sinx=1/4, x in quadrant II. Find the value of sinx/2

sinx=1/4, x in 2nd quadrant ⇒ π/2 < x < π and cosx is negative in 2nd quadrant
Using 1 - cosx = 2sin²x/2 ⇒ sinx/2 = ±√(1 - cosx)/2
We get, sinx/2 = ±√(1 - (-√15/4))/2 = ±√(4 + √15)/8
As π/2 < x < π ⇒ π/4 < x/2 < π/2 and sin is positive in 1st quadrant
∴ sinx/2 = √(8 + 2√15)/4
Using 1 + cosx = 2cos²x/2 ⇒ cosx/2 = ±√(1 + cosx)/2
We get cosx/2 = ±√(1 + (-√15/4))/2 = ±√(4 - √15)/8 = ±√(8 - 2√15)/4
As π/2 < x < π ⇒ π/4 < x/2 < π/2 and cos is positive in first quadrant
∴ cosx/2 = √(8 - 2√15)/4
Using cosx = (1 - tan²x/2)/(1 + tan²x/2) ⇒ tanx/2 = ±√(1 - cosx)/(1 + cosx)
We get tanx/2 = ±√(1 - (-√15/4))/(1 + (-√15/4)) = ±√(4 + √15)/(4 - √15)
As π/2 < x < π ⇒ π/4 < x/2 < π/2 and tan is positive in first quadrant
∴ tanx/2 = √(4 + √15)/(4 - √15)