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

Question

Use the trigonometric substitution to rewrite the algebraic expression as a trigonometric function of θ,where $0 < \theta < \frac { \pi } { 2 }$ . $\sqrt { 11 - x ^ { 2 } } , x = \sqrt { 11 } \sin \theta$ ​
A) $- \sqrt { 11 } \cos \theta$
B) $\sqrt { 11 } \cos \theta$
C) $\sqrt { 11 } \sin \theta$
D) $11 \cos \theta$
E) $11 \sin \theta$

Quizplus · Accepted Answer

Given $x = \sqrt{11} \sin \theta$, substituting $x$ into the expression $\sqrt{11 - x^2}$ and using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$, we get $\sqrt{11 - 11\sin^2 \theta} = \sqrt{11(1 - \sin^2 \theta)} = \sqrt{11\cos^2 \theta} = \sqrt{11} \cos \theta$.

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