Is there a table to resolve the ambiguities of sin(A/2) and cos(A/2) given the value g=sin(A)

Question

Is there a table that can be employed to solve sin(A/2) and cos(A/2) given the value of sin(A)?

Given sin(A)=g, in what looks like a magic mirror of ambiguity

sin(A/2)=(1/2)*[(±)√(1+g)+(±)√(1+g)] where of course -1≤g≤1 so 1-g could be larger than 1+g

and

cos(A/2)=(1/2)*[(±)√(1+g)+(±)√(1+g)}

The ambiguity for sin(A/2) is foursome given the value of sin(A), the same for cos(A/2)

e.g. sin(A)=(√3)/2, angle A has two sets of solutions, of form (π/3)+2πk and of form (2π/3)+2πk [k=0,1,-1,2,-2,...]

consider the solutions A₁=π/3,A₂= 7π/3, A₃=2π/3, and A₄=8π/3 where sin(A)=√3)/2

sin(A₁/2)=1/2, sin(A₂/2)= -(1/2), sin(A₃/2)=√3)/2, sin(A₄/2)= -√3)/2 (wow, sin(A) can = sin(A/2) and A not zero)

so the ambiguity is foursome for sin(A/2) given sin(A)

still having sin(A)=(√3)/2, angle A still has two sets of solutions, of form (π/3)+2πk and of form (2π/3)+2πk

and still considering the solutions A₁=π/3,A₂= 7π/3, A₃=2π/3, and A₄=8π/3

cos(A₁/2)=√3)/2, cos(A₂/2)= -√3)/2, cos(A₃/2)=1)/2, cos(A₄/2)= -1)/2

so the ambiguity is also foursome for cos(A/2) given sin(A)

Answer 1

Algebraically the ambiguities arise from the derivation of sin(A/2) and cos(A/2)

it is that (sin(A/2)+cos(A/2))²=1+sin(A), sin(A/2)+cos(A/2)=±√(1+sin(A))

and (sin(A/2)-cos(A/2))²=1-sin(A), sin(A/2)-cos(A/2)=±√(1-sin(A))

adding, 2*sin(A/2)=±√(1+sin(A))+±√(1-sin(A))

it is also that (cos(A/2)+sin(A/2))²=1+sin(A) and (cos(A/2)-sin(A/2))²=1-sin(A)

adding 2*cos(A/2)=±√(1+sin(A))+±√(1-sin(A))

call ±√(1+sin(A))=R and call ±√(1-sin(A))=T, then the following table based on the half-angle value is valid. Use half of full angle if needed to make half angle. Once A/2 is determined, all 2πk additions/subtractions to A/2 are also valid for values of sin(A/2) and cos(A/2) , WHICH IMPLY 4πk additions/subtractions to A.

1.) 0 ≤ A/2 < π/4 for sin(A/2) use R=+, T= -, for cos (A/2) use R=+, T=+

2.) π/4 ≤ A/2 <3π/4 for sin(A/2) use R=+, T= +, for cos (A/2) use R=+, T= -

3.) 3π/4 ≤ A/2 <π for sin(A/2) use R= -, T=+, for cos (A/2) use R= -, T= -

4.) -π/4 ≤ A/2 <0 for sin(A/2) use R=+, T= -, for cos (A/2) use R=+, T=+

5.) -3π/4 ≤ A/2<-π/4 for sin(A/2) use R= -, T= -, for cos (A/2) use R=-, T=+

6.) -π ≤ A/2<-3π/4 for sin(A/2) use R= -, T=+, for cos (A/2) use R= -, T= -

very generally the tables can be built using the fact sin(A/2)+cos(/A/2)=+√2*sin(A/2+π/4)

which provides a range of A/2 satisfying either + or - for ±√(1+sin(A))

sin(A/2)-cos(/A/2)=+√2*sin(A/2-π/4) which provides a range of A/2 satisfying either + or - for ±√(1-sin(A))

note given cos(A), the ambiguity for sin(A/2) is only sin(A/2)=±√((1-cos(A))/2)

and for cos(A/2) is only cos(A/2)=±√((1+cos(A))/2)