To find the inverse sine, cosine, and tangent of \(\frac{\sqrt{3}}{3}\), we will evaluate each function one by one.
1. \( \sin^{-1}\left(\frac{\sqrt{3}}{3}\right) \)
The value of \(\sin \theta = \frac{\sqrt{3}}{3}\) corresponds to an angle where the sine is \(\frac{\sqrt{3}}{3}\). This occurs at \(\theta = \frac{\pi}{6}\) (or 30 degrees).
Thus, \[ \sin^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}. \]
2. \( \cos^{-1}\left(\frac{\sqrt{3}}{3}\right) \)
The value of \(\cos \theta = \frac{\sqrt{3}}{3}\) occurs at the angle \(\theta = \frac{\pi}{6}\) where \(\cos\) of \(\frac{\pi}{3}\) equals \(\frac{1}{2}\) and \(\cos\) of \(\frac{\pi}{4}\) equals \(\frac{\sqrt{2}}{2}\). For \(\cos\), we look for the complementary angle:
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}, \] and we find the corresponding angle where cosine equals \(\frac{\sqrt{3}}{3}\), which is actually: \[ \cos^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}, \] but to also find where this is consistent with conventional intervals, we can use: \[ \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{3} \text{ not in domain}, \text{ which maintains within } 0 \to \pi: \]
Thus, \[ \cos^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{3}. \]
3. \( \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) \)
The value of \(\tan \theta = \frac{\sqrt{3}}{3}\) occurs at \(\theta = \frac{\pi}{6}\) (or 30 degrees), since \(\tan\) of \(\frac{\pi}{6}\) (opposite over adjacent) gives this result.
Hence, \[ \tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}. \]
Final Answers:
- \(\sin^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}\)
- \(\cos^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{3}\)
- \(\tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}\)