# How do you multiply \sin ( - \frac { 50\pi } { 3} ) \cdot \cos \frac { 53\pi } { 6}?

Mar 2, 2018

$\frac{3}{4}$.

#### Explanation:

$\sin \left(- \frac{50}{3} \pi\right) \cos \left(\frac{53}{6} \pi\right)$,

$= - \sin \left(\frac{50}{3} \pi\right) \cos \left(\frac{53}{6} \pi\right) \ldots \ldots \left[\because , \sin \left(- x\right) = - \sin x\right]$,

$= - \sin \left(\left(\frac{51 - 1}{3}\right) \pi\right) \cos \left(\left(\frac{54 - 1}{6}\right) \pi\right)$,

$= - \sin \left(\left(\frac{51}{3} - \frac{1}{3}\right) \pi\right) \cos \left(\left(\frac{54}{6} - \frac{1}{6}\right) \pi\right)$,

$= - \sin \left(17 \pi - \frac{1}{3} \pi\right) \cos \left(9 \pi - \frac{1}{3} \pi\right)$.

Now, $\left(17 \pi - \frac{1}{3} \pi\right) \text{ is in QII, where } \sin > 0$,

$\therefore \sin \left(17 \pi - \frac{1}{3} \pi\right) = \sin \left(\frac{1}{3} \pi\right) = \frac{\sqrt{3}}{2}$.

Similarly, $\cos \left(9 \pi - \frac{1}{6} \pi\right) = - \cos \left(\frac{1}{6} \pi\right) = - \frac{\sqrt{3}}{2}$.

$\Rightarrow \sin \left(- \frac{50}{3} \pi\right) \cos \left(\frac{53}{6} \pi\right) = \left(- \frac{\sqrt{3}}{2}\right) \left(- \frac{\sqrt{3}}{2}\right) = \frac{3}{4}$.

Mar 3, 2018

First, simplify the 2 factors -->
$\sin \left(\frac{- 50 \pi}{3}\right) = \sin \left(- \frac{2 \pi}{3} - \frac{48 \pi}{3}\right) = \sin \left(\frac{- 2 \pi}{3} - 16 \pi\right) =$
$= \sin \left(\frac{- 2 \pi}{3}\right) = - \sin \left(\frac{2 \pi}{3}\right) = - \frac{\sqrt{3}}{2}$
$\cos \left(\frac{53 \pi}{6}\right) = \cos \left(\frac{5 \pi}{6} + \frac{48 \pi}{6}\right) = \cos \left(\frac{5 \pi}{6} + 8 \pi\right) =$
$= \cos \left(\frac{5 \pi}{6}\right) = - \frac{\sqrt{3}}{2}$
Finally,
$\sin \left(\frac{- 50 \pi}{3}\right) . \cos \left(\frac{53 \pi}{6}\right) = \left(- \frac{\sqrt{3}}{2}\right) \left(- \frac{\sqrt{3}}{2}\right) = \frac{3}{4}$