Inverse trigonometric Conversion

Question

Prove that :

$$\frac{1}{2} \tan^{-1}(2)+\frac{1}{3} \tan^{-1}(3)-2\tan^{-1}\left(\frac{1}{2}\right)-3\tan^{-1}\left(\frac{1}{3}\right)=-\frac{5 \pi}{8}+\frac{5}{6} \tan^{-1}(3)$$

Using the identities $\tan^{-1}(x)=\cot^{-1} (1/x)$ for $x>0$ and $\cot^{-1}(x)+\tan^{-1}(x)=\pi/2$ I have got $\frac{3}{2} \tan^{-1}(2)+\frac{10}{3} \tan^{-1}(3)- \frac{5 \pi}{2}$ but I am not able to get final result. Could someone help me with it?

First set up a clear path what you are trying to do. You have an equation you should prove. How can you do that? — mathreadler, Apr 25 '17 at 21:23
Use the formula used in my post here: https://math.stackexchange.com/questions/523625/showing-arctan-frac23-frac12-arctan-frac125 — lab bhattacharjee, Apr 26 '17 at 01:53

score 0 · Answer 1 · answered Apr 25 '17 at 21:50

Firstly, $$\frac 12\arctan2-2\arctan\frac 12=\frac 12\left(\frac{\pi}{2}-\arctan \frac 12\right)-2\arctan\frac 12=\frac{\pi}{4}-\frac 52\arctan \frac 12$$

Secondly, you should be able to show in a similar way that $$\frac 13\arctan3-3\arctan \frac 13-\frac 56\arctan 3=-\frac{\pi}{4}-\frac 52\arctan\frac 13$$

It then remains to obtain the result by using $$\arctan a+\arctan b=\arctan\left(\frac{a+b}{1-ab}\right)$$

score 0 · Answer 2 · answered Apr 25 '17 at 22:06

Start from the addition formula for arctangents: $$ \tan^{-1} x + \tan^{-1} y = \tan^{-1}\left( \frac{x+y}{1-xy} \right) $$

So $$ \tan^{-1} \left(\frac12 \right)+ \tan^{-1} \left(\frac13 \right) = \tan^{-1}\left( \frac{\frac12 + \frac13}{1-\frac12 \frac13} \right)= \tan^{-1}\frac{5/6}{5/6} = \frac{\pi}{4} $$ Then use $$ \tan^{-1} x = \frac{\pi}{2} -\tan^{-1} \left(\frac1x\right) $$ to write $$ \frac12 \tan^{-1}(2) +\left(\frac13-\frac56\right)\tan^{-1}(3) = \frac\pi4 - \tan^{-1} \left(\frac12\right) -\frac\pi4 + \tan^{-1} \left(\frac12\right) $$ or $$ \frac12 \tan^{-1}(2) -\frac12\tan^{-1}(3) = \tan^{-1} \left(\frac13\right)- \tan^{-1} \left(\frac12\right) $$ Your identity to be proved is $$ \frac12 \tan^{-1}(2) -\frac12\tan^{-1}(3) -2 \tan^{-1} \left(\frac12\right)-3 \tan^{-1} \left(\frac13\right) = -\frac58\pi$$ Replace the first two terms by $\tan^{-1} \left(\frac13\right)- \tan^{-1} \left(\frac12\right)$ and this becomes $$ -\frac52\left[ \tan^{-1} \left(\frac12\right)+ \tan^{-1} \left(\frac13\right)\right] =-\frac58\pi $$ and then using our addition formula for arctangents of $\frac12$ and $\frac13$, we see that this is an identity: $$ -\frac52\left[ \frac\pi4 \right]=-\frac58\pi $$

Inverse trigonometric Conversion

2 Answers2