How do you evaluate $e^( ( 5 pi)/4 i) - e^( ( 17 pi)/12 i)$ using trigonometric functions?

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Answer 1 · 2017-07-06T15:05:00

The answer is $=(sqrt6-3sqrt2)/4-i(sqrt6-sqrt2)/4$

We need

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

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

$sin(pi/4)=cos(pi/4)=sqrt2/2$

$cos(pi/6)=sqrt3/2$

$sin(pi/2)=1/2$

We apply Euler's Formula

$e^(ix)=cosx+isinx$

$e^(5/4pii)=cos(5/4pi)+isin(5/4pi)$

$=-cos(1/4pi)-isin(1/4pi)$

$=-sqrt2/2-isqrt2/2$

$e^(17/12pii)=cos(17/12pi)+isin(17/12pi)$

$=cos(15/12pi+2/12pi)+isin(15/12pi+2/12pi)$

$=cos(5/4pi)cos(1/6pi)-sin(5/4pi)sin(1/6pi)+i(sin(5/4pi)cos(1/6pi)+cos(5/4pi)sin(1/6pi))$

$=-sqrt2/2*sqrt3/2+sqrt2/2*1/2+i(-sqrt2/2*sqrt3/2-sqrt2/2*1/2)$

$=(sqrt2-sqrt6)/4+i(-sqrt6-sqrt2/4)$

Therefore,

$e^(5/4pii)-e^(17/12pii)=-sqrt2/2-isqrt2/2-((sqrt2-sqrt6)/4+i(-sqrt6-sqrt2)/4)$

$=(sqrt6-3sqrt2)/4-i(sqrt6-sqrt2)/4$

How do you evaluate e^( ( 5 pi)/4 i) - e^( ( 17 pi)/12 i) e^( ( 5 pi)/4 i) - e^( ( 17 pi)/12 i) using trigonometric functions?