Chapter 14.6, Problem 3WE

To determine

To find: the value of the inverse trigonometric function.

Answer to Problem 3WE

  ${\text{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\text{-}\frac{\text{π}}{\text{6}}$ .

Explanation of Solution

Given:

  ${\text{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ .

Concept used:

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

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

The range of ${\text{cos}}^{\text{-1}}\left(\text{x}\right)$ : $\left[\text{0,π}\right]$ .

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

  

Calculation:

  ${\text{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ .

  $\text{cosx=-}\frac{\sqrt{3}}{2}$ .

The range of the ${\text{cos}}^{\text{-1}}\left(\text{x}\right)$ lies in between $\text{x=0}$ radians and $\text{x=π}$ radians.

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

The one angle between $\text{x=0}$ radians and $\text{x=π}$ radians with the $\text{cosx=-}\frac{\sqrt{3}}{2}$ is:

From unit circle at ${\text{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\text{-}\frac{\text{π}}{\text{6}}$ .

Hence, ${\text{cos}}^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\text{-}\frac{\text{π}}{\text{6}}$ is the solution of the expression.