Chapter 5.1, Problem 89E

To determine

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

Answer to Problem 89E

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

Explanation of Solution

Given information :

The given algebraic expression:

  $\sqrt{2-x^2}$

The given trigonometric substitution:

  $x=\sqrt{2}\sin \theta$

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

Calculation :

As per problem,

The given expression:

  $\sqrt{2-x^2}$

Let $x=\sqrt{2}\sin \theta$

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

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