Let's solve each of these problems one by one.

Problem (a)

Evaluate: $\text{tan}^{-1} \left( -\frac{\sqrt{3}}{3} \right)$

The expression $\text{tan}^{-1}(x)$ represents the angle $\theta$ such that $\tan(\theta) = x$ and $\theta$ lies within the range $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

To find $\theta$ such that: $\tan(\theta) = -\frac{\sqrt{3}}{3}$

We recognize that: $\tan\left(-\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{3}$

Thus: $\text{tan}^{-1} \left( -\frac{\sqrt{3}}{3} \right) = -\frac{\pi}{6}$

Problem (b)

Evaluate: $\sin \left( \cos^{-1} \frac{3}{8} \right)$

Let $\theta = \cos^{-1} \frac{3}{8}$. This means: $\cos(\theta) = \frac{3}{8}$

We need to find $\sin(\theta)$. Using the Pythagorean identity: $\sin^2(\theta) + \cos^2(\theta) = 1$

Substitute $\cos(\theta) = \frac{3}{8}$: $\sin^2(\theta) + \left(\frac{3}{8}\right)^2 = 1$ $\sin^2(\theta) + \frac{9}{64} = 1$ $\sin^2(\theta) = 1 - \frac{9}{64}$ $\sin^2(\theta) = \frac{64 - 9}{64} = \frac{55}{64}$ $\sin(\theta) = \sqrt{\frac{55}{64}} = \frac{\sqrt{55}}{8}$

Therefore: $\sin \left( \cos^{-1} \frac{3}{8} \right) = \frac{\sqrt{55}}{8}$

Would you like further details or explanations on these solutions?

Here are some related questions for further practice:

1. What is $\sin^{-1} \left( \frac{1}{2} \right)$?
2. How would you find $\cos(\tan^{-1} x)$ for an arbitrary value of $x$?
3. What is the value of $\tan \left( \cos^{-1} \frac{4}{5} \right)$?
4. How does the Pythagorean identity apply to inverse trigonometric functions?
5. How do you determine the range of each inverse trigonometric function?

Tip: Remember that each inverse trigonometric function has a specific range for its principal values. This can help in determining the correct angle when solving inverse trigonometric expressions.