It is given that \( \cos \left( \frac{23\pi}{12} \right) = \frac{\sqrt{6} + \sqrt{2}}{4} \) - HSC - SSCE Mathematics Extension 1 - Question 1 - 2022 - Paper 1

To find ( \cos^{-1} \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) ), we recognize that this expression represents an angle whose cosine is ( \frac{\sqrt{6} + \sqrt{2}}{4} ).

From the provided information, we have: [ \cos \left( \frac{23\pi}{12} \right) = \frac{\sqrt{6} + \sqrt{2}}{4} ] This implies that ( \cos^{-1} \left( \frac{\sqrt{6} + \sqrt{2}}{4} \right) ) must equal ( \frac{23\pi}{12} ) or any angle coterminal with this value.

However, since the range of the ( \cos^{-1} ) function is limited to ( [0, \pi] ), we should consider equivalent angles. The equivalent angle can be found by subtracting ( 2\pi ) as necessary. Since ( \frac{23\pi}{12} ) is more than ( 2\pi ), we need to reduce it: [ \frac{23\pi}{12} - 2\pi = \frac{23\pi}{12} - \frac{24\pi}{12} = -\frac{\pi}{12} ] Thus, we can find a coterminal angle within the principal range by adding ( 2\pi ): [ -\frac{\pi}{12} + 2\pi = -\frac{\pi}{12} + \frac{24\pi}{12} = \frac{23\pi}{12} ] So, we can express the answer as ( \frac{\pi}{12} ) as our target value is within the principal range.