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 $0 < x < \frac { \pi } { 2 }$ . $\sqrt { 6 - x ^ { 2 } } , x = \sqrt { 6 } \sin \theta$  
A)6 sin $\theta $
B) $- \sqrt { 6 } \cos \theta$
C) $\sqrt { 6 } \sin \theta$
D) $\sqrt { 6 } \cos \theta$
E)6 cos $\theta $

Quizplus · Accepted Answer

Given $x = \sqrt{6} \sin \theta$, substituting $x$ into the expression $\sqrt{6 - x^2}$ and simplifying using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$ leads to $\sqrt{6 - 6\sin^2 \theta} = \sqrt{6(1 - \sin^2 \theta)} = \sqrt{6\cos^2 \theta} = \sqrt{6} \cos \theta$.

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