Chapter 9.7, Problem 7CFU

To determine

To find each product or quotient and express the result in rectangular form

Answer to Problem 7CFU

The product of equation $\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right).6\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right)$ is $3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)$

The rectangular form of the product $3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)$ is $-\frac{3\sqrt{3}}{2}-\frac{3}{2}i$

Explanation of Solution

Given information

The given form of equation is $\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right).6\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right)$

Formula used:for product

  $\begin{array}{l}{r}_1\left(\cos {\theta }_1+i\sin {\theta }_1\right).r_2\left(\cos {\theta }_2+i\sin {\theta }_2\right)=r_1{r}_2\left(\cos \left({\theta }_1+{\theta }_2\right)+i\sin \left({\theta }_1+{\theta }_2\right)\right),\\ \left[\text{for modulus}\right]\\ r=r_1{r}_{2,}\\ \left[\text{for amplitude}\right]\\ \theta ={\theta }_1+{\theta }_2\end{array}$

Calculation:

Firstly,finding the modulus and amplitude of the product.

  $\begin{array}{l}r=r_{1r}{r}_2=\frac{1}{2}\times 6=3\\ \theta ={\theta }_1+{\theta }_2\\ \theta =\frac{\pi }{3}+\frac{5\pi }{6}\\ \theta =\frac{7\pi }{6}\end{array}$

The product of $\frac{1}{2}\left(\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}\right).6\left(\cos \frac{5\pi }{6}+i\sin \frac{5\pi }{6}\right)$ is $3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)$

Now, find the rectangular form of the product is $\begin{array}{l}3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)=3\left(-\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)\left[\cos \frac{7\pi }{6}=-\frac{\sqrt{3}}{2},\sin \frac{7\pi }{6}=-\frac{1}{2}\right]\\ =-\frac{3}{2}\left(\sqrt{3}+1i\right)=-\frac{3\sqrt{3}}{2}-\frac{3}{2}i\end{array}$

The rectangular form of the product $3\left(\cos \frac{7\pi }{6}+i\sin \frac{7\pi }{6}\right)$ is $-\frac{3\sqrt{3}}{2}-\frac{3}{2}i$