Chapter 14.6, Problem 22WE

To determine

To find: the value of the trigonometric function without use of trigonometric and inverse trigonometric identities.

Answer to Problem 22WE

  $\text{sin}\left({\text{sin}}^{\text{-1}}{\text{u+sin}}^{\text{-1}}\text{v}\right)=\text{u}\sqrt{{\text{1-v}}^{\text{2}}}\text{+v}\sqrt{{\text{1-u}}^{\text{2}}}$ .

Explanation of Solution

Given:

  $\text{sin}\left({\text{sin}}^{\text{-1}}{\text{u+sin}}^{\text{-1}}\text{v}\right)$ .

Concept used:

  ${\text{cos}}^{\text{-1}}\left(\text{x}\right)\text{=θ}$ it implies $\left(\text{x}\right)\text{=cosθ}$ or $\text{cosθ=x}$ .

Trigonometric formula:

  $\text{sin}\left(\text{x+y}\right)\text{=sinxcosy+cosxsiny}$ .

  $\text{cosx=}\sqrt{{\text{1-sin}}^{\text{2}}\text{x}}$ .

For unit circle hypotenuse will be $\text{1}$ since the radius of unit circle is $\text{1}$ .

Calculation:

  $\text{sin}\left({\text{sin}}^{\text{-1}}{\text{u+sin}}^{\text{-1}}\text{v}\right)$............ $\left(\text{i}\right)$

  ${\text{sin}}^{\text{-1}}\text{u=x}$ which implies $\text{u=sinx}$ .

  ${\text{sin}}^{\text{-1}}\text{v=y}$ which implies $\text{v=siny}$ .

Putting the values in equation $\left(\text{i}\right)$

  $\text{sin}\left({\text{sin}}^{\text{-1}}{\text{u+sin}}^{\text{-1}}\text{v}\right)$ :

  $\text{sin}\left(\text{x+y}\right)$ .

  $\Rightarrow \text{sinxcosy+cosxsiny}$ .

  $\text{cosx=}\sqrt{{\text{1-sin}}^{\text{2}}\text{x}}\text{=}\sqrt{{\text{1-u}}^{\text{2}}}$ , $\text{cosy=}\sqrt{{\text{1-sin}}^{\text{2}}\text{y}}\text{=}\sqrt{{\text{1-v}}^{\text{2}}}$ .

  $\Rightarrow \text{u}\sqrt{{\text{1-v}}^{\text{2}}}\text{+v}\sqrt{{\text{1-u}}^{\text{2}}}$ .

Hence, $\text{sin}\left({\text{sin}}^{\text{-1}}{\text{u+sin}}^{\text{-1}}\text{v}\right)=\text{u}\sqrt{{\text{1-v}}^{\text{2}}}\text{+v}\sqrt{{\text{1-u}}^{\text{2}}}$ .