Hello.
trigonometric functions in exp/ln expressions with Eulero formulas:
>
> $cos(x)=\frac{e^{ix}+e^{-ix}}{2}$
> $sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
>
> with 'i' imaginary unit, and 'x' a real value (if I'm right, 'x' must
> be real).
>
> No. For every complex number z, we have cos z = (exp(iz) + exp(-iz)) / 2
and sin(x) = (exp(iz) - exp(-iz)) / (2i).
The idea of using Euler's formulae is good because this will give polynomial
equations to solve. Then you will have to implement the method to solve
exp(iZ) = W for W the solutions of the polynomial equations verified by
exp(iZ) . If iZ and W are complex numbers there are no genral solutions
(seek for the complex logarithm to know the reason).
Indeed here if we are working with real trigonometric functions, we'll be
only interest by the complex W corresponding to points on the trigonometric
circle, ie verifying abs(W) = 1.
This complex numbers are the only ones that give a real solution Z = arg W
[2 pi].
There one pitfull : if kx = arg W [2 pi] , thenyou will have to send x =
arg W [2 pi/k] .
Christophe.
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