Why does $\tan10^\circ+\tan20^\circ+\tan50^\circ$ equal $\tan60^\circ$?

Question

My calculator says that $\tan10^\circ+\tan20^\circ+\tan50^\circ=\tan60^\circ$, and this is confirmed to 15 decimal places by a more precise online calculator. So it looks plausible. However, I have not managed to prove the result, despite its apparent simplicity. A similar result was proved by MSE poster achille hui, using the formula $$n \cot n\theta = \sum_{k\,=0}^{n-1}\cot\!\left(\!\theta+\frac{k\pi}{n}\!\right) \;\;(n=1,2,...).$$ In terms of cotangents, the desired result is $\cot40^\circ+\cot70^\circ+\cot80^\circ=\cot30^\circ\;(=3\cot60^\circ)$. However, I can't see how to fit these angles to the formula with $n=3$. Is there another formula of this sort that does the trick?

A more general question arises: What other identities of the form $\tan\alpha+\tan\beta+\tan\gamma=\tan\delta\,$ hold, where the angles are all simple rational multiples of $\pi$?

Answer 1

Proof of this one....

Let $\theta = \pi/18$. We want to show $$ \tan(\theta)+\tan(2\theta)+\tan(5\theta)-\tan(6\theta)=0 \tag1$$ Using the addition formula for tangent, this is equivalent to $$ 2\,{\frac { \left( \left( \tan \left( \theta \right) \right) ^{2}+2 \,\tan \left( \theta \right) -1 \right) \left( \left( \tan \left( \theta \right) \right) ^{2}-2\,\tan \left( \theta \right) -1 \right) \left( 3\, \left( \tan \left( \theta \right) \right) ^{6}-27\, \left( \tan \left( \theta \right) \right) ^{4}+33\, \left( \tan \left( \theta \right) \right) ^{2}-1 \right) \tan \left( \theta \right) }{ \left( 1-10\, \left( \tan \left( \theta \right) \right) ^ {2}+5\, \left( \tan \left( \theta \right) \right) ^{4} \right) \left( \tan \left( \theta \right) -1 \right) \left( \tan \left( \theta \right) +1 \right) \left( \left( \tan \left( \theta \right) \right) ^{2}-4\,\tan \left( \theta \right) +1 \right) \left( \left( \tan \left( \theta \right) \right) ^{2}+4\,\tan \left( \theta \right) +1 \right) }} = 0 \tag2$$ So it suffices to prove that $$ 3\tan^6(\theta) - 27\tan^4(\theta) + 33\tan^2(\theta) - 1 = 0 \tag3$$

Starting with $$ \frac{1}{\sqrt{3}} = \tan(\pi/6) = \tan(3\theta) =\frac{3\tan(\theta) - \tan^3(\theta)}{1-3\tan^2(\theta)} $$ we may deduce (square both sides, then common denominator) $$ \frac{3\tan^6(\theta) - 27\tan^4(\theta) + 33\tan^2(\theta) - 1}{3(1-3\tan^2(\theta))^2} = 0 $$ as required.

Answer 2

We have that by summation formula

$$\tan10^\circ+\tan50^\circ =\tan 60^\circ(1-\tan10^\circ\tan50^\circ)$$

then

$$\tan10^\circ+\tan20^\circ+\tan50^\circ=\tan60^\circ \iff \tan20^\circ=\tan10^\circ\tan50^\circ\tan60^\circ \tag 1$$

we also have

$$\tan20^\circ =\frac{2\tan10^\circ}{1-\tan^2 10^\circ} \;\;\text{and}\;\;\tan50^\circ\tan60^\circ=\frac{\cos 10^\circ+\sin 20^\circ}{\cos 10^\circ-\sin 20^\circ}$$

therefore substituting in $(1)$ and simplifying we obtain

$$\frac{2}{1-\tan^2 10^\circ}=\frac{\cos 10^\circ+\sin 20^\circ}{\cos 10^\circ-\sin 20^\circ}$$

$$\frac{2\cos^2 10^\circ}{\cos^2 10^\circ-\sin^2 10^\circ}=\frac{\cos 10^\circ(1+2\sin 10^\circ)}{\cos 10^\circ(1-2\sin 10^\circ)}$$

$$\frac{2(1-\sin^2 10^\circ)}{1-2\sin^2 10^\circ}=\frac{1+2\sin 10^\circ}{1-2\sin 10^\circ}$$

$$2(1-\sin^2 10^\circ)(1-2\sin 10^\circ)=(1+2\sin 10^\circ)(1-2\sin^2 10^\circ)$$

$$8\sin^3 10^\circ-6\sin 10^\circ+1=0 $$

and using that $\sin^3 x= \frac14\left(3\sin x -\sin(3x)\right)$, the latter is finally equivalent to

$$\boxed {1-2\sin 30^\circ =0}$$

which is true.

