8
$\begingroup$

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

$\endgroup$
26
+50
$\begingroup$

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


Alternatively, study this diagram: diagram goes here

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Hey! That's something really creative. Never so that stuff before :D $\endgroup$ – Parth Thakkar Oct 12 '13 at 17:02
  • $\begingroup$ Could you elaborate the first one? I'm probably missing something obvious. $\endgroup$ – Lazar Ljubenović Oct 13 '13 at 8:38
  • 2
    $\begingroup$ @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. $\endgroup$ – hmakholm left over Monica Oct 13 '13 at 10:08
  • $\begingroup$ I knew I had a silly brainfart. Thanks. $\endgroup$ – Lazar Ljubenović Oct 13 '13 at 10:18
8
$\begingroup$

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 $\displaystyle x=\frac23$

| cite | improve this answer | |
$\endgroup$
4
$\begingroup$

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

Triangle

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 ;-)

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

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.