Sec 6.6 Inverse Trig: Problem 5 (1 point) Evaluate the following expressions. Your answer must be an angle in radians and in the interval \( \left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \). (a) \( \sin ^{-1}(-1)= \) \( \square \) (b) \( \sin ^{-7}\left(\frac{\sqrt{2}}{2}\right)= \) \( \square \) (c) \( \sin ^{-1}\left(\frac{x}{2}\right)= \) \( \square \) Note: Wou cun earn partiol credit on this problem.

Let's evaluate each expression step by step. ### (a) \( \sin^{-1}(-1) \) The function \( \sin^{-1}(x) \) (also known as arcsin) gives us the angle whose sine is \( x \). We need to find the angle \( \theta \) such that: \[ \sin(\theta) = -1 \] The sine function equals \(-1\) at: \[ \theta = -\frac{\pi}{2} \] Thus, \[ \sin^{-1}(-1) = -\frac{\pi}{2} \] ### (b) \( \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) \) Again, we are looking for the angle \( \theta \) such that: \[ \sin(\theta) = \frac{\sqrt{2}}{2} \] The sine function equals \(\frac{\sqrt{2}}{2}\) at: \[ \theta = \frac{\pi}{4} \] Thus, \[ \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \] ### (c) \( \sin^{-1}\left(\frac{x}{2}\right) \) For this expression, we are looking for the angle \( \theta \) such that: \[ \sin(\theta) = \frac{x}{2} \] The output of \( \sin^{-1}\left(\frac{x}{2}\right) \) will depend on the value of \( x \). The range of \( \sin^{-1}(x) \) is defined for \( x \) in the interval \([-1, 1]\). Therefore, \( \frac{x}{2} \) must also be in the interval \([-1, 1]\), which means \( x \) must be in the interval \([-2, 2]\). Thus, the answer for this part is: \[ \sin^{-1}\left(\frac{x}{2}\right) = \theta \quad \text{where } \theta \text{ is such that } \sin(\theta) = \frac{x}{2} \] ### Summary of Answers (a) \( \sin^{-1}(-1) = -\frac{\pi}{2} \) (b) \( \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) = \frac{\pi}{4} \) (c) \( \sin^{-1}\left(\frac{x}{2}\right) = \theta \) where \( \sin(\theta) = \frac{x}{2} \) and \( x \in [-2, 2] \)