How do you multiply #e^((pi)/12i ) * e^( pi i ) # in trigonometric form?

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Answer 1 · 2017-12-11T06:25:10

The answer is #=-((sqrt2+sqrt6))/4+i((sqrt2-sqrt6))/4#

We know that

#e^a* e^b=e^(a+b)#

#cos(a+b)=cosacosb-sinasinb#

#sin(a+b)=sinacosb+sinbcosa#

Therefore,

#e^(pi/12i)*e^(ipi)=e^(13/12ipi)#

According to Euler's Identity

#e^(13/12ipi)=cos(13/12pi)+isin(13/12pi)#

But,

#13/12pi=3/4pi+1/3pi#

Therefore,

#cos(13/12pi)=cos(3/4pi+1/3pi)=cos(3/4pi)cos(1/3pi)-sin(3/4pi)sin(1/3pi)#

# = (-sqrt2/2) * (1/2) -( sqrt2/2)*(sqrt3/2) #

#=-((sqrt2+sqrt6))/4#

#sin(13/12pi)=sin(3/4pi+1/3pi)=sin(3/4pi)cos(1/3pi)+cos(3/4pi)sin(1/3pi)#

# = (sqrt2/2) * (1/2) +( -sqrt2/2)*(sqrt3/2) #

#=((sqrt2-sqrt6))/4#

So,

#e^(13/12ipi)=-((sqrt2+sqrt6))/4+i((sqrt2-sqrt6))/4#