0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $b$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.