Chapter 14.6, Problem 16WE

To determine

To find: the value of the inverse trigonometric function.

Answer to Problem 16WE

The value of the expression is $\text{cos}\left({\sin }^{-1}\frac{1}{2}+{\cos }^{-1}\frac{1}{2}\right)=0$ .

Explanation of Solution

Given:

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

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.

Trigonometric formula:

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

According to the inverse trigonometry:

  ${\text{sin}}^{\text{-1}}{\text{x+cos}}^{\text{-1}}\text{x=}\frac{\text{π}}{\text{2}}$ .

Calculation:

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

  ${\text{sin}}^{\text{-1}}\frac{\text{1}}{\text{2}}\text{=x}$ …………… $\left(\text{ii}\right)$

which implies $\text{sinx}=\frac{\text{1}}{\text{2}}$ .

  ${\text{cos}}^{\text{-1}}\frac{\text{1}}{\text{2}}\text{=y}$ …………………….. $\left(\text{iii}\right)$

which implies $\text{cosy}=\frac{\text{1}}{\text{2}}$ .

From the unit circle the angle of the given range can be calculated:

  $\text{sinx}=\frac{\text{1}}{\text{2}}$ at $\text{x=30}$ .

  $\text{cosy}=\frac{\text{1}}{\text{2}}$ at $\text{y=60}$ .

Putting value of $\text{x and y}$ in equation $\left(\text{ii}\right)$ and $\left(\text{iii}\right)$ :

  ${\text{sin}}^{\text{-1}}\frac{\text{1}}{\text{2}}\text{=30}$ and ${\text{cos}}^{\text{-1}}\frac{\text{1}}{\text{2}}\text{=60}$ .

Putting those $\text{2}$ values in equation $\left(\text{i}\right)$

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

  $\text{cos}{\left(\text{30+60}\right)}^{\text{0}}$ .

  $\Rightarrow \text{cos}{\left(90\right)}^{\text{0}}=0$ .

Or

According to the inverse trigonometry:

  ${\text{sin}}^{\text{-1}}{\text{x+cos}}^{\text{-1}}\text{x=}\frac{\text{π}}{\text{2}}$ .

  $\text{cos}\left(\frac{\text{π}}{\text{2}}\right)\text{=0}$ .

Hence, the value of the expression is $\text{cos}\left({\sin }^{-1}\frac{1}{2}+{\cos }^{-1}\frac{1}{2}\right)=0$ .