Q2:

Find the principal values of the following: $\cos^{-1} (\frac{\sqrt{3}}{2})$

Step 1: Establishing the Theoretical Range

The inverse cosine function $\cos^{-1}(x)$ is defined such that its principal value branch lies within the interval $[0, \pi]$. Let $y$ be the principal value of $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$. Thus, we define the relation:

$y = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

This is equivalent to the trigonometric equation:

$\cos(y) = \frac{\sqrt{3}}{2}$, where $y \in [0, \pi]$ [Per the definition of the Principal Value Branch of Inverse Cosine].

Step 2: Analytical Identification of the Angle

We evaluate the specific value of the cosine function within the unit circle geometry. Recalling the standard trigonometric table values for acute angles:

• $\cos(0) = 1$
• $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
• $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$
• $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$
• $\cos\left(\frac{\pi}{2}\right) = 0$

By comparison, $\cos(y) = \frac{\sqrt{3}}{2}$ matches the identity $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$.

Step 3: Verification of Domain and Range

Since $\frac{\pi}{6} \approx 0.5236$ radians, and the defined range is $[0, \pi] \approx [0, 3.14159]$, it is clear that $\frac{\pi}{6} \in [0, \pi]$. [By the Property of Inverse Trigonometric Functions: $\cos^{-1}(\cos \theta) = \theta$ for $\theta \in [0, \pi]$].

Step 4: Geometric Representation (Unit Circle)

Step 5: Final Conclusion

As the value $y = \frac{\pi}{6}$ satisfies both the trigonometric equation $\cos(y) = \frac{\sqrt{3}}{2}$ and the principal branch constraint $y \in [0, \pi]$, we conclude the result.

Final Solution: The principal value is $\frac{\pi}{6}$ radians.