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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part. Find the exact value of the expression. Exercise (a) $\sin^{-1}(\frac{-\sqrt{3}}{2})$ Part 1 of 2 If $y = \sin^{-1}x$, then $\sin y = x$, $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$. Therefore, to find $y = \sin^{-1}(\frac{-\sqrt{3}}{2})$, we must find an angle $y$ whose sine is $-\frac{\sqrt{3}}{2}$. There are many possible angles with this sine, but the range of $y = \sin^{-1}x$ is restricted to $[\frac{-\pi}{2}, \frac{\pi}{2}]$, and so $y$ must be in this interval. Part 2 of 2 The angle $y$ whose sine is $-\frac{\sqrt{3}}{2}$ and which is in the interval $[\frac{-\pi}{2}, \frac{\pi}{2}]$ is $y = $ radians. Submit Skip (you cannot come back)

          This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.
Find the exact value of the expression.
Exercise (a)
$\sin^{-1}(\frac{-\sqrt{3}}{2})$
Part 1 of 2
If $y = \sin^{-1}x$, then $\sin y = x$, $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$. Therefore, to find $y = \sin^{-1}(\frac{-\sqrt{3}}{2})$, we must find an angle $y$ whose sine is $-\frac{\sqrt{3}}{2}$.
There are many possible angles with this sine, but the range of $y = \sin^{-1}x$ is restricted to $[\frac{-\pi}{2}, \frac{\pi}{2}]$, and so $y$ must be in this interval.
Part 2 of 2
The angle $y$ whose sine is $-\frac{\sqrt{3}}{2}$ and which is in the interval $[\frac{-\pi}{2}, \frac{\pi}{2}]$ is $y = $ radians.
Submit Skip (you cannot come back)

This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped part, and you will not be able to come back to the skipped part.
Find the exact value of the expression.
Exercise (a)
sin^-1((-√(3))/(2))
Part 1 of 2
If y = sin^-1x, then sin y = x, -(π)/(2)≤ y ≤(π)/(2). Therefore, to find y = sin^-1((-√(3))/(2)), we must find an angle y whose sine is -(√(3))/(2).
There are many possible angles with this sine, but the range of y = sin^-1x is restricted to [(-π)/(2), (π)/(2)], and so y must be in this interval.
Part 2 of 2
The angle y whose sine is -(√(3))/(2) and which is in the interval [(-π)/(2), (π)/(2)] is y = radians.
Submit Skip (you cannot come back)

Added by Jose Manuel D.

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