Chapter 14.6, Problem 21WE

To determine

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

Answer to Problem 21WE

  $\text{sin}\left({\text{cos}}^{\text{-1}}\text{x}\right)=\sqrt{{\text{1-x}}^{\text{2}}}$ .

Explanation of Solution

Given:

  $\text{sin}\left({\text{cos}}^{\text{-1}}\text{x}\right)$ .

Concept used:

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

Pythagoras theorem state that:

  ${\text{H}}^{\text{2}}{\text{=P}}^{\text{2}}{\text{+B}}^{\text{2}}$ .

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

Calculation:

  $\text{sin}\left({\text{2sin}}^{\text{-1}}\text{x}\right)$ ….. $\left(\text{i}\right)$

  $\text{siny=x}$ .

  ${\text{y=sin}}^{\text{-1}}\text{x}$ .

Put the value in equation $\left(\text{i}\right)$ :

  $\text{sin}\left({\text{2sin}}^{\text{-1}}\text{x}\right)=\text{sin}\left(\text{2y}\right)$ .

Trigonometric identities:

  $\text{sin2y=2sinycosy}$ .

  $\text{cosy=}\sqrt{{\text{1-sin}}^{\text{2}}\text{y}}$ .

  $\Rightarrow \text{sin}\left({\text{2sin}}^{\text{-1}}\text{x}\right)=\text{2sinycosy}$ .

  $\Rightarrow \text{2siny}\sqrt{{\text{1-sin}}^{\text{2}}\text{y}}$ .

Since, $\text{siny=x}$ the equation will be:

  $\Rightarrow \text{2x}\sqrt{{\text{1-x}}^2}$ .

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