Questions: Find the exact value of each expression, (a) sin^(-1)(-1/2) (b) cos^(-1)(-sqrt(3)/2) (c) tan^(-1)(-sqrt(3)/3)

Solution Steps

To find the exact values of the given inverse trigonometric functions, we need to determine the angles whose sine, cosine, and tangent values match the given expressions. These angles should be within the principal range of the inverse trigonometric functions.

To find \( \sin^{-1}\left(-\frac{1}{2}\right) \), we determine the angle \( \theta \) such that \( \sin(\theta) = -\frac{1}{2} \). The angle that satisfies this condition in the principal range is: \[ \theta = -\frac{\pi}{6} \quad \text{or} \quad -30^\circ \] Thus, we have: \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \quad \text{or} \quad -30^\circ \]

Next, we find \( \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) \). The angle \( \phi \) that satisfies \( \cos(\phi) = -\frac{\sqrt{3}}{2} \) in the principal range is: \[ \phi = \frac{5\pi}{6} \quad \text{or} \quad 150^\circ \] Thus, we have: \[ \cos^{-1}\left(-\frac{\sqrt{3}}{2}\right) = \frac{5\pi}{6} \quad \text{or} \quad 150^\circ \]

Finally, we find \( \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right) \). The angle \( \psi \) that satisfies \( \tan(\psi) = -\frac{\sqrt{3}}{3} \) is: \[ \psi = -\frac{\pi}{6} \quad \text{or} \quad -30^\circ \] Thus, we have: \[ \tan^{-1}\left(-\frac{\sqrt{3}}{3}\right) = -\frac{\pi}{6} \quad \text{or} \quad -30^\circ \]

Final Answer