# A Tangent Identity

Here's a slightly non-obvious relation between some angles, that I ran into accidentally.

Claim: (angles below given in degrees)

$\tan(10) \tan(50) = \tan(20) \tan(30)$

Verification

Claim is equivalent to

$\sin(10)\sin(50) \cos(20) \cos(30) - \sin(20)\sin(30) \cos(10) \cos(50) = 0$

the left hand side is equal to (by angle addition formula)

$\frac12 \left[ \cos(30 + 20) \cos(50-10) - \cos(30-20) \cos(50+10) \right]$

so the claim is equivalent to

$\cos(50) \cos(40) - \cos(10) \cos(60) = 0$

Observe that $\cos(50) = \sin(40)$ and vice versa, so we can write

$\cos(50) \cos(40) = \frac12 \left[ \cos(50) \cos(40) + \sin(50) \sin(40) \right] = \frac12 \cos(50 - 40)$

again using angle addition, and finally since $\cos(60) = \frac12$, the claim is proved.

Generalization for any $\gamma$ (measured in degrees)

$\tan(\gamma - 15) \tan(75-\gamma) = \tan(45-\gamma) \tan(3\gamma - 45)$

our original claim happens to be the case $\gamma = 25$. ##### Willie WY Wong
###### Assistant Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.