question on lorentz transformation equations

[demonelite123]

i am reading Lillian R. Lieber's book on the einstein theory of relativity and i am a bit confused on page 65. she wants to take the equations:
x=x'cosθ - y'sinθ
y=x'sinθ + y'cosθ

and compare them to:
x'=β(x-vt)
t'=β(t-vx/c2)

she takes c as one so:
x'=β(x-vt)
t'=β(t-vx)

she solves for x and t and gets:
x=β(x'+vt')
t=β(t'+vx')

then she replaces t with iτ and t' with iτ' and she gets:
x=β(x'+vt')
iτ = iβτ' + βvx'

x=β(x'+vt')
τ=βτ' + iβvx'

next is the part i am confused about. she sets β = cosθ and -iβv = sinθ. this nicely turns the Lorentz equations into:
x=x'cosθ - τ'sinθ
τ=x'sinθ + τ'cosθ

what i don't understand is how did she choose -iβv = sinθ? it works out all nicely in the end but how did she know that sinθ had to equal -iβv? was it arbitrary as a result of trial and error or did she use β = cosθ in order to figure out that -iβv = sinθ?

[starthaus]

next is the part i am confused about. she sets β = cosθ and -iβv = sinθ. this nicely turns the Lorentz equations into:
x=x'cosθ - τ'sinθ
τ=x'sinθ + τ'cosθ

what i don't understand is how did she choose -iβv = sinθ? it works out all nicely in the end but how did she know that sinθ had to equal -iβv? was it arbitrary as a result of trial and error or did she use β = cosθ in order to figure out that -iβv = sinθ?

This can't be, are you sure that the author is not setting:

\beta=cosh(\theta) and \beta v= sinh (\theta). Note the use of hyperbolic trigonometry as in hete (http://www.physicsforums.com/blog.php?b=1911)

[jcsd]

This can't be, are you sure that the author is not setting:

\beta=cosh(\theta) and \beta v= sinh (\theta). Note the use of hyperbolic trigonometry as in hete (http://www.physicsforums.com/blog.php?b=1911)

But you can use the substitutions sinh x = -isin ix and cosh x = cos ix.

Anyway the answer to demonelite, thoguh I haven't been looked at myself, is that your question should be answered by considering what θ is.