Better way to solve \cos(\frac{\gamma '}{2})=g_0 and P

Michael Maggard

Michael Maggard

Answered question


Better way to solve cos(γ2)=g0 and eiβsin(γ2)=g1 for γ,β?
Where g0 is real, and g1 is some complex number. I tried
γ=2cos1(g0),   cos(β)sin(γ2)=Re(g1) (or) sin(β)sin(γ2)=Im(g1)
Then for each γ, there should be two corresponding β s. However, it turned out that my solution doesn't quite work

Answer & Explanation

Lynne Trussell

Lynne Trussell

Beginner2022-01-03Added 32 answers

Matthew Rodriguez

Matthew Rodriguez

Beginner2022-01-04Added 32 answers

Solve one for sinγ2 , solve the other for cosγ2, and then use sin2x+cos2x=1


Expert2022-01-08Added 669 answers

Clearly γ±2arccosg0(mod4π), which is consistent with sinγ2=±|g1| if g02+|g1|2=1, in which case eiβ=sgng1 (except for the case g0=±1,g1=0, where β is arbitrary). This constrains β to either the even or odd multiples of π

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