Please explain how L(1-cosφ) (line 2) changes to L[1-(1-(R/L)^2sin^2 θ)½] (line 3)

1) X = R + L – Rcos θ – Lsin φ
2) X = R (1-cosθ) + L(1-cosφ)
3) X = R (1-cosθ) + L[1-(1-(R/L)^2sin^2 θ)½]
Where cosφ = (L2 – R2sin2θ)½/L = (1-(Rsinθ/L)^2)½

Kind regards

Question

Please explain how L(1-cosφ) (line 2) changes to L[1-(1-(R/L)^2sin^2  θ)½] (line 3)



1) X = R + L – Rcos θ – Lsin φ
2) X = R (1-cosθ) + L(1-cosφ)
3) X = R (1-cosθ) + L[1-(1-(R/L)^2sin^2  θ)½]
Where cosφ = (L2 – R2sin2θ)½/L = (1-(Rsinθ/L)^2)½

Kind regards

anonymous · Answer

it's just because they say that 

cosφ = (1-(Rsinθ/L)^2)½

$$\left( \frac{ Rsin(\theta) }{ L } \right)^2 = \frac{ R^2 }{ L^2 } \sin^2 \theta$$

anonymous · Answer

Thank you. So in other words  (1-(Rsinθ/L)^2)½ is a trig identity of cosφ.

anonymous · Answer

I wouldn't call it a "trig" identity since phi and theta are undefined.

But they do have a definite relationship

anonymous · Answer

Thanks Euler. Where do I look it up? Haven't seen it before.