# How do you simplify (sin theta + cos theta) ^2 ?

Oct 18, 2015

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

#### Explanation:

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

We know that ${\sin}^{2} \left(\theta\right) + {\cos}^{2} \left(\theta\right) = 1$ so

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

However, $2 \sin \left(\theta\right) \cos \left(\theta\right) = \sin \left(2 \theta\right)$ so we can say

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