Chapter 14.6, Problem 15WE

To determine

To find: the value of the inverse trigonometric function.

Answer to Problem 15WE

The value of the expression is $\text{sin}\left({\text{cos}}^{-1}\left(-\frac{2}{5}\right)\right)=\frac{\sqrt{21}}{5}$ .

Explanation of Solution

Given:

  $\text{sin}\left({\text{cos}}^{-1}\left(-\frac{2}{5}\right)\right)$ .

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.

Trigonometric formula:

  $\text{sinθ=}\sqrt{{\text{1-cos}}^{\text{2}}\text{θ}}$ .

Calculation:

  $\text{sin}\left({\text{cos}}^{-1}\left(-\frac{2}{5}\right)\right)$ .

  ${\text{cos}}^{\text{-1}}\left(\text{-}\frac{2}{\text{5}}\right)\text{=θ}$............$\left(\text{i}\right)$

  $\text{-}\frac{2}{\text{5}}\text{=cosθ}$ .

  $\text{sinθ=}\sqrt{{\text{1-cos}}^{\text{2}}\text{θ}}$ .

  $\text{sinθ=}\sqrt{\text{1-}{\left(\text{-}\frac{2}{\text{5}}\right)}^{\text{2}}}$ .

  $\text{sinθ=}\sqrt{\frac{21}{25}}$ .

  ${\text{θ=sin}}^{-1}\frac{\sqrt{21}}{5}$............ $\left(\text{ii}\right)$

Comparing equation $\left(\text{i}\right)$ and $\left(\text{ii}\right)$

  ${\text{cos}}^{\text{-1}}\left(\text{-}\frac{2}{\text{5}}\right){\text{=sin}}^{-1}\frac{\sqrt{21}}{5}$ .

Putting in the given expression $\text{sin}\left({\text{cos}}^{-1}\left(-\frac{2}{5}\right)\right)$ :

  $\Rightarrow \sin \left({\text{sin}}^{-1}\frac{\sqrt{21}}{5}\right)$ .

  $\Rightarrow \frac{\sqrt{21}}{5}$ .

Hence, the value of the expression is $\text{sin}\left({\text{cos}}^{-1}\left(-\frac{2}{5}\right)\right)=\frac{\sqrt{21}}{5}$ .