How do you simplify #5(cospi+isinpi)*2(cos((3pi)/4)+isin((3pi)/4))# and express the result in rectangular form?

Question

1941 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-17T11:30:01

# 5sqrt2(1-i).#

#{5(cospi+isinpi)}{2(cos(3pi/4)+isin(3pi/4))}#

#={5(-1+0i)}{2(-cos(pi/4)+isin(pi/4))}#

#=-5{2(-1/sqrt2+1/sqrt2*i)}#

#=5sqrt2(1-i),#

Otherwise, knowing that,

#r(cosalpha+isinalpha)*R(cosbeta+isinbeta)=(rR)(cos(alpha+beta)+isin(alpha+beta)),#

we have, the Reqd. Value

#=(5xx2){cos(pi+3pi/4)+isin(pi+3pi/4)}#

#=10{cos(2pi-pi/4)+isin(2pi-pi/4)}#

#=10{cospi/4-isinpi/4}#

#=10(1/sqrt2-i/sqrt2)#

#=5sqrt2(1-i),# as derived earlier!

Enjoy Maths.!