# How do you express cos( (5 pi)/6 ) * cos (( 17 pi) /12 )  without using products of trigonometric functions?

Use SUM
$\frac{1}{2} \cdot \cos \left(\frac{\pi}{4}\right) + \frac{1}{2} \cdot \cos \left(\frac{7 \pi}{12}\right)$

#### Explanation:

The given: $\cos \left(\frac{5 \pi}{6}\right) \cdot \cos \left(\frac{17 \pi}{12}\right)$

Derivation: Use $\textcolor{red}{\text{Sum and Difference Formulas}}$
$\textcolor{b l u e}{\text{cos (x+y)=cos x cos y-sin x sin y" " }}$first equation
$\textcolor{b l u e}{\text{cos (x-y)=cos x cos y+sin x sin y" " " }}$second equation

$\cos \left(x + y\right) = \cos x \cos y - \cancel{\sin x \sin y}$
$\cos \left(x - y\right) = \cos x \cos y + \cancel{\sin x \sin y}$

th result is

$\cos \left(x + y\right) + \cos \left(x - y\right) = 2 \cdot \cos x \cdot \cos y$

divide both sides by 2 , the result is

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

Now, Let $x = \frac{5 \pi}{6}$ and $y = \frac{17 \pi}{12}$ then use

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

$\cos \left(\frac{5 \pi}{6}\right) \cdot \cos \left(\frac{17 \pi}{12}\right) =$
$\frac{1}{2} \cdot \cos \left(\frac{5 \pi}{6} + \frac{17 \pi}{12}\right) + \frac{1}{2} \cdot \cos \left(\frac{5 \pi}{6} - \frac{17 \pi}{12}\right)$

$= \frac{1}{2} \cdot \cos \left(\frac{27 \pi}{12}\right) + \frac{1}{2} \cdot \cos \left(\frac{- 7 \pi}{12}\right)$

Take note: $\cos \left(\frac{- 7 \pi}{12}\right) = \cos \left(\frac{+ 7 \pi}{12}\right)$

Also $\cos \left(\frac{27 \pi}{12}\right) = \cos \left(\frac{24 \pi}{12} + \frac{3 \pi}{12}\right) = \cos \left(2 \pi + \frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right)$

further simplification

$\frac{1}{2} \cdot \cos \left(\frac{\pi}{4}\right) + \frac{1}{2} \cdot \cos \left(\frac{7 \pi}{12}\right)$

God bless America ....