Question

How do you evaluate `e^( ( 7 pi)/4 i) - e^( ( 4 pi)/3 i)` using trigonometric functions?

Answer 1

`e^((7pi)/4i)-e^((4pi)/3i)=(1+sqrt(2))/2+i((-sqrt(2)+sqrt(3))/2)`

Explanation:

`e^(ix)=cosx+isinx`

Here, we have `e^((7pi)/4i)-e^((4pi)/3i)=cos((7pi)/4)+isin((7pi)/4)-(cos((4pi)/3)+isin((4pi)/3))=cos((7pi)/4)+isin((7pi)/4)-cos((4pi)/3)-isin((4pi)/3)`

`cos((7pi)/4)=sqrt(2)/2`
`sin((7pi)/4)=-sqrt(2)/2`
`cos((4pi)/3)=-1/2`
`sin((4pi)/3)=-sqrt(3)/2`

Therefore, `e^((7pi)/4i)-e^((4pi)/3i)=sqrt(2)/2-isqrt(2)/2+1/2+isqrt(3)/2=(1+sqrt(2))/2+i((-sqrt(2)+sqrt(3))/2)`