# How do you express sin(pi/ 4 ) * sin( ( 19 pi) / 12 )  without using products of trigonometric functions?

Apr 4, 2016

(sqrt2/2)cos (pi/12)

#### Explanation:

$P = \sin \left(\frac{\pi}{12}\right) . \sin \left(\frac{19 \pi}{12}\right)$
Trig table --> $\sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
Trig unit circle and property of complementary arcs give -->
$\sin \left(\frac{19 \pi}{12}\right) = \sin \left(\frac{7 \pi}{12} + \pi\right) = - \sin \left(\frac{7 \pi}{12}\right) =$
$= - \sin \left(\frac{\pi}{12} + \frac{\pi}{2}\right) = \cos \left(\frac{\pi}{12}\right)$
Finally,
$P = \left(\frac{\sqrt{2}}{2}\right) \cos \left(\frac{\pi}{12}\right)$

Apr 4, 2016

sin(pi/4)sin((19pi)/12)=-1/2 cos ((11pi)/6) +1/2 cos ((4pi)/3))=(-sqrt3-1)/4

#### Explanation:

$- 2 \sin A \sin B = \cos \left(A + B\right) - \cos \left(A - B\right)$

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

$A = \frac{\pi}{4} , B = \frac{19 \pi}{12}$

$\sin \left(\frac{\pi}{4}\right) \sin \left(\frac{19 \pi}{12}\right) = - \frac{1}{2} \left(\cos \left(\frac{\pi}{4} + \frac{19 \pi}{12}\right) - \cos \left(\frac{\pi}{4} - \frac{19 \pi}{12}\right)\right)$

$= - \frac{1}{2} \left(\cos \left(\frac{22 \pi}{12}\right) - \cos \left(- \frac{16 \pi}{12}\right)\right)$

$= - \frac{1}{2} \left(\cos \left(\frac{11 \pi}{6}\right) - \cos \left(\frac{4 \pi}{3}\right)\right)$

=-1/2 cos ((11pi)/6) +1/2 cos ((4pi)/3))

$= - \frac{1}{2} \left(\frac{\sqrt{3}}{2}\right) + \frac{1}{2} \left(- \frac{1}{2}\right)$

$= - \frac{\sqrt{3}}{4} - \frac{1}{4} = \frac{- \sqrt{3} - 1}{4}$

sin(pi/4)sin((19pi)/12)=-1/2 cos ((11pi)/6) +1/2 cos ((4pi)/3))=(-sqrt3-1)/4