# How do you use the product to sum formulas to write 4cos(pi/3)sin((5pi)/6) as a sum or difference?

Jul 8, 2017

$1.$

#### Explanation:

$4 \cos \left(\frac{\pi}{3}\right) \sin \left(5 \frac{\pi}{6}\right) = 4 \cos \left(\frac{\pi}{3}\right) \sin \left(\pi - \frac{\pi}{6}\right) ,$

$= 4 \cos \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{6}\right) ,$

$= 2 \left\{2 \cos \left(\frac{\pi}{3}\right) \sin \left(\frac{\pi}{6}\right)\right\} ,$

$= 2 \left\{\sin \left(\frac{\pi}{3} + \frac{\pi}{6}\right) - \sin \left(\frac{\pi}{3} - \frac{\pi}{6}\right)\right\} ,$

$= 2 \left\{\sin \left(\frac{\pi}{2}\right) - \sin \left(\frac{\pi}{6}\right)\right\} ,$

$= 2 \left(1 - \frac{1}{2}\right) ,$

$= 1.$

Jul 9, 2017

$2 \left[\cos \left(\frac{2 \pi}{3}\right) - \cos \pi\right]$

#### Explanation:

$y = 4 \cos \left(\frac{\pi}{3}\right) \sin \left(\frac{5 \pi}{6}\right)$
Because,
$\cos \left(\frac{\pi}{3}\right) = \sin \left(\frac{\pi}{2} - \frac{\pi}{3}\right) = \sin \left(\frac{\pi}{6}\right)$
Therefor,
$y = 4 \sin \left(\frac{\pi}{6}\right) . \sin \left(\frac{5 \pi}{6}\right)$
Apply the trig identity:
$\sin a . \sin b = \left(\frac{1}{2}\right) \left[\cos \left(a - b\right) - \cos \left(a + b\right)\right]$
In this case:
$y = 2 \left[\cos \left(\frac{2 \pi}{3}\right) - \cos \pi\right]$
That is the answer. However, we can go further, knowing
$\cos \left(\frac{2 \pi}{3}\right) = - \frac{1}{2}$ and $\cos \pi = - 1$.
$y = 2 \left(- \frac{1}{2} + 1\right) = \frac{2}{2} = 1$