# Convert 5cos((5pi)/3)cos((5pi)/3) as a sum of trigonometric ratios?

##### 1 Answer
Mar 29, 2017

$5 \cos \left(\frac{5 \pi}{3}\right) \cos \left(\frac{5 \pi}{3}\right) = \frac{5}{2} \left(\cos \left(\frac{10 \pi}{3}\right) + \cos 0\right)$

= $\frac{5}{2} \left(\cos \left(\frac{10 \pi}{3}\right) + 1\right)$

#### Explanation:

As $\cos \left(A + B\right) = \cos A \cos B - \sin A \sin B$ and

$\cos \left(A - B\right) = \cos A \cos B + \sin A \sin B$

adding the two we get $\cos \left(A + B\right) + \cos \left(A - B\right) = 2 \cos A \cos B$

Hence $5 \cos \left(\frac{5 \pi}{3}\right) \cos \left(\frac{5 \pi}{3}\right)$

$= \frac{5}{2} \times 2 \cos \left(\frac{5 \pi}{3}\right) \cos \left(\frac{5 \pi}{3}\right)$

$= \frac{5}{2} \times \left(\cos \left(\frac{5 \pi}{3} + \frac{5 \pi}{3}\right) + \cos \left(\frac{5 \pi}{3} - \frac{5 \pi}{3}\right)\right)$

$= \frac{5}{2} \times \left(\cos \left(\frac{10 \pi}{3}\right) + \cos 0\right)$, but $\cos 0 = 1$

Hence $5 \cos \left(\frac{5 \pi}{3}\right) \cos \left(\frac{5 \pi}{3}\right) = \frac{5}{2} \times \left(\cos \left(\frac{10 \pi}{3}\right) + 1\right)$