Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where $- \frac { \pi } { 2 }

Question

Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where $- \frac { \pi } { 2 } < \theta < \frac { \pi } { 2 }$ .Then find sin θ and cos θ.​ $3 \sqrt { 3 } = \sqrt { 81 - 9 x ^ { 2 } } , x = 3 \cos \theta$ ​
A)​ $\sin \theta = 3 ; \sin \theta = \frac { \sqrt { 3 } } { 3 } ; \cos \theta = \frac { \sqrt { 6 } } { 3 }$
B)​ $3 \sin \theta = 3 ; \sin \theta = 0 ; \cos \theta = 1$
C)​ $\sin \theta = \sqrt { 3 } ; \sin \theta = \frac { \sqrt { 3 } } { 3 } ; \cos \theta = \frac { \sqrt { 6 } } { 3 }$
D)​ $9 \sin \theta = 3 \sqrt { 3 } ; \sin \theta = \frac { \sqrt { 3 } } { 3 } ; \cos \theta = \frac { \sqrt { 6 } } { 3 }$
E)​ $3 \cos \theta = 3 ; \sin \theta = 1 ; \cos \theta = 0$

Quizplus · Accepted Answer

Given $x = 3\cos\theta$, substituting $x$ into the expression $3\sqrt{3} = \sqrt{81 - 9x^2}$ gives $3\sqrt{3} = \sqrt{81 - 9(3\cos\theta)^2}$. Simplifying, we get $3\sqrt{3} = \sqrt{81 - 81\cos^2\theta}$, which simplifies to $3\sqrt{3} = 3\sqrt{3}\sin\theta$, leading to $\sin\theta = \frac{\sqrt{3}}{3}$. Using the Pythagorean identity, $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \left(\frac{\sqrt{3}}{3}\right)^2} = \frac{\sqrt{6}}{3}$.

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