It is given that $ abla 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

We know that $abla cos \left( \frac{23\pi}{12} \right) = \frac{\sqrt{6} + \sqrt{2}}{4}.$ Therefore, to find $\theta$ such that $cos(\theta) = \frac{\sqrt{6} + \sqrt{2}}{4}$, we can analyze the angle.

The angle $\theta$ can be expressed in terms of fractions of $\pi$. Since $\frac{23\pi}{12}$ is greater than $\pi$, we need to identify its equivalent reference angle in the unit circle.