# 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 \(\cos ^{2}\alpha + \cos ^{2}\beta +\cos ^{2}\gamma =1\) then appears as a form of the Pythagorean theorem. It came to my attention that in the following reincarnation

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

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

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

### Proof 1

Multiply the identity by 2 and regroup:

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

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

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

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

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

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

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

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

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

### Proof 2

As was suggested by a comment below the expression $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1\,$ cann be converted into

$(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)=0\,$

with a required implication. This follows from a known identity

$\cos^2\alpha+\cos^2\beta+\cos^2\gamma+2\cos\alpha\cos\beta\cos\gamma=1\,$

and the Law of Cosines:

$\displaystyle \cos\gamma=\frac{a^2+b^2-c^2}{2ab}, \ldots$

### Proof 3

This proof is by mit Itagi.

If the triangle is right, WLOG let $\alpha=90^o$. Thus,

$\displaystyle \begin{align} &\frac{\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma}{\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma} \\ &=\frac{1+\sin^2\beta+\sin^2(90^o-\beta)}{\cos^2\beta+\cos^2(90^o-\beta)} \\ &=\frac{1+\sin^2\beta+\cos^2\beta}{\cos^2\beta+\sin^2\beta}=2 \end{align}$

For the converse, let $d$ be the diameter of the circumcircle. Using sine rule,

$\displaystyle \begin{align} &\frac{\sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma}{\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma}=2 \\ \Rightarrow& \frac{(a^2+b^2+c^2)/d^2}{3-(a^2+b^2+c^2)/d^2}=2 \\ \Rightarrow& (a^2+b^2+c^2)=2d^2 \\ \Rightarrow& (a^2+b^2+c^2)=\frac{a^2+b^2+c^2}{(1+\cos A\cos B\cos C)} \\ \Rightarrow& \cos A\cos B\cos C = 0. \end{align}$

Thus, one of the cosine terms goes to zero and that angle is a right angle.

### References

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

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