# Question #9e27a

Mar 22, 2015

Regardless of the unit you use for the angle, the following relations hold:

$\setminus \sin \left(a + b\right) = \setminus \sin \left(a\right) \setminus \cos \left(b\right) + \setminus \cos \left(b\right) \setminus \sin \left(a\right)$
$\setminus \sin \left(a - b\right) = \setminus \sin \left(a\right) \setminus \cos \left(b\right) - \setminus \cos \left(b\right) \setminus \sin \left(a\right)$
$\setminus \cos \left(a + b\right) = \setminus \cos \left(a\right) \setminus \cos \left(b\right) - \setminus \sin \left(a\right) \setminus \sin \left(b\right)$
$\setminus \cos \left(a - b\right) = \setminus \cos \left(a\right) \setminus \cos \left(b\right) + \setminus \sin \left(a\right) \setminus \sin \left(b\right)$

(you can check them out here )

You only need to recognize the right case:
$\setminus \sin \left(45\right) \setminus \cos \left(15\right) + \setminus \cos \left(45\right) \setminus \sin \left(15\right)$ is an expression of the form $\setminus \sin \left(a\right) \setminus \cos \left(b\right) + \setminus \cos \left(b\right) \setminus \sin \left(a\right)$, which is the sine of the sum of the angles, so
$\setminus \sin \left(45\right) \setminus \cos \left(15\right) + \setminus \cos \left(45\right) \setminus \sin \left(15\right) = \setminus \sin \left(45 + 15\right) = \setminus \sin \left(60\right) = \setminus \frac{\sqrt{3}}{2}$