Question

How do you evaluate `e^( ( 11 pi)/6 i) - e^( ( 9 pi)/8 i)` using trigonometric functions?

Answer 1

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

Explanation:

`e^(ix)=cos x + i sinx`.
`cos(2pi-x)=cos x, sin (2pi-x)=-sin x, cos(pi+x)=-cos x and sin(pi+x)=-sin x`.

So, `e^(11pi/6i)=cos (11pi/6)+i sin(11pi/6)=cos(2pi-pi/6)+i sin(2pi-pi/6)=cos (pi/6)-i sin (pi/6)=sqrt3/2-i(1/2)`.
`e^(9pi/8)=cos (9pi/8)+i sin(9pi/8)=cos(pi+pi/8)+i sin(pi+pi/8)=-cos (pi/8)-i sin( pi/8)`.
`cos A = sqrt((1+cos 2A)/2), sin A=sqrt((1-cos 2A)/2)`

So, `cos(pi/8)=sqrt((1+cos (pi/4))/2) = sqrt((sqrt2+1)/(2sqrt2))`.
`sin(pi/8)=sqrt((1-cos (pi/4))/2) = sqrt((sqrt2-1)/(2sqrt2))`.

Npw, `e^(9pi/8)=sqrt((sqrt2+1)/(2sqrt2))-i sqrt((sqrt2-1)/(2sqrt2))`.

So, the given expression =`(sqrt3/2+sqrt((sqrt2+1)/(2sqrt2)))-i(1/2-sqrt((sqrt2-1)/(2sqrt2)))`