If $x = 6 \sin \theta$, use trigonometric substitution to write $\sqrt { 36 - x ^ { 2 } }$ as a trigonometric function of $\theta$, where $0

Question

If $x = 6 \sin \theta$, use trigonometric substitution to write $\sqrt { 36 - x ^ { 2 } }$ as a trigonometric function of $\theta$, where $0 < \theta < \frac { \pi } { 2 }$.
A) $6 \csc \theta$
B) $6 \sec \theta$
C) $6 \sin \theta$
D) $6 \cos \theta$
E) $6 \tan \theta$

Quizplus · Accepted Answer

Given $x = 6 \sin \theta$, we have $\sqrt{36 - x^2} = \sqrt{36 - (6\sin\theta)^2} = \sqrt{36(1 - \sin^2\theta)} = \sqrt{36\cos^2\theta} = 6\cos\theta$, using the Pythagorean identity $\sin^2\theta + \cos^2\theta = 1$.

If x=6sin⁡θx = 6 \sin \thetax=6sinθ , Use Trigonometric Substitution to Write

If $x = 6 \sin \theta$ , Use Trigonometric Substitution to Write