# How do you simplify 2(cos((3pi)/4)+isin((3pi)/4))*sqrt2(cos(pi/2)+isin(pi/2)) and express the result in rectangular form?

Jan 11, 2017

$2 \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right) \cdot \sqrt{2} \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right) = - 2 - 2 i$

#### Explanation:

$2 \left(\cos \left(\frac{3 \pi}{4}\right) + i \sin \left(\frac{3 \pi}{4}\right)\right) \cdot \sqrt{2} \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)$

= $2 \sqrt{2} \left\{\cos \left(\frac{3 \pi}{4}\right) \cos \left(\frac{\pi}{2}\right) + i \cos \left(\frac{3 \pi}{4}\right) \sin \left(\frac{\pi}{2}\right) + i \cos \left(\frac{\pi}{2}\right) \sin \left(\frac{3 \pi}{4}\right) + {i}^{2} \sin \left(\frac{3 \pi}{4}\right) \sin \left(\frac{\pi}{2}\right)\right\}$

= $2 \sqrt{2} \left\{\cos \left(\frac{3 \pi}{4}\right) \cos \left(\frac{\pi}{2}\right) - \sin \left(\frac{3 \pi}{4}\right) \sin \left(\frac{\pi}{2}\right) + i \left(\cos \left(\frac{3 \pi}{4}\right) \sin \left(\frac{\pi}{2}\right) + i \cos \left(\frac{\pi}{2}\right) \sin \left(\frac{3 \pi}{4}\right)\right)\right\}$

= $2 \sqrt{2} \left\{\cos \left(\frac{3 \pi}{4} + \frac{\pi}{2}\right) + i \sin \left(\frac{3 \pi}{4} + \frac{\pi}{2}\right)\right\}$

= $2 \sqrt{2} \left(\cos \left(\frac{5 \pi}{4}\right) + i \sin \left(\frac{5 \pi}{4}\right)\right)$

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

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

= $- 2 - 2 i$