Multily #3(cos(pi/3)+isin(pi/3))# and #4(cos(pi/6)+isin(pi/6))#?

1 Answer
Jan 30, 2018

#3(cos(pi/3)+isin(pi/3))4(cos(pi/6)+isin(pi/6))=12i#

Explanation:

When we multiply two numbers in polar form, say

#r_1(cosalpha+isinalpha)# and #r_2(cosbeta+isinbeta)#, we get

#r_1r_2(cosalphacosbeta+isinalphaacosbeta+icosalphasinbeta+i^2sinalphasinbeta)#

= #r_1r_2(cosalphacosbeta+isinalphaacosbeta+icosalphasinbeta-sinalphasinbeta)#

= #r_1r_2(cosalphacosbeta-sinalphasinbeta+i(sinalphacosbeta+cosalphasinbeta))#

= #r_1r_2(cos(alpha+beta)+isin(alpha+beta))#

Hence #3(cos(pi/3)+isin(pi/3))4(cos(pi/6)+isin(pi/6))#

= #12(cos(pi/3+pi/6)+isin(pi/3+pi/6))#

= #12(cos(pi/2)+isin(pi/2))#

= #12(0+ixx1)#

= #12i#