How do you evaluate  e^( ( 7 pi)/4 i) - e^( ( 13 pi)/12 i) using trigonometric functions?

1 Answer
Dec 3, 2016

$\setminus \frac{\sqrt{3} + 3}{2 \sqrt{2}} - i \setminus \frac{\sqrt{3} + 3}{2 \sqrt{2}}$

Explanation:

Since ${e}^{i \theta} = \cos \left(\theta\right) + i \sin \left(\theta\right)$, your expression becomes

$\cos \left(\frac{7 \pi}{4}\right) + i \sin \left(\frac{7 \pi}{4}\right) - \left(\cos \left(\frac{13 \pi}{12}\right) + i \sin \left(\frac{13 \pi}{12}\right)\right)$

Now let's work with the angles: since $\frac{7 \pi}{4} = 2 \pi - \frac{\pi}{4} = - \frac{\pi}{4}$, we have that

• $\cos \left(\frac{7 \pi}{4}\right) = \cos \left(- \frac{\pi}{4}\right) = \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$
• $\sin \left(\frac{7 \pi}{4}\right) = \sin \left(- \frac{\pi}{4}\right) = - \sin \left(\frac{\pi}{4}\right) = - \frac{\sqrt{2}}{2}$

On the other hand, we have that $\frac{13 \pi}{12} = \pi + \frac{\pi}{12}$, and so the transformations are

• $\cos \left(\frac{13 \pi}{12}\right) = \cos \left(\pi + \frac{\pi}{12}\right) = - \cos \left(\frac{\pi}{12}\right) = - \setminus \frac{\sqrt{3} + 1}{2 \sqrt{2}}$
• $\sin \left(\frac{13 \pi}{12}\right) = \sin \left(\pi + \frac{\pi}{12}\right) = - \sin \left(\frac{\pi}{12}\right) = - \setminus \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

So, your expression becomes

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

Which simplifies into

$\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} + \setminus \frac{\sqrt{3} + 1}{2 \sqrt{2}} - i \setminus \frac{\sqrt{3} - 1}{2 \sqrt{2}}$

You can write it with just one denominator:

$\setminus \frac{\sqrt{3} + 3}{2 \sqrt{2}} - i \setminus \frac{\sqrt{3} + 3}{2 \sqrt{2}}$