# How do you use the sum to product formulas to write the sum or difference cos(theta+pi/2)-cos(theta-pi/2) as a product?

##### 1 Answer
Mar 3, 2017

$- 2 \sin \theta .$

#### Explanation:

We know that, $\cos A - \cos B = - 2 \sin \left(\frac{A + B}{2}\right) \sin \left(\frac{A - B}{2}\right) .$

Here, we have, $A = \theta + \frac{\pi}{2} , \mathmr{and} , B = \theta - \frac{\pi}{2.}$

$\therefore \frac{A + B}{2} = \frac{1}{2} \left\{\left(\theta + \frac{\pi}{2}\right) + \left(\theta - \frac{\pi}{2}\right)\right\} = \theta ,$ and,

$\frac{A - B}{2} = \frac{1}{2} \left\{\left(\theta + \frac{\pi}{2}\right) - \left(\theta - \frac{\pi}{2}\right)\right\} = \frac{\pi}{2.}$

$\Rightarrow \text{The Expression=} - 2 \left(\sin \theta\right) \left(\sin \left(\frac{\pi}{2}\right)\right) = - 2 \sin \theta .$

Alternatively,

$\cos \left(\theta + \frac{\pi}{2}\right) = - \sin \theta , \mathmr{and} , \cos \left(\theta - \frac{\pi}{2}\right) = \sin \theta ,$ hence,

$\text{The Exp.=} - \sin \theta - \sin \theta = - 2 \sin \theta ,$ as before!

Enjoy Maths.!