# How do you express cos( (4 pi)/3 ) * cos (( pi) /6 )  without using products of trigonometric functions?

Apr 1, 2016

$- \frac{\sqrt{3}}{2}$

#### Explanation:

$P = \cos \left(\frac{4 \pi}{3}\right) . \cos \left(\frac{\pi}{6}\right)$
Trig table --> $\cos \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$
$\cos \left(\frac{4 \pi}{3}\right) = \cos \left(\frac{\pi}{3} + \frac{3 \pi}{3}\right) = \cos \left(\frac{\pi}{3} + \pi\right) = - \cos \left(\frac{\pi}{3}\right) = - \frac{1}{2}$
$P = \left(- \frac{1}{2}\right) \left(\frac{\sqrt{3}}{2}\right) = - \frac{\sqrt{3}}{4}$

Apr 4, 2016

$\cos \left(\frac{4 \pi}{3}\right) \cos \left(\frac{\pi}{6}\right) = \frac{1}{2} \cos \left(\frac{9 \pi}{6}\right) + \frac{1}{2} \cos \left(\frac{7 \pi}{6}\right) = - \frac{\sqrt{3}}{4}$

#### Explanation:

$2 \cos A \cos B = \cos \left(A + B\right) + \cos \left(A - B\right)$
$\cos A \cos B = \frac{1}{2} \left(\cos \left(A + B\right) + \cos \left(A - B\right)\right)$
$A = \frac{4 \pi}{3} , B = \frac{\pi}{6}$
$\cos \left(\frac{4 \pi}{3}\right) \cos \left(\frac{\pi}{6}\right) = \frac{1}{2} \left(\cos \left(\frac{4 \pi}{3} + \frac{\pi}{6}\right) + \cos \left(\frac{4 \pi}{3} - \frac{\pi}{6}\right)\right)$
$= \frac{1}{2} \left(\cos \left(\frac{9 \pi}{6}\right) + \cos \left(\frac{7 \pi}{6}\right)\right)$
$= \frac{1}{2} \cos \left(\frac{9 \pi}{6}\right) + \frac{1}{2} \cos \left(\frac{7 \pi}{6}\right)$
$= \frac{1}{2} \left(0\right) + \frac{1}{2} \left(- \frac{\sqrt{3}}{2}\right)$
$= - \frac{\sqrt{3}}{4}$
$\cos \left(\frac{4 \pi}{3}\right) \cos \left(\frac{\pi}{6}\right) = \frac{1}{2} \cos \left(\frac{9 \pi}{6}\right) + \frac{1}{2} \cos \left(\frac{7 \pi}{6}\right) = - \frac{\sqrt{3}}{4}$