In trigonometric form the the Pythagorean theorem appears as cos^{2}\alpha + sin^{2}\alpha = 1. Here \alpha is one of the acute angles of a right triangle. If the other acute angle is \beta then, because sin\alpha = cos(\pi/2 - \alpha) we also have cos^{2}\alpha + cos^{2}\beta = 1,  for the two acute angles \alpha and \beta of a right triangle. Since the third angle, say, \gamma is right, cos\gamma = 0 so that in a right triangle

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

The converse is also true: if the above identity holds for the angles of a triangle, the triangle is necessarily right. Indeed, this follows directly from a more general identity

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

that holds whenever \alpha + \beta + \gamma = \pi.

The latter is an immediate consequence of an identity proved elsewhere:

cos{2}\alpha + cos{2}\beta + cos{2}\gamma + 1 = -4cos\alpha cos\beta cos\gamma.

(Use the addition formula for cosine, cos(\theta+\omega) = cos\theta cos\omega - sin\theta sin\omega, which, for \theta = \omega, gives cos2\theta = cos^{2}\theta - sin^{2}\theta = 2cos^{2}\theta - 1. )

Now, as we see, cos^{2}\alpha + cos^{2}\beta + cos^{2}\gamma = 1 implies cos\alpha cos\beta cos\gamma = 0  so that, for one of the angles, cosine equals zero, meaning that the angle is right.

The identity cos^{2}\alpha + cos^{2}\beta + cos^{2}\gamma = 1 is equivalent to sin^{2}\alpha + sin^{2}\beta + sin^{2}\gamma = 2, which then is also a property of the right triangles. Proving the latter was one of the 2007 Ir IMO problems. It was also suggested (but did not get it) by Poland for the 1967 IMO.