# How to proof $\arcsin(\sqrt{2*b*(\sqrt{1 + b^2} - b})) = \arccos(\sqrt{1 + b^2} - b)$

In a publication [1, p. 89] this equality is stated. Unfortunately, I seem unable to proof it myself

$\arcsin(\sqrt{2*b*(\sqrt{1 + b^2} - b})) = \arccos(\sqrt{1 + b^2} - b)$

The original formulation is

$\cos\theta = \sqrt{1 + b^2} - b$

$\sin^2\theta = 2*b*(\sqrt{1 + b^2} - b)$

Can anybody shed some light on this?

[1] Zongfu Dai. Particle-bubble heterocoagulation. PhD thesis, University of South Australia, Ian Wark Research Institute, 1998.

Start from $$(\sqrt{1+x^2}-x)^2 = 1-2x\sqrt{1+x^2}+x^2$$ So we find the key relation $$1-(\sqrt{1+x^2}-x)^2 = 2x\sqrt{1+x^2}-x^2=2x(\sqrt{1+x^2}-x)$$ Then use $\sin^2 + \cos^2 = 1$ to say that $$\cos^{-1}(t) = \sin^{-1}(\sqrt{1-t^2})$$ (true for all $|t|\leq 1$). If we use this and substitue $t = \sqrt{1+x^2}-x$ then the key relation says $$\cos^{-1}(\sqrt{1+x^2}-x) = \sin^{-1}(1-(\sqrt{1+x^2}-x)^2 ) = \sin^{-1}\left(2x(\sqrt{1+x^2}-x)\right)$$ which is your relationship, with $x$ replacing $b$.