Answer 3

Recall the cotangent identities: $$n\cot(n\theta) = \sum_{k=0}^{n-1} \cot\left(\theta + \frac{k}{n} \times 180^\circ\right)\quad\text{ for }n = 2,3,\ldots$$

The equality at hand can be deduced using $n = 2$ and $3$ version of these identites. Notice

$$\cot(\theta + 90^\circ) = \tan(90^\circ - (\theta + 90^\circ)) = -\tan\theta$$ The $n = 2$ version of cotangent identity tell us

$$2\cot(2\theta) = \cot\theta - \tan\theta \implies \tan\theta = \cot\theta - 2\cot(2\theta)$$ As a result,

$$\begin{align} &\; \tan(10^\circ) + \tan(20^\circ) + \tan(50^\circ)\\ = &\; \cot(80^\circ) + (\cot(20^\circ) - 2\cot(40^\circ)) + \cot(40^\circ)\\ = &\; \cot(20^\circ) + \cot(80^\circ) - \cot(40^\circ)\\ = &\; \cot(20^\circ) + \cot(80^\circ) + \cot(140^\circ)\\ = &\; \cot(20^\circ) + \cot(20^\circ + 60^\circ) + \cot(20^\circ + 120^\circ) \end{align} $$ Apply the $n = 3$ version of the contangent identity, last expression evaluates to

$$3\cot(3\times 20^\circ) = 3\cot(60^\circ) = \sqrt{3} = \tan(60^\circ)$$ and we are done.

Answer 4

One has $$\tan10^\circ+\tan20^\circ+\tan50^\circ=\tan60^\circ$$ Because $20=2\times10$ and $50=60-10$, $$\tan10^\circ+\frac{2\tan10^\circ}{1-\tan^2(10^\circ)}+\frac{\tan(60^\circ)+\tan(10^\circ)}{1-\tan(60^\circ)\tan(10^\circ)}=\tan60^\circ$$ Puting $x=\tan(10^\circ)$ we have $$x+\frac{2x}{1-x^2}+\frac{\sqrt3-x}{1+\sqrt3x}=\sqrt3$$ from which $$\sqrt3x^3-3x^2-3\sqrt3 x+1=0$$ this equation has three real roots and one of them is $\boxed{\tan(10^\circ)}$ the other two being $\tan(70^\circ)$ and $\tan(-50^\circ)$.

Answer 5

We need to show that $x:=\tan60^\circ-\tan50^\circ-\tan10^\circ-\tan20^\circ$ is zero. Rearranging the formula for the tangent of $60^\circ-50^\circ\,(=10^\circ)$ gives $$\tan60^\circ-\tan50^\circ=(1+\tan60^\circ\tan50^\circ)\tan10^\circ.$$ Hence $x=\tan60^\circ\tan50^\circ\tan10^\circ-\tan20^\circ$, and therefore $$x\times4\cos10^\circ\cos20^\circ\cos50^\circ\cos60^\circ$$ $$=4\sin10^\circ\cos20^\circ\sin50^\circ\sin60^\circ-4\cos10^\circ\sin20^\circ\cos50^\circ\cos60^\circ$$ $$=(2\sin60^\circ\sin10^\circ)(2\sin50^\circ\cos20^\circ)-(2\cos60^\circ\cos10^\circ)(2\cos50^\circ\sin20^\circ)$$ $$=(\cos50^\circ-\cos70^\circ)(\sin70^\circ+\sin30^\circ)-(\cos50^\circ+\cos70^\circ)(\sin70^\circ-\sin30^\circ)$$ $$=2\cos50^\circ\sin30^\circ-2\cos70^\circ\sin70^\circ=\cos50^\circ-\sin140^\circ=0.$$