you have to prove sin(theta)=(2z)/(1+z^2)) where z=tan(theta/2)?

Question

Please show step by step how solve this.

Answer 1

Prove the identity.

sinθ = (2z)/(1 + z²)

Let's work with the right hand side.

Right Hand Side = (2z)/(1 + z²) = 2tan(θ/2) / [1 + tan²(θ/2)]

= 2tan(θ/2) / [sec²(θ/2)] = 2[sin(θ/2) / cos(θ/2)] * [cos²(θ/2)]

= 2[sin(θ/2)*cos(θ/2)] = sinθ = Left Hand Side

Answer 2

z = tan(theta/2)
we know that tan(theta) =
2 tan(theta/2) / (1-tan^2(theta/2))
tan (theta) = 2z / (1-z^2)
2z is the perpendicular and (1-z^2) is the base
so HYpotenuse is 2z^2 + (1-z^2) ^2 = (1+z^2) ^2
sin (theta) = perpendicular/hypotenuse = 2z/(1+z^2) ^2