Evaluate the expression. sin − 1 ( 3 2 ) \sin^{-1}\left(\frac{\sqrt3}{2}\right) sin − 1 ( 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"> )

Here we're asked to evaluate the expression the inverse sine of the square root of 3 over 2. So we can think of this as the sine of what angle is going to give me root 3 over 2, and we want to find the angle for which that is true. Now looking within my interval, negative pi over 2 to pi over 2, because I'm taking the inverse sine of a positive value, I know that my solution, my angle, has to be in quadrant 1. So of these angles in quadrant 1, which one gives me a sine of root 3 over 2? Now remember, sine is your y values when dealing with the unit circle, and here at pi over 3 I know that the sine is root 3 over 2, so that represents my solution. Pi over 3. And we are completely good to go here. Thanks for watching. And let me know if you have questions.

Evaluate the expression. sin ⁡ − 1 ( 3 2 ) \sin^{-1}\left(\frac{\sqrt3}{2}\right) sin − 1 ( 2 3 <path d="M95,702 c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14 c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54 c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10 s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429 c69,-144,104.5,-217.7,106.5,-221 l0 -0 c5.3,-9.3,12,-14,20,-14 H400000v40H845.2724 s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7 c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z M834 80h400000v40h-400000z"> )

− π 3 -\frac{\pi}{3} − 3 π

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Inverse Sine, Cosine, & Tangent

Trigonometry

5. Inverse Trigonometric Functions and Basic Trigonometric Equations

Inverse Sine, Cosine, & Tangent

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Multiple Choice

Evaluate the expression.
$\sin^{-1}\left(\frac{\sqrt3}{2}\right)$

$\frac{\pi}{6}$

$\frac{3\pi}{2}$

$\frac{\pi}{3}$

$-\frac{\pi}{3}$

Verified step by step guidance

Understand that $ \sin^{-1} $ or arcsin is the inverse function of sine, which means it returns the angle whose sine is the given value.

Recognize that $ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) $ asks for the angle whose sine is $ \frac{\sqrt{3}}{2} $.

Recall that the sine of $ \frac{\pi}{3} $ is $ \frac{\sqrt{3}}{2} $. Therefore, $ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} $.

Consider the range of the arcsin function, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. This means the principal value of $ \sin^{-1} $ will be within this range.

Conclude that the correct angle that satisfies the expression $ \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) $ is $ \frac{\pi}{3} $, as it falls within the principal range of arcsin.