Question

How do you evaluate `e^( ( 7 pi)/6 i) - e^( ( 19 pi)/8 i)` using trigonometric functions?

Answer 1

`2sin((29pi)/24)e^(((25pi)/24)i)`

Explanation:

we know euler's theorem states that `e^(itheta)=costheta+isintheta`
so,`e^(7pi/6i)-e^(19pi/8 i)`
=`cos(7pi/6)+isin(7pi/6)-(cos(19pi/8)+isin(19pi/8))`
=`cos(7pi/6)-cos(19pi/8)+i(sin(7pi/6)-sin(19pi/8))`
now `cosc-cosd=2sin((c+d)/2) sin((d-c)/2)`
`sinc-sind=2cos((c+d)/2) sin((c-d)/2)`
so we get `cos((7pi)/6)-cos((19pi)/8)`
=`2sin((85pi)/24)sin((29pi)/24)`
again similarly we get `i(sin((7pi)/6)-sin((19pi)/8))`
=`2icos((85pi)/24)sin((29pi)/24)`
so,`e^((7pi)/6)-e^((19pi)/8)`=`2sin((29pi)/24)(sin((85pi)/24)-icos((85pi)/24))`
=`2sin((29pi)/24)[sin(2pi+((37pi)/24))-icos(2pi+((37pi)/24))]`
=`2sin((29pi)/24)[sin((37pi)/24)-icos((37pi)/24)]`
=`2sin((29pi)/24)[cos((25pi)/24)+isin((25pi)/24)]`
=`2sin((29pi)/24)e^(((25pi)/24)i)`.