Reese Munoz

2022-02-01

How do you simplify $2\left(\mathrm{cos}\left(\frac{3\pi }{4}\right)+i\mathrm{sin}\left(\frac{3\pi }{4}\right)\right)\cdot \sqrt{2}\left(\mathrm{cos}\left(\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{\pi }{2}\right)\right)$ and express the result in rectangular form?

### Answer & Explanation

Micheal Hensley

$2\left(\mathrm{cos}\left(\frac{3\pi }{4}\right)+i\mathrm{sin}\left(\frac{3\pi }{4}\right)\right)\cdot \sqrt{2}\left(\mathrm{cos}\left(\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{\pi }{2}\right)\right)$ $=2\sqrt{2}\left\{\mathrm{cos}\left(\frac{3\pi }{4}\right)\mathrm{cos}\left(\frac{\pi }{2}\right)+i\mathrm{cos}\left(\frac{3\pi }{4}\right)\mathrm{sin}\left(\frac{\pi }{2}\right)+i\mathrm{cos}\left(\frac{\pi }{2}\right)\mathrm{sin}\left(\frac{3\pi }{4}\right)+{i}^{2}\mathrm{sin}\left(\frac{3\pi }{4}\right)\mathrm{sin}\left(\frac{3\pi }{4}\right)\mathrm{sin}\left(\frac{\pi }{2}\right)\right\}$ $=2\sqrt{2}\left\{\mathrm{cos}\left(\frac{3\pi }{4}\right)\mathrm{cos}\left(\frac{\pi }{2}\right)-\mathrm{sin}\left(\frac{3\pi }{4}\right)\mathrm{sin}\left(\frac{\pi }{2}\right)+i\left(\mathrm{cos}\left(\frac{3\pi }{4}\right)\mathrm{sin}\left(\frac{\pi }{2}\right)+i\mathrm{cos}\left(\frac{\pi }{2}\right)\mathrm{sin}\left(\frac{3\pi }{4}\right)\right)\right\}$ $=2\sqrt{2}\left\{\mathrm{cos}\left(\frac{3\pi }{4}+\frac{\pi }{2}\right)+i\mathrm{sin}\left(\frac{3\pi }{4}+\frac{\pi }{2}\right)\right\}$ $=2\sqrt{2}\left(\mathrm{cos}\left(\frac{5\pi }{4}\right)+i\mathrm{sin}\left(\frac{5\pi }{4}\right)\right)$ $2\sqrt{2}\left(-\mathrm{cos}\left(\frac{\pi }{4}\right)-i\mathrm{sin}\left(\frac{\pi }{4}\right)\right)$ $=2\sqrt{2}\left(-\frac{1}{\sqrt{2}}-i\frac{1}{\sqrt{2}}\right)$ $=-2-2i$