Chapter 5.2, Problem 58E

To determine

To verify: The given trigonometric identity.

Explanation of Solution

Given information:

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

Formula used:

Conversion rules, $\cos x=\frac{\text{base}}{\text{hypotenuse}}$ .

Calculation:

Consider the function, $\cos \left({\sin }^{-1}x\right)=\sqrt{1-x^2}$ .

Rewrite the left hand side of the equation as,

  $\cos \left({\sin }^{-1}x\right)=\cos \left({\sin }^{-1}\frac{x}{1}\right)$

Recall that $\sin x=\frac{\text{perpendicular}}{\text{hypotenuse}}$

Construct a right angle triangle with perpendicular p as x and hypotenuse h as 1.

Since, $h^2=p^2+b^2$ , where b is base of triangle.

So, base b of triangle is $\sqrt{1-x^2}$ .

  

Recall that $\cos x=\frac{\text{base}}{\text{hypotenuse}}$ ,

Therefore, $\cos x=\frac{\sqrt{1-x^2}}{1}$

Now,

  $\begin{array}{c}\cos \left({\sin }^{-1}x\right)=\cos \left({\sin }^{-1}\frac{x}{1}\right)\\ =\cos \left({\cos }^{-1}\frac{\sqrt{1-x^2}}{1}\right)\\ =\sqrt{1-x^2}\end{array}$

Since, trigonometric and inverse trigonometric functions cancel each other.

Thus, left hand side and right hand side is equal, therefore, the trigonometric identity $\cos \left({\sin }^{-1}x\right)=\sqrt{1-x^2}$ is verified.