More Trigonometric Proofs of the Pythagorean Theorem

Unquestionably, \(\mbox{cos}^{2}\theta + \mbox{sin}^{2}\theta = 1.\) is the most fundamental identity in Euclidean trigonometry. The identity is equivalent to the Pythagorean theorem and this is probably why Elisha Loomis and many others believed that no trigonometric proof of the Pythagorean theorem is possible. The claim - unless specified further - is in fact illogical. Since the two statements are equivalent, proving one validates the other. The proper question is then whether it is possible to prove \(\mbox{cos}^{2}\theta + \mbox{sin}^{2}\theta = 1.\) without a recourse to the Pythagorean theorem. Jason Zimba has given an affirmative answer to this question, followed by another proof by David Houston (an earlier version) and Luc Gheysens (a more recent variant.)

What makes the two proofs "trigonometric"? J. Zimba's proof relies on the formulas for sine and cosine of the difference of two angles, first showing that either can be obtained without the Pythagorean theorem. The proofs by Houston and Gheysens makes use of the sine formula for the area of a triangle. In my view, both formulas illustrate an essential application of trigonometry and this fall under the rubric of "Trigonometric Proofs". What in my view would not be an essential use of trigonometry? I can answer this question metaphorically, as I had in an online discussion in the mathfuture google group

If all the usage of numbers was confined to numbering houses, would you call this "using arithmetic"?

This page was prompted by Michel Paul's post and David Chandler's remark at the mathfuture correspondence.

After several iteration Michel reduced his proof to a few words:

  1. Consider a point \((1, \theta)\).
  2. \(\mbox{cos}\theta\) and \(\mbox{sin}\theta\) are the lengths of the legs of a right triangle.
  3. Their projections onto the hypotenuse have lengths \(\mbox{cos}^{2}\theta\) and \(\mbox{sin}^{2}\theta\).
  4. Therefore, \(\mbox{cos}^{2}\theta + \mbox{sin}^{2}\theta = 1.\)

My initial reaction was to object: the proof is easily reformulated in terms of ratios. Let \(a\), \(b\) be the legs, and \(c\) the hypotenuse of a right triangle. Then the altitude cuts the triangle into two smaller ones, with all three similar. This makes projections of the legs on the hypotenuse equal to \(a^{2}/c\) and \(b^{2}/c\), adding which proves the Pythagorean theorem. My further objection was similar to a remark made by Andrius Kulikauskas:

A right triangle is made up of two right triangles which have the same shape as it does.

This is a fundamental property of the right triangles that Euclid used in his second proof of the Pythagorean theorem. On the same foundation are built other proofs, e.g., several by Geoffrey Margrave.

What undercut my determined objection to designating Michel's proof as "trigonometric" was a post by David Chandler:

Trigonometry is essentially a high level method of dealing with ratios. As such trig functions allow the mind to conceive of aspects of the problem that would be burried in detail otherwise. My use of cosines to represent projections of the legs onto the hypotenuse allowed me to visualize while driving what would have reduce to an algebraic proof with mean proportionals with more pieces to juggle. The two approaches may be equivalent, formally, but they are different conceptually.

I have not even suggested to "agree to disagree". I caught myself thinking that David may have been right. When I think in terms of similar triangles, ratios come naturally. But an equally valid view of projections on the hypotenuse is better expressed in trigonometric terms - I accept that. It's all in the eye of the beholder.

Having written that, I thought that an observation made by John Molokach - as much subject to controversy as Michel's proof - is worth bringing along.

Consider the indefinite integral \(I=\int\mbox{sin}(x)\cdot\mbox{cos}(x)dx\), or better yet \(2I=\int\mbox{sin}(2x)dx\). \(I\) can be calculated by two different substitutions:

\(I=\int\mbox{sin}(x)\cdot\mbox{cos}(x)dx = \int\mbox{sin}(x)d(\mbox{sin}(x)) = \frac{1}{2}\mbox{sin}^{2}(x)+C\) and

\(I=\int\mbox{sin}(x)\cdot\mbox{cos}(x)dx = -\int\mbox{cos}(x)d(\mbox{cos}(x)) = -\frac{1}{2}\mbox{cos}^{2}(x)+C\).

Comparing the two results we get

\(\mbox{sin}^{2}(x)+\mbox{cos}^{2}(x)=C\),

where C is a constant whose value is easily found by taking \(x=\pi/4\), to boot,

\(\big(\frac{\sqrt{2}}{2}\big)^{2}+\big(\frac{\sqrt{2}}{2}\big)^{2}=\frac{1}{2}+\frac{1}{2}=1\).

It was argued elsewhere that development of calculus does not require the 2-dimensional distance, a.k.a., the Pythagorean formula. Which certainly makes the derivation another valid proof of the Pythagorean theorem, and undoubtedly a "trigonometric proof" at that.

References

  1. E. S. Loomis, The Pythagorean Proposition, NCTM, 1968

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