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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?

Answer 1

Sadie Ackerman

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

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#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.