How do you evaluate $e^( ( 7 pi)/4 i) - e^( ( 5 pi)/12 i)$ using trigonometric functions?

Question

1405 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-22T17:22:12

The answer is $=1/4(3sqrt2-sqrt6)+i(1/4(3sqrt2+sqrt6))$

We apply Euler's formula

$e^(i x)=cosx +i sinx$

So,

$e^(7/4pi i)-e^(5/12pi i)=$

$cos(7/4pi)+isin(7/4pi)-cos(5/12pi)-isin(5/12pi)$

We calculate separately

$cos(7/4pi)=cos(pi+3/4pi)=cos(pi)cos(3/4pi)-sin(pi)sin(3/4pi)$

$=-1*-sqrt2/2-0=sqrt2/2$

$sin(7/4pi)=sin(pi+3/4pi)=sin(pi)cos(3/4pi)+sin(3/4pi)cos(pi)$

$=0+sqrt2/2*-1=-sqrt2/2$

$cos(5/12pi)=cos(3/12pi+2/12pi)=cos(1/4pi+1/6pi)=cos(1/4pi)cos(1/6pi)-sin(1/4pi)sin(1/6pi)$

$=sqrt2/2*sqrt3/2-sqrt2/2*1/2=1/4(sqrt6-sqrt2)$

$sin(5/12pi)=sin(3/12pi+2/12pi)=sin(1/4pi)cos(1/6pi)+sin(1/6pi)cos(1/4pi)$

$=sqrt2/2*sqrt3/2+sqrt2/2*1/2=1/4(sqrt6+sqrt2)$

$e^(7/4pi i)-e^(5/12pi i)=(sqrt2/2-1/4(sqrt6-sqrt2))-i(sqrt2/2+1/4(sqrt6+sqrt2))$

$=1/4(3sqrt2-sqrt6)+i(1/4(3sqrt2+sqrt6))$

How do you evaluate e^( ( 7 pi)/4 i) - e^( ( 5 pi)/12 i) e^( ( 7 pi)/4 i) - e^( ( 5 pi)/12 i) using trigonometric functions?