How do you write the trigonometric form into a complex number in standard form 6(cos((5pi)/12)+isin((5pi)/12))?

Mar 25, 2018

The complex number in standard form is $\frac{3 \sqrt{6} - 3 \sqrt{2}}{2} + \frac{3 \sqrt{6} + 3 \sqrt{2}}{2} i$.

Explanation:

To convert from trig form to standard form, simply compute the trig functions' values and expand the multiplication.

First, let's find the $\sin$ and $\cos$ of $\frac{5 \pi}{12}$ using the respective angle-sum identities:

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

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

We can figure out that $\frac{5 \pi}{12}$ is the sum of $\frac{\pi}{6}$ and $\frac{\pi}{4}$:

$\textcolor{w h i t e}{=} \frac{\pi}{6} + \frac{\pi}{4}$

$= \frac{2 \pi}{12} + \frac{\pi}{4}$

$= \frac{2 \pi}{12} + \frac{3 \pi}{12}$

$= \frac{2 \pi + 3 \pi}{12}$

$= \frac{5 \pi}{12}$

Now we can use those angle sum formulae.

Here is a unit circle to remind us of some trig values:

Here's computing $\sin \left(\frac{5 \pi}{12}\right)$:

$\textcolor{w h i t e}{=} \sin \left(\frac{5 \pi}{12}\right)$

$= \sin \left(\frac{2 \pi}{12} + \frac{3 \pi}{12}\right)$

$= \sin \left(\frac{\pi}{6} + \frac{\pi}{4}\right)$

$= \sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}\right)$

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

$= \frac{\sqrt{2}}{4} + \frac{\sqrt{6}}{4}$

$= \frac{\sqrt{6} + \sqrt{2}}{4}$

Here's computing $\cos \left(\frac{5 \pi}{12}\right)$:

$\textcolor{w h i t e}{=} \cos \left(\frac{5 \pi}{12}\right)$

$= \cos \left(\frac{2 \pi}{12} + \frac{3 \pi}{12}\right)$

$= \cos \left(\frac{\pi}{6} + \frac{\pi}{4}\right)$

$= \cos \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{4}\right) - \sin \left(\frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}\right)$

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

$= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4}$

$= \frac{\sqrt{6} - \sqrt{2}}{4}$

Now we can plug in the values to the trig form of the complex number:

$\textcolor{w h i t e}{=} 6 \left(\cos \left(\frac{5 \pi}{12}\right) + i \sin \left(\frac{5 \pi}{12}\right)\right)$

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

$= 6 \cdot \frac{\sqrt{6} - \sqrt{2}}{4} + 6 \cdot i \cdot \frac{\sqrt{6} + \sqrt{2}}{4}$

$= 3 \cdot \frac{\sqrt{6} - \sqrt{2}}{2} + 3 \cdot i \cdot \frac{\sqrt{6} + \sqrt{2}}{2}$

$= \frac{3 \sqrt{6} - 3 \sqrt{2}}{2} + \frac{3 \sqrt{6} + 3 \sqrt{2}}{2} i$

That's it. Hope this helped!