# How do you use the product to sum formulas to write 6sin(pi/4)cos(pi/4) as a sum or difference?

Feb 21, 2018

The expression is equal to $3 \sin \left(\frac{\pi}{2}\right)$.

#### Explanation:

Using this formula:

$\textcolor{w h i t e}{=} \sin \left(2 x\right) = 2 \sin x \cos x$

We can do it backwards:

$\textcolor{w h i t e}{=} 6 \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)$

$= 3 \left(2 \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)\right)$

$= 3 \left(\sin \left(2 \cdot \frac{\pi}{4}\right)\right)$

$= 3 \sin \left(\frac{\pi}{2}\right)$

Hope this was the answer you were looking for!

Feb 21, 2018

See below.

#### Explanation:

Identity.

$\textcolor{red}{\boldsymbol{2 \sin A \cos B = \sin \left(A + B\right) + \sin \left(A - B\right)}}$

$3 \left(2 \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)\right) = 3 \left(\sin \left(\frac{\pi}{4} + \frac{\pi}{4}\right) + \sin \left(\frac{\pi}{4} - \frac{\pi}{4}\right)\right)$

$3 \left(2 \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)\right) = 3 \sin \left(\frac{\pi}{4} + \frac{\pi}{4}\right) + 3 \sin \left(\frac{\pi}{4} - \frac{\pi}{4}\right)$

$3 \left(2 \sin \left(\frac{\pi}{4}\right) \cos \left(\frac{\pi}{4}\right)\right) = 3 \sin \left(\frac{\pi}{2}\right) + 3 \sin \left(0\right)$

This just simplifies to:

$3 \sin \left(\frac{\pi}{2}\right)$