v={\cos }^{-1}y$ .

So, using addition formula, we get

  $\begin{array}{c}\sin \left(u+v\right)=\sin u\cos v+\cos u\sin v\\ \sin \left({\sin }^{-1}x+{\cos }^{-1}y\right)=\sin \left({\sin }^{-1}x\right)\cos \left({\cos }^{-1}y\right)+\cos \left({\sin }^{-1}x\right)\sin \left({\cos }^{-1}y\right)\end{array}$ …… (1)

Now, $u={\sin }^{-1}x$

  $\begin{array}{c}\sin u=x\\ \cos u=\sqrt{1-{\sin }^{2}u}\\ =\sqrt{1-x^2}\\ \cos \left({\sin }^{-1}x\right)=\sqrt{1-x^2}\\ \sin \left({\sin }^{-1}x\right)=\sqrt{1-{\left(\sqrt{1-x^2}\right)}^2}\\ =\sqrt{x^2}\\ =x\end{array}$

Further $v={\cos }^{-1}y$

  $\begin{array}{c}\cos v=y\\ \cos \left({\cos }^{-1}y\right)=y\\ \sin v=\sqrt{1-{\cos }^{2}v}\\ =\sqrt{1-y^2}\\ \cos \left({\sin }^{-1}y\right)=\sqrt{1-y^2}\end{array}$

Now, substitute the values in equation (1).

  $\begin{array}{c}\sin \left({\sin }^{-1}x+{\cos }^{-1}y\right)=\sin \left({\sin }^{-1}x\right)\cos \left({\cos }^{-1}y\right)+\cos \left({\sin }^{-1}x\right)\sin \left({\cos }^{-1}y\right)\\ =x\times y+\sqrt{1-x^2}\times \sqrt{1-y^2}\\ =xy+\sqrt{1-x^2}\sqrt{1-y^2}\end{array}$