Trigonometric Proofs of the Pythagorean Theorem

A trigonometric proof derives a² + b² = c² from facts about sine and cosine. The danger is circular reasoning: the identity sin²θ + cos²θ = 1 is normally proved from the Pythagorean theorem, so it cannot be used to prove it. Valid proofs avoid that identity and use only right-triangle ratios, angle-difference formulas or the law of sines.

Two proofs below pass that test: Jason Zimba's short algebraic argument from 2009 and the "waffle cone" proof that two New Orleans high school students announced in 2023.

Why trigonometric proofs looked circular

In a right triangle with hypotenuse c and an acute angle θ, sine and cosine are ratios of sides: sinθ is the opposite leg over the hypotenuse and cosθ is the adjacent leg over the hypotenuse. These ratios depend only on θ, because all right triangles with the angle θ are similar. Nothing so far uses the Pythagorean theorem.

The trouble starts with the most famous identity in trigonometry. With legs a (opposite θ) and b (adjacent):

The last step is the Pythagorean theorem, divided by c². So the Pythagorean identity is the theorem in disguise, and using it to prove the theorem assumes what you are trying to prove. The same goes for the unit-circle equation x² + y² = 1, the distance formula and the usual proof of the law of cosines: all are normally built on the theorem.

That is why Elisha Loomis, in his influential collection The Pythagorean Proposition (1927), stated that there could be no trigonometric proofs at all, since the basic formulas of trigonometry depend on the theorem. The flaw in that argument is that not every trigonometric fact depends on it.

Which trig facts are safe to use?

Jason Zimba's proof (2009)

In 2009 Jason Zimba, then a physicist and mathematician at Bennington College, published a short paper in Forum Geometricorum titled "On the Possibility of Trigonometric Proofs of the Pythagorean Theorem". He first showed that the subtraction formulas can be proved for acute angles from a diagram of right triangles, without the theorem:

These hold whenever 0° < y < x < 90°. The clever step is to apply the first formula with x and x − y in place of x and y, then expand both factors with the formulas again:

The terms cosxsinxsiny cancel. Since cosy is a ratio of two lengths, it is not zero, so dividing by it gives cos²x + sin²x = 1 for every acute angle x (choose any y between 0° and x, such as x/2). In a right triangle with acute angle x, sinx = a/c and cosx = b/c, so a²/c² + b²/c² = 1, which is a² + b² = c².

The proof surprised many who had accepted Loomis's claim. Alexander Bogomolny, whose website Cut the Knot collects more than a hundred proofs of the theorem, added it and wrote: "I happily admit to being in the wrong."

Jackson and Johnson's proof (2023–2024)

Ne'Kiya Jackson and Calcea Johnson were students at St. Mary's Academy, a high school in New Orleans, when a bonus question in a math contest asked for a new proof of the Pythagorean theorem. Their answer used trigonometry. They presented it at a meeting of the American Mathematical Society in Atlanta in March 2023. In 2024 they published the work in The American Mathematical Monthly as "Five or Ten New Proofs of the Pythagorean Theorem": five new proofs, plus a method that produces five more.

Their proofs rest on the law of sines, which does not depend on the theorem, combined in the first proof with an infinite geometric series. What follows is a simplified version of the idea behind that first proof, in our own notation. The paper's lettering and figures differ, and it handles details we skip.

The construction

1. Start with a right triangle ABC, right angle at C and angle α at A, with the legs unequal: a < b, so α < 45°.
2. Reflect the triangle across the leg AC to get the point B′. Triangle ABB′ is isosceles, with AB = AB′ = c, base BB′ = 2a and angle 2α at A.
3. Draw the perpendicular to AB at B. Because 2α < 90°, it meets the line AB′ at a point D, making a right triangle ABD with angle 2α at A.
4. Repeat inside. From B′ draw the perpendicular to AD, meeting BD at E. Triangle EB′D has a right angle at B′ and shares the angle at D with ABD, so it is a smaller copy of ABD. Do the same inside it, and so on forever.

How big is each copy? In triangle BB′E the angles at B and B′ are both α, so EB = EB′. Its altitude from E cuts it into two right triangles similar to ABC, each with a (half of BB′) as the leg next to the angle α. So EB′ = EB = a · c over b. In the copy EB′D, the side EB′ plays the role of the side AB = c of triangle ABD, so the copy is triangle ABD shrunk by the factor

Two infinite series

Each copy of ABD adds one segment to the side AD and one to the side BD, and each copy is r times the size of the previous one. Adding all the segments gives two geometric series:

Both sums use 1 + r² + r⁴ + ⋯ = 1 over 1 − r², which is valid because r < 1.

The angle 2α, measured two ways

From the series. Triangle ABD has a right angle at B, so sin2α is the opposite side over the hypotenuse. Multiplying the top and bottom by b²:

From the law of sines. In triangle ABB′, the side BB′ = 2a is opposite the angle 2α and the side AB′ = c is opposite the angle 90° − α at B. In triangle ABC, sin(90° − α) = sinB = b/c. So

Compare. The two expressions for sin2α have the same numerator, so their denominators are equal: a² + b² = c². With legs 1 and 2, for example, r = 1 over 2 and the series give sin2α = 4 over 5; the law of sines then forces c² = 5 = 1² + 2².

Why this is not circular

• Sine is used only as a ratio in right triangles, which needs similar triangles, not the theorem.
• The law of sines is proved by dropping an altitude and writing its length two ways.
• The infinite process is legitimate: after n copies, what is left of AD and BD are sides of a copy shrunk by rⁿ, which tends to zero, so the partial sums really do approach AD and BD.
• Nowhere does the argument use sin²θ + cos²θ = 1 or the law of cosines.

The construction needs unequal legs. If a > b, swap the names of the legs. If a = b, then 2α = 90°, the perpendicular at B runs parallel to AB′ and the cone never closes, so the isosceles right triangle needs a separate argument.

What these proofs do and don't show

They show that Loomis was wrong: trigonometry can prove the theorem, provided you track which facts you use. They do not make the theorem independent of geometry, because sine and cosine are defined through similar triangles, which belong to flat, Euclidean geometry.

Some headlines called the students' work the first trigonometric proof, or something long thought impossible. That overstates it: Zimba's proof appeared in 2009, and others followed, such as Nuno Luzia's in 2015. What stands out is several genuinely new proofs by high school students, published in a leading journal, with an inventive use of the law of sines and infinite series. For the everyday question of when to reach for the theorem and when for trig, see Pythagorean theorem or trigonometry? and for geometric proofs, the proofs overview and Einstein's proof.