# showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$

I'm having problem with showing that: $$\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$$

I would need some help in the right direction

$(3+2i)^2=5+12i$, so $\arctan\frac23+\arctan\frac23=\arctan\frac{12}{5}$.

Alternatively, study this diagram: • Hey! That's something really creative. Never so that stuff before :D Oct 12, 2013 at 17:02
• Could you elaborate the first one? I'm probably missing something obvious. Oct 13, 2013 at 8:38
• @Lazar: It's just a way to write the second one in algebraic guise. $\arctan \frac ab$ is the argument of $b+ai$ when $b>0$, and arguments are additive when multiplying complex numbers. Oct 13, 2013 at 10:08
• I knew I had a silly brainfart. Thanks. Oct 13, 2013 at 10:18

From the Article $240,$ Ex$-5$ of Plane Trigonometry(by Loney),

$$\arctan x+\arctan y=\begin{cases} \arctan\frac{x+y}{1-xy} &\mbox{if } xy<1\\ \pi+\arctan\frac{x+y}{1-xy} & \mbox{if } xy>1\end{cases}$$

$$\implies 2\arctan x=\arctan \frac{2x}{1-x^2}\text{ if }x^2<1$$

Here $\displaystyle x=\frac23$

There is a nice geometrical way of showing that. Your statement can be shown with the bisector theorem (http://en.wikipedia.org/wiki/Angle_bisector_theorem)... Check the picture Sorry but I have quickly drawn it. Now your problem is equivalent to say that $\arctan(\frac{2}{3}) = \alpha$ since $\beta = \arctan(\frac{12}{5})$. For the bisector theorem it is true that $$\frac{BC}{CD} = \frac{5}{13}$$ now we know that $BC+CD=12$ and solving the two equations we find that $BC=10/3$ that is exactly equivalent saying $$\arctan(\alpha) = \frac{2}{3}$$ that is exactly what we wanted to demonstrate.

This was fun ;-)

The basic concept of inverse trigonometry is as follows - Arctan is same as tan^-1 with in the specified values.

tan^-1 x + tan ^-1 y = tan ^ -1 (x+ y/1−xy) if xy<1 π+ tan ^ -1 (x+y/1−xy) if xy>1 and x>0 and y>0 -π + tan ^ -1 (x + y/1-xy) if xy>1 and x<0 and y

This concept will solve your proof above by instantly putting the given value of x.