Evaluate the following expressions. (a) \( \sin \left(\sin ^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right)=\square \) (b) \( \tan \left(\tan ^{-1}\left(-\frac{1}{2}\right)\right)=\square \)

To evaluate the expressions, we can use the properties of inverse trigonometric functions. (a) For the expression \( \sin \left(\sin ^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right) \): - The function \( \sin^{-1} \) (also known as arcsin) takes a value and gives an angle whose sine is that value. The range of \( \sin^{-1} \) is \([- \frac{\pi}{2}, \frac{\pi}{2}]\). - Thus, \( \sin^{-1}\left(-\frac{\sqrt{2}}{2}\right) \) corresponds to the angle \( -\frac{\pi}{4} \) radians, or \( -45^\circ \), because \(\sin(-45^\circ) = -\frac{\sqrt{2}}{2}\). - Then, taking the sine of \( -\frac{\pi}{4} \) gives: \[ \sin\left(-\frac{\pi}{4}\right) = -\frac{\sqrt{2}}{2}. \] So, \(\sin \left(\sin ^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right) = -\frac{\sqrt{2}}{2}\). (b) For the expression \( \tan \left(\tan ^{-1}\left(-\frac{1}{2}\right)\right) \): - The function \( \tan^{-1} \) (also known as arctan) gives an angle whose tangent is \( -\frac{1}{2} \). The range of \( \tan^{-1} \) is \((- \frac{\pi}{2}, \frac{\pi}{2})\). - Therefore, \( \tan^{-1}\left(-\frac{1}{2}\right) \) results in an angle \( \theta \) such that \( \tan(\theta) = -\frac{1}{2} \). - And since \(\tan(\theta)\) is just the fractional value corresponding to that angle, we find that: \[ \tan\left(\tan^{-1}\left(-\frac{1}{2}\right)\right) = -\frac{1}{2}. \] Thus, the evaluations are: (a) \( \sin \left(\sin ^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right) = -\frac{\sqrt{2}}{2} \) (b) \( \tan \left(\tan ^{-1}\left(-\frac{1}{2}\right)\right) = -\frac{1}{2} \)