Let's solve each part one by one. We'll determine the exact value of each inverse trigonometric expression in radians.

(a) $\sin^{-1} \left( -\frac{\sqrt{2}}{2} \right)$

The expression asks for the angle $x$ such that: $\sin(x) = -\frac{\sqrt{2}}{2}$ The principal range of the inverse sine function, $\sin^{-1}(x)$, is $[- \frac{\pi}{2}, \frac{\pi}{2}]$, where sine values are defined from $-1$ to $1$.

The angle that satisfies $\sin(x) = -\frac{\sqrt{2}}{2}$ in this range is: $x = -\frac{\pi}{4}$ Thus, the answer is: $\sin^{-1} \left( -\frac{\sqrt{2}}{2} \right) = -\frac{\pi}{4}$

(b) $\cos^{-1} \left( -\frac{\sqrt{2}}{2} \right)$

This expression asks for the angle $y$ such that: $\cos(y) = -\frac{\sqrt{2}}{2}$ The principal range of the inverse cosine function, $\cos^{-1}(x)$, is $[0, \pi]$.

The angle that satisfies $\cos(y) = -\frac{\sqrt{2}}{2}$ in this range is: $y = \frac{3\pi}{4}$ Thus, the answer is: $\cos^{-1} \left( -\frac{\sqrt{2}}{2} \right) = \frac{3\pi}{4}$

(c) $\tan^{-1}(-1)$

This expression asks for the angle $z$ such that: $\tan(z) = -1$ The principal range of the inverse tangent function, $\tan^{-1}(x)$, is $(- \frac{\pi}{2}, \frac{\pi}{2})$.

The angle that satisfies $\tan(z) = -1$ in this range is: $z = -\frac{\pi}{4}$ Thus, the answer is: $\tan^{-1}(-1) = -\frac{\pi}{4}$

Summary of Answers

• (a) $\sin^{-1} \left( -\frac{\sqrt{2}}{2} \right) = -\frac{\pi}{4}$
• (b) $\cos^{-1} \left( -\frac{\sqrt{2}}{2} \right) = \frac{3\pi}{4}$
• (c) $\tan^{-1}(-1) = -\frac{\pi}{4}$

Would you like further details or have any questions?

Related Questions:

1. How do we determine the principal range for each inverse trigonometric function?
2. Why is it necessary to use radians instead of degrees in inverse trigonometric functions?
3. What are the values of inverse trigonometric functions for other common angles, like $\frac{\sqrt{3}}{2}$?
4. How would the answers change if we were solving for inverse trigonometric functions outside their principal ranges?
5. Can we apply similar steps to solve inverse hyperbolic functions?

Tip:

Always keep in mind the principal ranges for inverse trigonometric functions to ensure the solutions are within the correct interval.