Chapter 14.6, Problem 8OE

To determine

To find: the value of the inverse trigonometric function.

Answer to Problem 8OE

  ${\text{sin}}^{\text{-1}}\frac{\sqrt{\text{2}}}{\text{2}}\text{=}\frac{\text{π}}{\text{4}}$ .

Explanation of Solution

Given:

  ${\text{sin}}^{\text{-1}}\frac{\sqrt{\text{2}}}{\text{2}}$ .

Concept used:

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

The domain of ${\text{sin}}^{\text{-1}}\left(\text{x}\right)$ : $\left[-1,1\right]$ .

The range of ${\text{sin}}^{\text{-1}}\left(\text{x}\right)$ : $\left[\text{-}\frac{\text{π}}{\text{2}}\text{,}\frac{\text{π}}{\text{2}}\right]$ .

From Unit circle the angle of the trigonometric value can be measure accurately.

  

Calculation:

  ${\text{sin}}^{\text{-1}}\frac{\sqrt{\text{2}}}{\text{2}}$ .

  $\text{sinx=}\frac{\sqrt{\text{2}}}{\text{2}}$

The range of the ${\text{sin}}^{\text{-1}}\left(\text{x}\right)$ lies in between $\text{x=-}\frac{\text{π}}{\text{2}}$ radians and $\text{x=}\frac{\text{π}}{\text{2}}$ radians.

The range of the ${\text{sin}}^{\text{-1}}\left(\text{x}\right)$ is the domain of the $\text{sinx}$ .

The one angle between $\text{x=-}\frac{\text{π}}{\text{2}}$ radians and $\text{x=}\frac{\text{π}}{\text{2}}$ radians with the $\text{sinx=}\frac{\sqrt{\text{2}}}{\text{2}}$ is:

From the unit circle:

  ${\text{sin}}^{\text{-1}}\frac{\sqrt{\text{2}}}{\text{2}}\text{=}\frac{\text{π}}{\text{4}}$ .

Hence, the value of expression is at ${\text{sin}}^{\text{-1}}\frac{\sqrt{\text{2}}}{\text{2}}\text{=}\frac{\text{π}}{\text{4}}$ .