How do you simplify 4(cos(pi/3)+isin(pi/3))*7(cos((2pi)/3)+isin((2pi)/3)) and express the result in rectangular form?
1 Answer
Dec 31, 2016
Explanation:
Note that:
e^(itheta) = cos theta + i sin theta
So we have:
(cos alpha + i sin alpha)(cos beta + i sin beta) = e^(ialpha)*e^(ibeta)
color(white)((cos alpha + i sin alpha)(cos beta + i sin beta)) = e^(i(alpha+beta))
color(white)((cos alpha + i sin alpha)(cos beta + i sin beta)) = cos(alpha+beta) + i sin(alpha+beta)
So in our example:
4(cos(pi/3)+isin(pi/3))*7(cos((2pi)/3)+isin((2pi)/3))
= 28(cos(pi/3+(2pi)/3)+isin(pi/3+(2pi)/3))
= 28(cos(pi)+isin(pi))
= 28(-1+i*0)
= -28
