Another Face and Proof of a Trigonometric Identity

Problem 2 at the 2007 Irish Mathematical Olympiad required to prove that the identity \(\mbox{sin}^{2}\alpha + \mbox{sin}^{2}\beta +\mbox{sin}^{2}\gamma =2\), where \(\alpha +\beta +\gamma = 180^{\circ}\), holds if and only if one of the angles \(\alpha\), \(\beta\), or \(\gamma\) is right. The equivalent identity \(\mbox{cos}^{2}\alpha + \mbox{cos}^{2}\beta +\mbox{cos}^{2}\gamma =1\) then appears as a form of the Pythagorean theorem. It came to my attention that in the following reincarnation

\(\displaystyle\frac{\mbox{sin}^{2}\alpha + \mbox{sin}^{2}\beta +\mbox{sin}^{2}\gamma}{\mbox{cos}^{2}\alpha + \mbox{cos}^{2}\beta +\mbox{cos}^{2}\gamma}=2\)

it was suggested by Poland at the 1967 International Mathematical Olympiad [The IMO Compendium, p. 46].

The equivalence of the three formulations is obvious. Thus, the proof below covers all three.

Assume \(\alpha +\beta +\gamma = 180^{\circ}\). Then the identity

\(\mbox{cos}^{2}\alpha + \mbox{cos}^{2}\beta +\mbox{cos}^{2}\gamma =1\)

Holds if and only if one of the angles \(\alpha\), \(\beta\), or \(\gamma\) is right.

Proof

Multiply the identity by 2 and regroup:

\((2\mbox{cos}^{2}\alpha -1) + (2\mbox{cos}^{2}\beta -1) +2\mbox{cos}^{2}\gamma =0\).

By the double argument formulas, \(2\mbox{cos}^{2}t -1=\mbox{cos}(2t)\), this is converted to

\(\mbox{cos}^{2}2\alpha + \mbox{cos}^{2}2\beta +2\mbox{cos}^{2}\gamma =0\).

Next apply the edition formulas \(\mbox{cos}(2s)\mbox{cos}(2t)=2\mbox{cos}(s-t)\mbox{cos}(s+t)\) to obtain

\(\mbox{cos}(\alpha -\beta)\mbox{cos}(\alpha +\beta) +\mbox{cos}^{2}\gamma =0\).

Taking into account that, by the stipulation of the problem, \(\mbox{cos}\gamma = -\mbox{cos}(\alpha +\beta)\), we factor the latter

\(\mbox{cos}\gamma\space (\mbox{cos}(\alpha -\beta)-\mbox{cos}\gamma) =0\).

It follows that either \(\mbox{cos}\gamma =0\) or \(\mbox{cos}(\alpha -\beta) -\mbox{cos}\gamma =\mbox{cos}(\alpha -\beta) +\mbox{cos}(\alpha +\beta) =0\). In the former case, \(\gamma\) is right.

In the latter case, \(\mbox{cos}(\alpha -\beta) +\mbox{cos}(\alpha +\beta)=2\space\mbox{cos}\alpha\space\mbox{cos}\beta =0\), implying that either \(\alpha\) or \(\beta\) is right.

References

D. Djukic et al, The IMO Compendium, Springer, 2011 (Second edition)

Trigonometry

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