Tangent triple sum identity — 28-MAR-2025
That video opened by introducing first the specific example of tan(59°)+tan(60°)+tan(61°) = tan(59°)tan(60°)tan(61°), and then later generalizes the result to any three angles of a triangle.
If three angles form a euclidean triangle (sum to 180°=π), then they have this property where the sum of their tangents is the product of their tangents!
$$\small{a+b+c=\pi}\implies\tan(a)+\tan(b)+\tan(c)=\tan(a)\tan(b)\tan(c)$$
Algebraic proof (adapted from Dr. Barker):
$$\small\begin{aligned}\tan(a)+\tan(b)+\tan(c)&=\tan(a)+\tan(b)+\tan(\pi-(a+b))\\
&=\tan(a)+\tan(b)+\tan(-(a+b))^{\href{https://proofwiki.org/wiki/Tangent_of_Angle_plus_Straight_Angle}{\text{iii.}}}\\
&=\tan(a)+\tan(b)-\tan(a+b)^{\href{https://proofwiki.org/wiki/Tangent_Function_is_Odd}{\text{ii.}}}\\
&=\tan(a)+\tan(b)-{\tan(a)+\tan(b)\over1-\tan(a)\tan(b)}^{\href{https://proofwiki.org/wiki/Tangent_of_Sum}{\text{i.}}}\\
&={(\tan(a)+\tan(b))}{(1-\tan(a)\tan(b))\over1-\tan(a)\tan(b)}-{\tan(a)+\tan(b)\over1-\tan(a)\tan(b)}\\
&={(\tan(a)+\tan(b))(1-\tan(a)\tan(b))-(\tan(a)+\tan(b))\over1-\tan(a)\tan(b)}\\
&={\tan(a)+\tan(b)+(\tan(a)+\tan(b))(-\tan(a)\tan(b))-\tan(a)-\tan(b)\over1-\tan(a)\tan(b)}\\
&={-(\tan(a)\tan(b))(\tan(a)+\tan(b))\over1-\tan(a)\tan(b)}\\
&=\tan(a)\tan(b)\left(-{\tan(a)+\tan(b)\over1-\tan(a)\tan(b)}\right)\\
&=\tan(a)\tan(b)(-\tan(a+b))^{\href{https://proofwiki.org/wiki/Tangent_of_Sum}{\text{i.}}}\\
&=\tan(a)\tan(b)\tan(-(a+b))^{\href{https://proofwiki.org/wiki/Tangent_Function_is_Odd}{\text{ii.}}}\\
&=\tan(a)\tan(b)\tan(\pi-(a+b))^{\href{https://proofwiki.org/wiki/Tangent_of_Angle_plus_Straight_Angle}{\text{iii.}}}\\
&=\tan(a)\tan(b)\tan(c)
\end{aligned}$$
Properties of the tangent function used:
- Tangent of a sum
- Tangent is odd
- Tangent is periodic with a period of π
As can be seen above, I dug into
Pr∞fWiki to find citations for the various properties leveraged in Dr. Barker's proof of this property.
Algebraic proof (adapted from Pr∞fWiki):
$$\small\begin{aligned}
{\tan(a)+\tan(b)+\tan(c)-\tan(a)\tan(b)\tan(c)\over1-\tan(b)\tan(c)-\tan(c)\tan(a)-\tan(a)\tan(b)}&=\tan(a+b+c)^{\href{https://proofwiki.org/wiki/Tangent_of_Sum_of_Three_Angles}{\text{i.}}}\\
&=\tan(\pi)\\
&=0^{\href{https://proofwiki.org/wiki/Tangent_of_Straight_Angle}{\text{ii.}}}\\
\tan(a)+\tan(b)+\tan(c)-\tan(a)\tan(b)\tan(c)&=0\\
\tan(a)+\tan(b)+\tan(c)&=\tan(a)\tan(b)\tan(c)
\end{aligned}$$
Key properties used:
- Tangent of a sum of three angles
- tan(π) = 0
a + b + c = π is leveraged for the second step as the hypothesis.
Dr. Barker also remarked toward the end of the video that they would love to see a geometric proof of this property, especially considering that the sole hypothesis needed to prove it is that the angles form a euclidean triangle. I share this sentiment, and may attempt to find a nice geometric argument for this and update this page.
Additionally, after introducing this to a friend of mine, they were worried that the proof adapted from Dr. Barker might fail in the case of tan(a)tan(b) = 1, because 1 - tan(a)tan(b) is divided through the denominator. Shortly after, they produced a proof that this case can't occur where tan(c) is defined anyway.
That proof justifying that the problematic case is already excluded, adapted from my friend Loïc Taleb:
$$\begin{aligned}
\tan(a)\tan(b)=1&\implies{\sin(a)\over\cos(a)}{\sin(b)\over\cos(b)}=1\\
&\implies{\sin(a)\sin(b)}={\cos(b)\cos(b)}\\
&\implies0={\cos(b)\cos(b)}-{\sin(a)\sin(b)}\\
&\implies0=\cos(a+b)\\
&\implies\exists k\in\mathbb Z,\ a+b={\pi\over2}+k\pi\\
&\implies\exists k\in\mathbb Z,\ c={\pi\over2}+k\pi\\
&\implies\tan(c)\text{ isn't defined.}
\end{aligned}$$