+0

Check for me plz!

0
157
3

Check for me, that I did it right hopefully?? :)

May 15, 2020

#1
+9015
+2

$$\phantom{=\qquad}(\ 3\sqrt6\ \text{cis}(\frac{\pi}{8})\ )(\ 2\sqrt5\ \text{cis}(\frac{7\pi}{6})\ )\\~\\ {=\qquad}6\sqrt{30}(\ \text{cis}(\frac{\pi}{8})\ )(\ \text{cis}(\frac{7\pi}{6})\ )\\~\\ {=\qquad}6\sqrt{30}(\ \cos(\frac{\pi}{8})+i\sin(\frac{\pi}{8})\ )(\ \cos(\frac{7\pi}{6})+i\sin(\frac{7\pi}{6})\ )\\~\\ {=\qquad}6\sqrt{30}(\ \cos(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ +\ i\cos(\frac{\pi}{8})\sin(\frac{7\pi}{6})\ +\ i\sin(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ -\ \sin(\frac{\pi}{8})\sin(\frac{7\pi}{6}) \ )\\~\\ {=\qquad}6\sqrt{30}(\ \cos(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ -\ \sin(\frac{\pi}{8})\sin(\frac{7\pi}{6}) \ +\ i\cos(\frac{\pi}{8})\sin(\frac{7\pi}{6})\ +\ i\sin(\frac{\pi}{8})\cos(\frac{7\pi}{6}) \ )\\~\\ {=\qquad}6\sqrt{30}(\ { \color{blue} \cos(\frac{\pi}{8})\cos(\frac{7\pi}{6})\ -\ \sin(\frac{\pi}{8})\sin(\frac{7\pi}{6}) } \ +\ i {\color{purple}(\ \cos(\frac{\pi}{8})\sin(\frac{7\pi}{6})\ +\ \sin(\frac{\pi}{8})\cos(\frac{7\pi}{6}) \ )} \ )\\~\\ {=\qquad}6\sqrt{30}(\ { \color{blue} \cos(\frac{\pi}{8}+\frac{7\pi}{6}) } \ +\ i\ {\color{purple} \sin(\frac{\pi}{8}+\frac{7\pi}{6}) } \ )\\~\\ {=\qquad}6\sqrt{30}(\ { \cos(\frac{31\pi}{24}) } \ +\ i\ { \sin(\frac{31\pi}{24}) } \ )\\~\\ {=\qquad}6\sqrt{30}\ \text{cis}(\ \frac{31\pi}{24}\ )$$

Check

May 15, 2020
#2
+21953
+3

You can use this formula for multiplying:  r1cis(A1) · r2cis(A2)  =  (r1·r2)cis(A1 + A2).

In your problem, you are correct in multiplying the two constants, but you should have added the angles, not multiplied them.

May 15, 2020
#3
+111456
+1

THX, hectictar  and geno.....I learned something new today    !!!!!

CPhill  May 15, 2020