Indeed. Since we get
(1) -> normalize sqrt(1-sin(z)^2)
z 2
tan(-) - 1
2
(1) -----------
z 2
tan(-) + 1
2
*but*
Type: Expression(Integer)
(2) -> normalize cos(z)
z 2
- tan(-) + 1
2
(2) -------------
z 2
tan(-) + 1
2
it's accountable. BTW OpenAxiom gives a correct answer.
> But also
>
> (1) -> complexNormalize acos(cos(z))
>
> (1) - z
>
>
>
>
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