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

Answer:

#4(cos(pi/3)+isin(pi/3))*7(cos((2pi)/3)+isin((2pi)/3)) = -28#

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#