To solve these problems, let’s analyze each part carefully.

Part (a)

We need to find the exact value of: $\cos^{-1} \left( \cos \frac{\pi}{4} \right)$

1. First, evaluate $\cos \frac{\pi}{4}$: $\cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}$

2. Now, the expression becomes: $\cos^{-1} \left( \frac{\sqrt{2}}{2} \right)$

3. We know that $\cos^{-1} \left( \frac{\sqrt{2}}{2} \right)$ is the angle $\theta$ in the range $[0, \pi]$ for which $\cos \theta = \frac{\sqrt{2}}{2}$. This occurs when: $\theta = \frac{\pi}{4}$

Therefore, the answer to part (a) is: $\cos^{-1} \left( \cos \frac{\pi}{4} \right) = \frac{\pi}{4}$

Part (b)

We need to find the exact value of: $\sin^{-1} \left( \cos \frac{5\pi}{6} \right)$

1. First, evaluate $\cos \frac{5\pi}{6}$: $\cos \frac{5\pi}{6} = -\frac{\sqrt{3}}{2}$

2. Now, the expression becomes: $\sin^{-1} \left( -\frac{\sqrt{3}}{2} \right)$

3. We know that $\sin^{-1} \left( -\frac{\sqrt{3}}{2} \right)$ is the angle $\theta$ in the range $[- \frac{\pi}{2}, \frac{\pi}{2}]$ for which $\sin \theta = -\frac{\sqrt{3}}{2}$. This occurs when: $\theta = -\frac{\pi}{3}$

Therefore, the answer to part (b) is: $\sin^{-1} \left( \cos \frac{5\pi}{6} \right) = -\frac{\pi}{3}$

Summary of Results

• (a) $\cos^{-1} \left( \cos \frac{\pi}{4} \right) = \frac{\pi}{4}$
• (b) $\sin^{-1} \left( \cos \frac{5\pi}{6} \right) = -\frac{\pi}{3}$

Would you like further explanations on any part?

Additional Questions for Practice

1. How would you solve $\cos^{-1} \left( \cos \frac{3\pi}{4} \right)$?
2. What is $\sin^{-1} \left( \cos \frac{7\pi}{6} \right)$?
3. If $\cos^{-1} (x) = \frac{\pi}{3}$, what is $x$?
4. What angle satisfies $\sin^{-1} \left( -\frac{1}{2} \right)$?
5. Can you determine $\cos^{-1} \left( -\cos \frac{\pi}{3} \right)$?

Tip

Always remember that $\cos^{-1}$ has a range of $[0, \pi]$ and $\sin^{-1}$ has a range of $[- \frac{\pi}{2}, \frac{\pi}{2}]$, which can help in finding exact values of angles in inverse trigonometric functions.