Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of $\theta $ , where $0

Question

Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of $\theta $ , where $0 < x < \frac { \pi } { 2 }$ . 
 $\sqrt { 16 - x ^ { 2 } } , x = 4 \cos \theta$

A)16 cos $\theta $ 
B)16 sin $\theta $ 
C)4 sin $\theta $ 
D)- 4 sin $\theta $ 
E)4 cos $\theta $

Quizplus · Accepted Answer

Given $x = 4\cos\theta$, substituting $x$ into the expression $\sqrt{16 - x^2}$ gives $\sqrt{16 - (4\cos\theta)^2} = \sqrt{16 - 16\cos^2\theta} = \sqrt{16(1 - \cos^2\theta)}$. Using the Pythagorean identity $1 - \cos^2\theta = \sin^2\theta$, we get $\sqrt{16\sin^2\theta} = 4\sin\theta$.

Use the Trigonometric Substitution to Rewrite the Algebraic Expression as a Trigonometric