# Question #d444c

Feb 4, 2018

see explanation.

#### Explanation:

${\cos}^{2} \left(\theta\right) \ne \frac{{\cos}^{2} \left(\theta\right) + 1}{2}$ in general. There are solutions to the equation, though:

${\cos}^{2} \left(\theta\right) = \frac{{\cos}^{2} \left(\theta\right) + 1}{2}$

$2 {\cos}^{2} \left(\theta\right) = {\cos}^{2} \left(\theta\right) + 1$

${\cos}^{2} \left(\theta\right) = 1$

$\cos \left(\theta\right) = 1$ or $\cos \left(\theta\right) = - 1$

$\cos \left(\theta\right) = 1 \rightarrow \theta = 2 \pi \cdot n$, $n \in \mathbb{Z}$

$\cos \left(\theta\right) = - 1 \rightarrow \theta = \pi + 2 \pi \cdot n$, $n \in \mathbb{Z}$.

However! I suspect there's a typo because the following is true:

${\cos}^{2} \left(\theta\right) = \frac{\cos \left(2 \theta\right) + 1}{2}$

This is a power reducing formula that's quite useful and comes from the double angle formula for cosine:

$\cos \left(2 \theta\right) = 2 {\cos}^{2} \left(\theta\right) - 1$

If we solve this for ${\cos}^{2} \left(\theta\right)$,

$\cos \left(2 \theta\right) + 1 = 2 {\cos}^{2} \left(\theta\right)$

$\frac{\cos \left(2 \theta\right) + 1}{2} = {\cos}^{2} \left(\theta\right)$

Where the double angle formula comes from is a bit of a rabbit hole, so I'm going to stop there.