# A Trigonometric Puzzle

A puzzle that came to me via @realityminus3, who credits it to @manuelcj89:

- $\sin(A) + \sin(B) + \sin(C) = 0$
- $\cos(A) + \cos(B) + \cos(C) = 0$
Find $\cos(A-B)$.

There’s something pretty about that puzzle. Interestingly, my approach differed substantially from all of my Trusted And Respected Friends’.

Spoilers below the line.

## What my friends did

Everyone else took broadly the same approach, using trigonometric identities:

- $-\sin(C) = \sin(A) + \sin(B)$
- $-\cos(C) = \cos(A) + \cos(B)$

Squaring both:

- $\sin^2(C) = \sin^2(A) + 2\sin(A)\sin(B) + \sin^2(B)$
- $\cos^2(C) = \cos^2(A) + 2\cos(A)\cos(B) + \cos^2(B)$

And adding together:

- $1 = 1 + 2(\sin(A)\sin(B) + \cos(A)\cos(B)) + 1$
- $-1 = 2\cos(A-B)$

So $\cos(A-B) = -\frac{1}{2}$.

Now, I like that. It’s just not how I did it.

## An imaginary diversion

What *I* did was to multiply the sine equation by $i$ and add, giving $\br{\cos(A) + i\sin(A)} + \br{\cos(B) + i\sin(B)} + \br{\cos(C) + i\sin(C)} = 0$.

Those three terms are imaginary numbers on the unit circle; by symmetry, the only possible configuration is an equilateral triangle.

That makes angle $A\hat{O}B = \pm \frac{2}{3}\pi$, the cosine of which is $-\frac{1}{2}$.

- Thanks to @realityminus3, @ajk_44 and @dragon_dodo for being Trusted And Respected Friends.