Chapter 5.1, Problem 80E

To determine

To express: The given algebraic expression as a trigonometric function of $\theta$ .

Answer to Problem 80E

The expression $\sqrt{4-x^2}$ is equal to $(2\sin \theta )$ , when $x=2\cos \theta$ .

Explanation of Solution

Given information:

The given algebraic expression:

  $\sqrt{4-x^2}$

The given trigonometric substitution:

  $x=2\cos \theta$

Where $0<\theta <\pi /2$

Calculation:

As per problem,

The given expression:

  $\sqrt{4-x^2}$

Let $x=2\cos \theta$

  $\begin{array}{l}\sqrt{4-x^2}=\sqrt{4-{(2\cos \theta )}^2}\text{ [Substitute 2cos}\theta \text{ for }x]\\ =\sqrt{4-4{\cos }^2\theta }\text{ [Rule of exponents]}\\ =\sqrt{4(1-{\cos }^2\theta )}\text{ [Factor]}\\ =\sqrt{4{\sin }^2\theta }\text{ [Pythagorean identity: }{\sin }^2\theta \text{=}1-{\text{cos}}^2\theta \text{]}\\ =2\sin \theta \text{ [sin}\theta \text{>0 for 0<}\theta \text{<}\pi /2\text{]}\end{array}$

Hence, the expression $\sqrt{4-x^2}$ is equal to $(2\sin \theta )$ ,when $x=2\cos \theta$ .