Trigonometric Proof of the Pythagorean Theorem

Elisha Loomis, myself and no doubt many others believed and still believe that no trigonometric proof of the Pythagorean theorem is possible. This belief stems from the assumption that any such proof would rely on the most fundamental of trigonometric identities sin²α + cos²α = 1 which is nothing but a reformulation of the Pythagorean theorem proper. Thus, as the common argument goes, relying on it would lead inevitably to the vicious circle of circular reasoning. Now, Jason Zimba has showed that the theorem can be derived from the subtraction formulas for sine and cosine without a recourse to sin²α + cos²α = 1.

I happily admit to being in the wrong.

Here are the two subtraction formulas that J. Zimba employs in his derivation of the Pythagorean theorem:

cos (α - β) = cos α cos β + sin α sin β,
sin (α - β) = sin α cos β - cos α sin β.

Note that the common definitions of sine and cosine as well as the proofs of the two identities always restrict α and β to the positive values less than 90°. For this reason, simply setting α = β in the first of these leads to cos 0 = cos²α + sin²α. However, the fact that cos 0 = 1 is not a consequence of more basic trigonometry but a definition by convenience. As such, it could not be reasonably used in a proof of such a fundamental result as the Pythagorean theorem. Instead, assume that 0 < y < x < 90°. This makes 0 < x - y < 90°. Then

which implies sin²x + cos²x = 1 since, by the definition, cos y, being the ratio of a leg of a right triangle to its hypotenuse, is never zero. (As was already mentioned, the base definition of the trigonometric functions always assumes that the argument is a positive angle less than 90°.)

References

1. E. S. Loomis, The Pythagorean Proposition, NCTM, 1968
2. J. Zimba, On the Possibility of Trigonometric Proofs of the Pythagorean Theorem, Forum Geometricorum, Volume 9 (2009) 275-278

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