# How do you evaluate  e^( (3 pi)/2 i) - e^( ( pi)/12 i) using trigonometric functions?

Jun 10, 2016

${e}^{\frac{3 \pi}{2} i} - {e}^{\frac{\pi}{12} i} = - \left(1 + \cos \left(\frac{\pi}{12}\right)\right) + i \sin \left(\frac{\pi}{12}\right)$

#### Explanation:

We can write ${e}^{i \theta} = \cos \theta + i \sin \theta$

Hence ${e}^{\frac{3 \pi}{2} i} - {e}^{\frac{\pi}{12} i}$

= $\left[\cos \left(\frac{3 \pi}{2}\right) + i \sin \left(\frac{3 \pi}{2}\right)\right] - \left[\cos \left(\frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{12}\right)\right]$

But $\cos \left(\frac{3 \pi}{2}\right) = 0$ and $\sin \left(\frac{3 \pi}{2}\right) = - 1$

Hence ${e}^{\frac{3 \pi}{2} i} - {e}^{\frac{\pi}{12} i}$

= $\left[0 - 1\right] - \left[\cos \left(\frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{12}\right)\right]$

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