# Question #01a2a

Jan 30, 2018

$\frac{1}{2} \left(\cos \left(5 \theta\right) - \cos \left(9 \theta\right)\right)$

#### Explanation:

Using the sum and difference formulas for cosine we have:

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

If we subtract those formulas:

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

so

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

so

$\sin \left(2 \theta\right) \sin \left(7 \theta\right) = \frac{1}{2} \left(\cos \left(2 \theta - 7 \theta\right) - \cos \left(2 \theta + 7 \theta\right)\right)$

$= \frac{1}{2} \left(\cos \left(- 5 \theta\right) - \cos \left(9 \theta\right)\right)$

since cosine is even we can rewrite this as:

$= \frac{1}{2} \left(\cos \left(5 \theta\right) - \cos \left(9 \theta\right)\right)$