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

Feb 14, 2017

$\sin \left(x + y\right) \cos \left(x - y\right) = \frac{1}{2} \left(\sin 2 x + \sin 2 y\right)$

#### Explanation:

As $\cos A = \sin \left(\frac{\pi}{2} - A\right)$

$\sin \left(x + y\right) \cos \left(x - y\right)$

= $\sin \left(x + y\right) \sin \left(\frac{\pi}{2} - \left(x - y\right)\right)$

= $\sin \left(x + y\right) \sin \left(\frac{\pi}{2} - x + y\right)$

Now as $2 \sin A \sin B = \cos \left(A - B\right) - \cos \left(A + B\right)$

$\sin \left(x + y\right) \sin \left(\frac{\pi}{2} - x + y\right)$

= $\frac{1}{2} \left[\cos \left(x + y - \left(\frac{\pi}{2} - x + y\right)\right) - \cos \left(x + y + \left(\frac{\pi}{2} - x + y\right)\right)\right]$

= $\frac{1}{2} \left[\cos \left(x + y - \frac{\pi}{2} + x - y\right) - \cos \left(x + y + \frac{\pi}{2} - x + y\right)\right]$

= $\frac{1}{2} \left[\cos \left(2 x - \frac{\pi}{2}\right) - \cos \left(2 y + \frac{\pi}{2}\right)\right]$

= $\frac{1}{2} \left[\cos \left(\frac{\pi}{2} - 2 x\right) - \cos \left(\frac{\pi}{2} + 2 y\right)\right]$

• as $\cos \left(- A\right) = \cos A$

= $\frac{1}{2} \left(\sin 2 x + \frac{1}{2} \sin 2 y\right)$

• as $\cos \left(\frac{\pi}{2} - A\right) = \sin A$ and $\cos \left(\frac{\pi}{2} + A\right) = - \sin A$