Question

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

Answer 1

`{(sqrt(2+sqrt2)-sqrt3)-i(sqrt(2-sqrt2)+1)}/2.`

Explanation:

We have, `e^(itheta)=costheta+isintheta.`

`:.e^((7pi/6)i)-e^((7pi/8)i)=cos7pi/6+isin7pi/6-cos7pi/8-isin7pi/8=cos(pi+pi/6)+isin(pi+pi/6)-cos(pi-pi/8)-isin(pi-pi/8)=-cos(pi/6)-isin(pi/6)+cos(pi/8)-isin(pi/8)=-sqrt3/2-i/2+(sqrt(2+sqrt2))/2-i(sqrt(2-sqrt2))/2=(sqrt(2+sqrt2))/2-sqrt3/2-i(sqrt(2-sqrt2))/2-i/2={(sqrt(2+sqrt2)-sqrt3)-i(sqrt(2-sqrt2)+1)}/2.`