# How do you express sin(pi/ 8 ) * cos(( ( 5 pi) / 4 )  without using products of trigonometric functions?

May 21, 2016

$- \left(\frac{\sqrt{2}}{2}\right) \sin \left(\frac{\pi}{8}\right)$

#### Explanation:

$P = \sin \left(\frac{\pi}{8}\right) . \cos \left(\frac{5 \pi}{4}\right)$
Trig table and property of supplementary arcs -->
$\cos \left(\frac{5 \pi}{4}\right) = \cos \left(\frac{\pi}{4} + \pi\right) = - \cos \left(\frac{\pi}{4}\right) = - \frac{\sqrt{2}}{2.}$
Therefor, P can be expressed as
$P = - \left(\frac{\sqrt{2}}{2}\right) \sin \left(\frac{\pi}{8}\right)$

Note. We can evaluate P by finding exact value of sin (pi/8), using the trig identity:
$\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} = 1 - 2 {\sin}^{2} \left(\frac{\pi}{8}\right)$

May 21, 2016

$\sin \left(\frac{\pi}{8}\right) \cos \left(\frac{5 \pi}{4}\right) = - \frac{1}{4} \sqrt{4 - 2 \sqrt{2}}$

#### Explanation:

$\sin \left(\frac{\pi}{8}\right) \cos \left(\frac{5 \pi}{4}\right)$

Let us first calculate

$\cos \left(\frac{5 \pi}{4}\right) = \cos \left(\frac{\pi}{4} + \pi\right) = - \cos \left(\frac{\pi}{4}\right) = - \frac{1}{\sqrt{2}}$

For $\sin \left(\frac{\pi}{8}\right)$, let us use the identity $\cos 2 \theta = 1 - 2 {\sin}^{2} \theta$

Hence, $\cos \left(\frac{\pi}{4}\right) = 1 - 2 {\sin}^{2} \left(\frac{\pi}{8}\right)$ or

$\frac{1}{\sqrt{2}} = 1 - 2 {\sin}^{2} \left(\frac{\pi}{8}\right)$ or

$2 {\sin}^{2} \left(\frac{\pi}{8}\right) = 1 - \frac{1}{\sqrt{2}} = 1 - \frac{\sqrt{2}}{2} = \frac{2 - \sqrt{2}}{2}$

or ${\sin}^{2} \left(\frac{\pi}{8}\right) = \frac{2 - \sqrt{2}}{4}$

or $\sin \left(\frac{\pi}{8}\right) = \frac{1}{2} \sqrt{2 - \sqrt{2}}$

Hence, $\sin \left(\frac{\pi}{8}\right) \cos \left(\frac{5 \pi}{4}\right) = \frac{1}{2} \sqrt{2 - \sqrt{2}} \times - \frac{\sqrt{2}}{2}$

or $\sin \left(\frac{\pi}{8}\right) \cos \left(\frac{5 \pi}{4}\right) = - \frac{1}{4} \sqrt{4 - 2 \sqrt{2}}$