1
$\begingroup$

Is it possible to prove

$$\cos^{-1}\left[\frac{\cos(a) + \cos(b)}{1 + \cos(a)\cos(b)}\right] = 2\tan^{-1}\left[\tan\left(\frac{a}{2}\right)\tan\left(\frac{b}{2}\right)\right]$$

with the help of integration?

I know how to prove it without integration but can't develop any approach to solve it using integration.

$\endgroup$
  • $\begingroup$ I think you cant as integration operator would just get cancelled a d if integration of someone gives another then they arent equal as its area under the curve for the integral function $\endgroup$ – Archis Welankar Feb 25 '16 at 6:59
  • $\begingroup$ Are there any restrictions on $a$ and $b$? The left side is always positive, $range(\arccos)=[0,\pi]$, but if $a·b<0$ then the right side is negative. $\endgroup$ – LutzL Feb 25 '16 at 11:25
1
$\begingroup$

If $b=0$ then the equation reduces to $$ \arccos(1)=2\arctan(0) $$ which is true. Then compute the derivatives of both sides for fixed $a$ and variable $b$ and hopefully both will be the same.


In other news, $$ \arccos x=2\arctan y \iff y=\tan(\tfrac12\arccos x)=\frac{\sin\arccos x}{1+\cos\arccos x}=\frac{\sqrt{1-x^2}}{1+x}=\sqrt{\frac{1-x}{1+x}} $$ or $$ x=\frac{1-y^2}{1+y^2} $$

$\endgroup$

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.