# How do you multiply e^((pi)/12i ) * e^( pi i )  in trigonometric form?

##### 1 Answer
Dec 11, 2017

The answer is $= - \frac{\left(\sqrt{2} + \sqrt{6}\right)}{4} + i \frac{\left(\sqrt{2} - \sqrt{6}\right)}{4}$

#### Explanation:

We know that

${e}^{a} \cdot {e}^{b} = {e}^{a + b}$

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

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

Therefore,

${e}^{\frac{\pi}{12} i} \cdot {e}^{i \pi} = {e}^{\frac{13}{12} i \pi}$

According to Euler's Identity

${e}^{\frac{13}{12} i \pi} = \cos \left(\frac{13}{12} \pi\right) + i \sin \left(\frac{13}{12} \pi\right)$

But,

$\frac{13}{12} \pi = \frac{3}{4} \pi + \frac{1}{3} \pi$

Therefore,

$\cos \left(\frac{13}{12} \pi\right) = \cos \left(\frac{3}{4} \pi + \frac{1}{3} \pi\right) = \cos \left(\frac{3}{4} \pi\right) \cos \left(\frac{1}{3} \pi\right) - \sin \left(\frac{3}{4} \pi\right) \sin \left(\frac{1}{3} \pi\right)$

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

$= - \frac{\left(\sqrt{2} + \sqrt{6}\right)}{4}$

$\sin \left(\frac{13}{12} \pi\right) = \sin \left(\frac{3}{4} \pi + \frac{1}{3} \pi\right) = \sin \left(\frac{3}{4} \pi\right) \cos \left(\frac{1}{3} \pi\right) + \cos \left(\frac{3}{4} \pi\right) \sin \left(\frac{1}{3} \pi\right)$

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

$= \frac{\left(\sqrt{2} - \sqrt{6}\right)}{4}$

So,

${e}^{\frac{13}{12} i \pi} = - \frac{\left(\sqrt{2} + \sqrt{6}\right)}{4} + i \frac{\left(\sqrt{2} - \sqrt{6}\right)}{4}$