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.