Chapter 5.1, Problem 85E

To determine

To write: The algebraic expression as a trigonometric function of $\theta$ using trigonometric substitution, where $0<\theta <\frac{\pi }{2}$ .

Answer to Problem 85E

The required function is $7\cos \theta$ .

Explanation of Solution

Given information:

The algebraic expression: $\sqrt{49-x^2}$ , $x=7\sin \theta$

Formula used:

Trigonometric identity:

  ${\cos }^2\theta +{\sin }^2\theta =1\Rightarrow 1-{\sin }^2\theta ={\cos }^2\theta$

Calculation:

The given algebraic expression is $\sqrt{49-x^2}$ and $x=7\sin \theta$ .

Substitute $7\sin \theta$ for x in the above expression.

  $\begin{array}{c}\sqrt{49-x^2}=\sqrt{49-\left(7{\sin }^2\theta \right)}\\ =\sqrt{49-49{\sin }^2\theta }\\ =\sqrt{49\left(1-{\sin }^2\theta \right)}\end{array}$

Use the formula $1-{\sin }^2\theta ={\cos }^2\theta$ on the above expression.

  $\begin{array}{c}\sqrt{49\left(1-{\sin }^2\theta \right)}=\sqrt{49{\cos }^2\theta }\\ =\sqrt{7^2{\cos }^2\theta }\\ =\sqrt{{\left(7\cos \theta \right)}^2}\\ =7\cos \theta \end{array}$

Hence, the required function is $7\cos \theta$ .