# How do you express cos( (5 pi)/6 ) * cos (( 7 pi) /12 )  without using products of trigonometric functions?

Mar 2, 2016

$P = \frac{\sqrt{3}}{4} \left(\sqrt{2 - \sqrt{3}}\right)$

#### Explanation:

Trig table --> $\cos \left(\frac{5 \pi}{6}\right) = - \frac{\sqrt{3}}{2.}$
Trig unit circle -->
$\cos \left(\frac{7 \pi}{12}\right) = \cos \left(\frac{\pi}{12} + \frac{\pi}{2}\right) = - \sin \left(\frac{\pi}{12}\right)$
Product $P = \frac{\sqrt{3}}{2.} \sin \left(\frac{\pi}{12}\right)$.
Find $\sin \left(\frac{\pi}{12}\right)$ by the identity: $\cos \left(\frac{\pi}{6}\right) = 1 - 2 {\sin}^{2} \left(\frac{\pi}{12}\right)$
$2 {\sin}^{2} \left(\frac{\pi}{12}\right) = 1 - \frac{\sqrt{3}}{2} = \frac{2 - \sqrt{3}}{2}$
${\sin}^{2} \left(\frac{\pi}{12}\right) = \frac{2 - \sqrt{3}}{4}$
Take square root of both sides:
$\sin \left(\frac{\pi}{12}\right) = \frac{\sqrt{2 - \sqrt{3}}}{2}$ (because $\sin \left(\frac{\pi}{12}\right)$ is positive)
Finally,
$P = \frac{\sqrt{3}}{4} \left(\sqrt{2 - \sqrt{3}}\right)$

Mar 2, 2016

$\frac{1}{2} \cdot \cos \left(\frac{\pi}{4}\right) - \frac{1}{2} \cdot \cos \left(5 \frac{\pi}{12}\right)$

#### Explanation:

As $\cos \left(A \pm B\right) = \cos A \cdot \cos B - + \sin A \cdot \sin B$, we can write

cosA*cosB=1/2[(cos(A+B)+cos(A-B)]

Hence, cos(5pi/6)â‹…cos(7pi/12) =

1/2[(cos(5pi/6+7pi/12)+cos(5pi/6-7pi/12)]#

= $\frac{1}{2} \left(\cos \left(17 \frac{\pi}{12}\right) + \cos \left(3 \frac{\pi}{12}\right)\right)$

= $\frac{1}{2} \left(- \cos \left(5 \frac{\pi}{12}\right) + \cos \left(\frac{\pi}{4}\right)\right)$

=$\frac{1}{2} \cdot \cos \left(\frac{\pi}{4}\right) - \frac{1}{2} \cdot \cos \left(5 \frac{\pi}{12}\right)$