It is given that $ \cos\left(\frac{23\pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4}.$ Which of the following is the value of $\cos^{-1}\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right)$? - HSC - SSCE Mathematics Extension 1 - Question 1 - 2022 - Paper 1

The cosine value corresponds to an angle of $\frac{23\pi}{12}$ in the fourth quadrant, where the cosine is positive. The inverse cosine function will provide the corresponding angle in the first range from $0$ to $\pi$.

Since we know that $\cos^{-1}\left(\cos(\theta)\right) = \theta$, the calculated inverse cosine is: $\cos^{-1}\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) = \frac{23\pi}{12}.$ However, as this angle is outside the standard range of $[0, \pi]$, we convert the angle within the allowed range. The equivalent angle within the range is obtained by subtracting $2\pi$.

To express $\frac{23\pi}{12}$ within the range of $0$ to $\pi$, we calculate: $\frac{23\pi}{12} - 2\pi = \frac{23\pi}{12} - \frac{24\pi}{12} = -\frac{\pi}{12}.$\text{ (This is not in the required domain.)} To adjust again into the required range: \text{Thus, } $\cos^{-1}\left(\frac{\sqrt{6} + \sqrt{2}}{4}\right) = \frac{\pi}{12}.$

From the options provided, the correct choice is therefore: C. $\frac{\pi}{12}$.