Let's solve each composition by evaluating the inside function first, then applying the outside function.

Problem 1

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

1. Evaluate the Inside Function: We start with $\cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)$.

   • The value $-\frac{\sqrt{2}}{2}$ corresponds to an angle in the range $[0, \pi]$ where the cosine of that angle is $-\frac{\sqrt{2}}{2}$.
   • This angle is $\frac{3\pi}{4}$ (since $\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$).

2. Apply the Outside Function: Now, we find $\sin\left(\frac{3\pi}{4}\right)$.

   • $\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$.

Thus, the answer to the first problem is: $\sin\left(\cos^{-1}\left(-\frac{\sqrt{2}}{2}\right)\right) = \frac{\sqrt{2}}{2}$

Problem 2

$\cos\left(\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)\right)$

1. Evaluate the Inside Function: We start with $\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)$.

   • The value $-\frac{\sqrt{3}}{3}$ corresponds to an angle whose tangent is $-\frac{\sqrt{3}}{3}$.
   • This angle is $-\frac{\pi}{6}$ (since $\tan\left(-\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{3}$).

2. Apply the Outside Function: Now, we find $\cos\left(-\frac{\pi}{6}\right)$.

   • $\cos\left(-\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

Thus, the answer to the second problem is: $\cos\left(\tan^{-1}\left(-\frac{\sqrt{3}}{3}\right)\right) = \frac{\sqrt{3}}{2}$

Would you like further explanation on these steps or details on trigonometric identities?

Here are 5 related questions to deepen your understanding:

1. How would you evaluate $\sin\left(\tan^{-1}\left(\frac{1}{\sqrt{3}}\right)\right)$?
2. What is the process for finding $\tan\left(\sin^{-1}\left(\frac{1}{2}\right)\right)$?
3. Can you explain why certain angles are chosen for inverse trigonometric functions?
4. How does the range of each inverse trigonometric function affect the solutions?
5. What is the significance of special angles in trigonometric compositions?

Tip: For compositions involving inverse trigonometric functions, always evaluate the inside function first and recall the principal ranges of these functions to find correct values.