# How do you write the complex number in standard form 8(cos(pi/12)+isin(pi/12))?

Jan 24, 2017

$8 \left(\cos \left(\frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{12}\right)\right) = \left(2 \sqrt{6} + 2 \sqrt{2}\right) + i \left(2 \sqrt{6} - 2 \sqrt{2}\right)$

#### Explanation:

To write this complex number in standard form, what is needed is value of $\cos \left(\frac{\pi}{12}\right) = \cos {15}^{\circ}$ and $\sin \left(\frac{\pi}{12}\right) = \sin {15}^{\circ}$

$\cos {15}^{\circ} = \cos \left({45}^{\circ} - {30}^{\circ}\right) = \cos {45}^{\circ} \cos {30}^{\circ} + \sin {45}^{\circ} \sin {30}^{\circ}$

= $\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} + 1}{2 \sqrt{2}}$

Similarly $\sin {15}^{\circ} = \sin \left({45}^{\circ} - {30}^{\circ}\right) = \sin {45}^{\circ} \cos {30}^{\circ} - \cos {45}^{\circ} \sin {30}^{\circ}$

= $\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}} \times \frac{1}{2} = \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

Hence, $8 \left(\cos \left(\frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{12}\right)\right)$

= $8 \frac{\sqrt{3} + 1}{2 \sqrt{2}} + i 8 \times \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

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

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