Question

Find the product. Write the product in rectangular form, using exact values.

Answer 1

20 cis 420o = 20 cis 60o

Explanation:

Answer 2

Explanation

We must find the product of complex numbers and write it in rectangular form i.e. like a+ib where a and b are real numbers. In order to do tthis we first need to write the values of the sine and cosine of 120° and 210°. For this purpose we can use a table displaying the values of the trigonometric functions for several angles. However, these tables usually display the values for angles between 0° and 90° like the following one:

However this table is still useful. 120° is an angle in the second quadrant which means that it has the same sine as an angle of the first quadrant that meets x=180°-120° and its cosine is the same but multiplied by -1. We have x=180°-120°=60° which means that:

`\begin{gathered} \sin120^(\circ)=\sin60^(\circ)=(√(3))/(2) \\ \cos120^(\circ)=-\cos60^(\circ)=-(1)/(2) \end{gathered}`

We can do something similar for 210°. It's in the third quadrant which means that both its sine and cosine are thoes of x=210°-180°=30° but multiplied by -1. Then we get:

`\begin{gathered} \sin210^(\circ)=-\sin30^(\circ)=-(1)/(2) \\ \cos210^(\circ)=-\cos30^(\circ)=-(√(3))/(2) \end{gathered}`

Now that we found the sine and cosine of both angles we can rewrite the product:

Then we get:

`\begin{gathered} \lbrack-2√(3)-2i\rbrack\cdot\lbrack-(5)/(2)+\mathrm{i}(5√(3))/(2)\rbrack=2√(3)\cdot(5)/(2)-i2√(3)\cdot(5√(3))/(2)+i2\cdot(5)/(2)-i\cdot i\cdot2\cdot(5√(3))/(2) \\ =5√(3)-i15+i5+5√(3)=10√(3)-10i \end{gathered}`Answer

Then the answer is:

`10√(3)-i10`