evaluate the following expressions. your answer must be an exact angle in radians and in the interval…

evaluate the following expressions. your answer must be an exact angle in radians and in the interval $\\left-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right$. example: enter pi/6 for $\\frac{\\pi}{6}$.\n(a) $\\sin^{-1}\\left(\\frac{\\sqrt{2}}{2}\\right)=$\n(b) $\\sin^{-1}(-1)=$\n(c) $\\sin^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right)=$

evaluate the following expressions. your answer must be an exact angle in radians and in the interval $\\left-\\frac{\\pi}{2},\\frac{\\pi}{2}\\right$. example: enter pi/6 for $\\frac{\\pi}{6}$.\n(a) $\\sin^{-1}\\left(\\frac{\\sqrt{2}}{2}\\right)=$\n(b) $\\sin^{-1}(-1)=$\n(c) $\\sin^{-1}\\left(-\\frac{\\sqrt{3}}{2}\\right)=$

Answer

Explanation:

Step1: Recall the range of inverse sine function

The range of (y = \sin^{-1}(x)) is (\left[-\frac{\pi}{2},\frac{\pi}{2}\right]).

Step2: Evaluate (\sin^{-1}\left(\frac{\sqrt{2}}{2}\right))

We know that (\sin\left(\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}) and (\frac{\pi}{4}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]). So (\sin^{-1}\left(\frac{\sqrt{2}}{2}\right)=\frac{\pi}{4}).

Step3: Evaluate (\sin^{-1}(- 1))

We know that (\sin\left(-\frac{\pi}{2}\right)=-1) and (-\frac{\pi}{2}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]). So (\sin^{-1}(-1)=-\frac{\pi}{2}).

Step4: Evaluate (\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right))

We know that (\sin\left(-\frac{\pi}{3}\right)=-\frac{\sqrt{3}}{2}) and (-\frac{\pi}{3}\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]). So (\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}).

Answer:

(a) (\frac{\pi}{4}) (b) (-\frac{\pi}{2}) (c) (-\frac{\pi}{3})