# How do you use the product to sum formulas to write cos(theta-pi)sin(theta+pi) as a sum or difference?

Jan 6, 2017

$\cos \left(\theta - \pi\right) \sin \left(\theta + \pi\right) = \frac{1}{2} \sin 2 \theta - 0 = \sin \theta \cos \theta$

#### Explanation:

We can use here the product formula

$\cos A \sin B = \frac{1}{2} \left[\sin \left(A + B\right) - \sin \left(A - B\right)\right]$

Hence $\cos \left(\theta - \pi\right) \sin \left(\theta + \pi\right)$

= $\frac{1}{2} \left[\sin \left(\left(\theta - \pi\right) + \left(\theta + \pi\right)\right) - \sin \left(\left(\theta - \pi\right) - \left(\theta + \pi\right)\right)\right]$

= $\frac{1}{2} \left[\sin \left(\theta - \pi + \theta + \pi\right) - \sin \left(\theta - \pi - \theta - \pi\right)\right]$

= 1/2sin2theta-sin(-2pi)] $-$ $\textcolor{red}{b u t}$ $\textcolor{red}{a s}$ $\sin \left(- 2 \pi\right) = 0$

= $\frac{1}{2} \sin 2 \theta - 0$

Here, we have written as a difference, but latter component is zero

and can be simplified further as

$\frac{1}{2} \sin 2 \theta = \frac{1}{2} \times 2 \sin \theta \cos \theta = \sin \theta \cos \theta$