Question

How do you evaluate `e^( ( 13 pi)/8 i) - e^( ( 7 pi)/12 i)` using trigonometric functions?

Answer 1

`e^((13pi)/8i)-e^((7pi)/12i)=(0.6415+0.042i)`

Explanation:

As `e^(itheta)=costheta+isintheta`, we have

`e^((13pi)/8i)=cos((13pi)/8)+isin((13pi)/8)` and

`e^((7pi)/12i)=cos((7pi)/12)+isin((7pi)/12)`

Hence, `e^((13pi)/8i)-e^((7pi)/12i)`

= `(cos((13pi)/8)+isin((13pi)/8))-(cos((7pi)/12)+isin((7pi)/12))`

As `cos(13pi/8)=0.3827`, `sin(13pi/8)=-0.9239`,

`cos(7pi/12)=-0.2588` and `sin((7pi)/12)=-0.9659`

`e^((13pi)/8i)-e^((7pi)/12i)`

= `(0.3827+i(-0.9239))-((-0.2588)+i(-0.9659)`

= `(0.6415+0.042i)`