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,

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

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