Trigonometric function?

∴A+C=π-B

∴sin[(A+C)/2]=sin[(π-B)/2]

=sin(π/2 - B/2)

From the inductive formula sin(π/2-α)=cosα:

=cos(B/2)

The original equation is: acos(B/2)=bsinA.

SinAcos(B/2)=sinBsinA comes from sine theorem.

In a triangle: Sina ≠0

∴ Both sides cut off Sina at the same time: COS (b/2) = SINB.

Multiply the formula from the angle: sin2α=2sinα? Cosα has:

cos(B/2)=2sin(B/2)？ cos(B/2)

2sin(B/2)？ cos(B/2) - cos(B/2)=0

cos(B/2)？ [2sin(B/2) - 1]=0

Then cos(B/2)=0 or 2sin(B/2)- 1=0.

∴B/2=π/2 or sin(B/2)= 1/2.

∫B is the inner angle of a triangle.

∴B=π (give up) or B/2=π/6.

Then B=π/3.