how do you apply taylor expansion for small numbers?

Question

f= frequency
f'= f(1 + (v cos theta/c))
f'= wavelength(wl')/c(speed of light)
f=wl/c
wl'= wl(1- (v cos theta/c))

how does the + change to - ?

Answer 1

This question is extremely confused/confusing! For one thing, the speed of light (exceedable only by warp drive) is wavelength TIMES frequency, not wavelength OVER frequency, as implied by two of your lines. Any Starfleet Academy student should know that.

Ignoring that, I THINK that the nub of your question is this: presumably v/c is small in the original question that you must have been asked. Call that small value x. Then, for small x, a BINOMIAL SERIES expansion (NOT a Taylor expansion) tells you that
(1 + x)^(-1) is approximately 1 - x, ignoring terms of order x^2 and higher.

Live long and prosper.

Answer 2

Your equation f=wl/c is wrong. It should be f*wl = c. In that case
f'*wl' = f(1 + (v cos theta/c)) * wl(1- (v cos theta/c)) =
f * wl * (1 - (v cos theta/c)^2) = f * wl = c
as long as v/c is much smaller than 1