Let's solve the inverse trigonometric expressions given in the image:

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

This is asking for the angle $\theta$ such that: $\sin(\theta) = -\frac{1}{2}$ We know that $\sin(\theta) = -\frac{1}{2}$ occurs at: $\theta = -\frac{\pi}{6} \quad \text{(in Quadrant IV)}$ So, the exact value of $\sin^{-1}\left( -\frac{1}{2} \right)$ is: $-\frac{\pi}{6} \, \text{radians}$

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

This is asking for the angle $\theta$ such that: $\cos(\theta) = \frac{1}{2}$ We know that $\cos(\theta) = \frac{1}{2}$ occurs at: $\theta = \frac{\pi}{3} \quad \text{(in Quadrant I)}$ So, the exact value of $\cos^{-1}\left( \frac{1}{2} \right)$ is: $\frac{\pi}{3} \, \text{radians}$

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

This is asking for the angle $\theta$ such that: $\tan(\theta) = \frac{\sqrt{3}}{3}$ We know that $\tan(\theta) = \frac{\sqrt{3}}{3}$ occurs at: $\theta = \frac{\pi}{6} \quad \text{(in Quadrant I)}$ So, the exact value of $\tan^{-1}\left( \frac{\sqrt{3}}{3} \right)$ is: $\frac{\pi}{6} \, \text{radians}$

Summary:

• (a) $\sin^{-1}\left( -\frac{1}{2} \right) = -\frac{\pi}{6}$
• (b) $\cos^{-1}\left( \frac{1}{2} \right) = \frac{\pi}{3}$
• (c) $\tan^{-1}\left( \frac{\sqrt{3}}{3} \right) = \frac{\pi}{6}$

Would you like any further details or clarifications?

Here are five related questions you can consider:

1. What is the range of the inverse sine function $\sin^{-1}(x)$?
2. How does the unit circle help in finding inverse trigonometric values?
3. What are the principal values for $\tan^{-1}(x)$?
4. How can you determine when inverse trigonometric values are undefined?
5. What are the differences in behavior between $\sin^{-1}(x)$, $\cos^{-1}(x)$, and $\tan^{-1}(x)$?

Tip: Remember that inverse trigonometric functions return principal values, typically within specific ranges, depending on the function!