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.


Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.


One comment on “A Trigonometric Puzzle

  • Benjamin

    Well if its any consolation, I saw it as a variant on your approach. The fact that all the cosines and sines balance out precisely defines a root of unity. Since there were 3 of them we’re looking at the the roots of x^3 – 1 and they form an equilateral triangle at 0, 2/3 pi and 4/3 pi.

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