Bobby Jones
New member
- Joined
- Jul 8, 2011
- Messages
- 41
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
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