Complex Hyperbolics

[Bobby Jones]

Please confirm if I've got the correct answer:

If:

Cosx = (e^jx +e^-jx)/2

Sinx = (e^jx - e^-jx)/2j

Therefore,

Question 1) 2SinxCosx = (e^2jx - e^-2jx)/2j

and

Question 2) cos^2 x - sin^2 x = (e^2jx + e^-2jx)/2


Cheers guys, I really appretiate your professional help

[mmm4444bot]

2 sinx cosx = (e^[2jx] - e^[-2jx])/(2j)


cos^2 x - sin^2 x = (e^[2jx] + e^[-2jx])/2


These both look good, to me. 8-)

[Subhotosh Khan]

Please confirm if I've got the correct answer:

If:

Cosx = (e^jx +e^-jx)/2

Sinx = (e^jx - e^-jx)/(2j)

Therefore,

Question 1) 2SinxCosx = sin(2x) = [e^(2xj) - e^-(2xj]/(2j) ...............Correct

and

Question 2) cos^2 x - sin^2 x = cos(2x) = [e^(2xj) + e^-(2xj)]/2...........correct


Cheers guys, I really appretiate your professional help

Watch those grouping symbols. Those change the "meaning" of your statement and prevent you from getting 100% marks in exam.