prove that tan((u+iv))/2=((sinu+isinhv)/cosu+ icoshv)
Hi there, I think there may be some typos in your question, specifically in the trig expression, but I’m not sure. I wasn’t able to get a proof of the (hyperbolic) trig identify you wrote down, but I do have a proof of something similar, which hopefully will be able to help you solve your own problem.
N.B. I changerd the sign of +i.sinhv to –i.sinhv, also i.coshv to coshv.
A solution to:
Prove that tan((u + iv)/2) = (sinu – i.sinhv) / (cosu + coshv).
Some useful identities
cos(x) = cosh(ix) and cosh(x) = cos(ix)
sin(x) = i.sinh(ix) and sinh(x) = i.sin(ix)
also,
sin(A) + sin(B) = 2,sin((A+B)/2).cos((A-B)/2)
cos(A) + cos(B) = 2.cos((A+B)/2).cos((A-B)/2)
using these identities in the rhs of the original expression,
(sinu – isinhv) / (cosu + coshv) = (sin(u) – i^2.sin(iv)) / (cos(u) + cos(iv))
(sinu – isinhv) / (cosu + coshv) = (sin(u) + sin(iv)) / (cos(u) + cos(iv))
(sinu – isinhv) / (cosu + coshv) = 2(sin((u + iv)/2).cos((u – iv)/2)) / 2(cos((u + iv)/2).cos((u – iv)/2))
Cancelling out cos((u – iv)/2) top and bottom,
(sinu – isinhv) / (cosu + coshv) = 2(sin((u + iv)/2)) / 2cos((u + iv)/2))
(sinu – isinhv) / (cosu + coshv) = tan((u + iv)/2)
or,
tan((u + iv)/2) = (sinu – isinhv) / (cosu + coshv).