Showing $\tan(A) + \tan(60^\circ+A) - \tan(60^\circ-A) = 3\tan(3A)$

Question

I am looking for an elegant and quick proof for the formula: $ \tan(A) + \tan(60^\circ+A) - \tan(60^\circ-A) = 3\tan(3A)$ I have looked through some proofs through pure trigonometry but was hoping was something involving complex numbers or geometry.

Answer 1

By the addition formula,

$$\text{LHS}:t+t_++t_-=t+\frac{t+\sqrt3}{1-\sqrt3t}+\frac{t-\sqrt3}{1+\sqrt3t}=3\frac{3t-t^3}{1-3t^2}$$

and

$$\text{RHS}:t_3=\frac{t+t_2}{1-t\,t_2}=\frac{t+\dfrac{2t}{1-t^2}}{1-t\dfrac{2t}{1-t^2}}=\frac{3t-t^3}{1-3t^2}.$$

I can't think of a simple geometric proof, as there would be an angle trisection.

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Note that

$$1+it_3\propto(1+it)^3=1+3it-3t^2-it^3\implies t_3=\frac{3t-t^3}{1-3t^2}.$$

Answer 2

I'm confident that in terms of text this is not the simplest way but i consider it elegant because it uses very few first principles and can also be easily automated. So if you ever have a trig identity and don't know where to go, this will ALWAYS work as a faithful faithful last resort.

The result we want to show is

$$ \tan(A) + \tan \left(\frac{\pi}{3}+A \right) - \tan \left( \frac{\pi}{3} - A \right) = 3 \tan \left( 3A\right)$$

We can utilize Euler's formula. So we have that $\tan(x) = \frac{\sin(x)}{\cos(x)} $ and because of the formula:

$$\sin(x) = \frac{e^{ix} - e^{-ix}}{2i}$$

$$\cos(x) = \frac{e^{ix} + e^{-ix}}{2}$$

So then it naturally follows that

$$ \tan(x) = \frac{e^{ix} - e^{-ix}}{i(e^{ix} + e^{-ix})} $$

Our identity then is to verify that $$ \tan(A) + \tan \left(\frac{\pi}{3}+A \right) - \tan \left( \frac{\pi}{3} - A \right) = 3 \tan \left( 3A\right)$$

But instead it will look like:

$$ \frac{e^{iA} - e^{-iA}}{i(e^{iA} + e^{-iA})} + \frac{e^{i(\frac{\pi}{3}+A)} - e^{-i(\frac{\pi}{3}+A)}}{i(e^{i(\frac{\pi}{3}+A)} + e^{-i(\frac{\pi}{3}+A)})} -\frac{e^{i(\frac{\pi}{3}-A)} - e^{-i(\frac{\pi}{3}-A)}}{i(e^{i(\frac{\pi}{3}-A)} + e^{-i(\frac{\pi}{3}-A)})} = 3\frac{e^{i3A} - e^{-i3A}}{i(e^{i3A} + e^{-3iA})} $$

We can make some simplifications [example the i on the denominators on all the fractions cancel out]

$$ \frac{e^{iA} - e^{-iA}}{e^{iA} + e^{-iA}} + \frac{e^{i(\frac{\pi}{3}+A)} - e^{-i(\frac{\pi}{3}+A)}}{e^{i(\frac{\pi}{3}+A)} + e^{-i(\frac{\pi}{3}+A)}} -\frac{e^{i(\frac{\pi}{3}-A)} - e^{-i(\frac{\pi}{3}-A)}}{e^{i(\frac{\pi}{3}-A)} + e^{-(\frac{\pi}{3}-A)}} = 3\frac{e^{i3A} - e^{-i3A}}{e^{i3A} + e^{-3iA}} $$

And now all that's left is to combine these 3 fractions. As a quick sanity check we verify that if we combine the 3 denominators everything matches, that is we want to see that

$$ (e^{iA} + e^{-iA})(e^{i(\frac{\pi}{3}+A)} + e^{-i(\frac{\pi}{3}+A)})(e^{i(\frac{\pi}{3}-A)} + e^{-(\frac{\pi}{3}-A)}) = e^{i3A} + e^{-3iA}$$

We do some good old FOIL on the right 2 products on the left hand side

$$ (e^{iA} + e^{-iA})(e^{2\frac{i\pi}{3}}+e^{2iA} + e^{-2iA} + e^{- 2\frac{i \pi}{3}})$$

And from here we evaluate the $e^{\frac{2\pi i}{3}}$ terms. Review this if unfamiliar with roots of unity and we have

$$ (e^{iA} + e^{-iA})(\frac{-1-i\sqrt{3}}{2}+e^{2iA} + e^{-2iA} + \frac{-1+i\sqrt{3}}{2}) $$

Which yields:

$$(e^{iA} + e^{-iA})(e^{2iA} + e^{-2iA} - 1 ) = e^{3iA} + e^{iA}+e^{-iA} + e^{-3iA} - e^{iA} - e^{-iA} = e^{3iA} - e^{-3iA} $$

As desired!

So now all of that is the sanity check... to do the numerator is considerably more work, but maybe with about 15-20 minutes of continuous algebra its doable.

What's nice, is ALL we used here were: complex numbers, multiplying exponents/simplifying fractions, and one formula.

This strategy in general can be utilized to verify ANY trig identity you want. So this strategy requires the least intelligence (there's almost no thinking involved just repeatedly simplifying and multiplying) but probably the most endurance.

Answer 3

Like Proof$\#2$ of Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$

$$\tan\left(y+r\cdot60^\circ\right); r=0,1,2$$ are the roots of $$\tan^3y-3\tan3y\tan^2y-3\tan y+\tan3y=0$$

Using Vieta's formula $$\tan y+\tan(60^\circ+y)+\tan(120^\circ+y)=\dfrac{3\tan3y}1$$

But $\tan(y+120^\circ)=\tan(180^\circ-(60^\circ-y))=-\tan(60^\circ-y)$

I believe this is how the problem came into being.

We can similarly prove the following :

$$\tan y\tan(60^\circ+y)\tan(120^\circ+y)=-\dfrac{\tan3y}1$$

$$\tan y\tan(60^\circ+y)+\tan(60^\circ+y)\tan(120^\circ+y)+\tan(120^\circ+y)\tan y=-\dfrac31$$