# How do you express sin(pi/ 6 ) * cos( ( 13 pi) / 8 )  without using products of trigonometric functions?

Mar 19, 2016

=>1/2*sqrt(1/2(1-1/sqrt2)

#### Explanation:

$\sin \left(\frac{\pi}{6}\right) \cdot \cos \left(3 \frac{\pi}{8}\right)$
=>sin(pi/6)*sqrt(1/2(1+cos(cancel2*3pi/cancel8^4))
since $\cos \theta = \sqrt{\frac{1}{2} \left(1 + \cos 2 \theta\right)}$
=>1/2*sqrt(1/2(1+cos(3pi/4))
=>1/2*sqrt(1/2(1+cos(pi-pi/4))
=>1/2*sqrt(1/2(1-cos(pi/4))
=>1/2*sqrt(1/2(1-1/sqrt2)

Mar 19, 2016

$\sin \frac{\frac{\pi}{8}}{2}$

#### Explanation:

P = sin (pi/6).cos ((13pi)/8)
sin (pi/6) = 1/2
On the trig unit circle,
$\cos \left(\frac{13 \pi}{8}\right) = \cos \left(- \frac{3 \pi}{8} + 2 \pi\right) = \cos \left(\frac{- 3 \pi}{8}\right) =$
$= \cos \left(\frac{3 \pi}{8}\right) = \cos \left(- \frac{\pi}{8} + \frac{\pi}{2}\right) = \sin \left(\frac{\pi}{8}\right) .$
Therefor, the product can be expressed as:
$P = \sin \frac{\frac{\pi}{8}}{2}$
Next, we can find the product P by evaluating $\sin \left(\frac{\pi}{8}\right)$