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#
