In any right-angled triangle, the area of the square whose side is the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares whose sides are the two legs (the two sides that meet at a right angle).

The theorem can be written as an equation relating the lengths of the sides $a$, $b$, and $c$, often called the Pythagorean equation: $$a^2 + b^2 = c^2$$

Can I prove Pythagoras' Theorem by the following way?

Actually, my question is: does it violate any rules of mathematics, or is it alright? Sorry, it may not be a valid question for this site. But I want to know. Thanks.

Like it

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    $\begingroup$ Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. $\endgroup$ – Zev Chonoles Jul 21 '13 at 17:16
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    $\begingroup$ The identity $\cos^2\theta + \sin^2\theta = 1$ is based on the Pythagorean Theorem. So I would consider this inappropriate. $\endgroup$ – Ted Shifrin Jul 21 '13 at 17:17
  • $\begingroup$ I know that The identity cos2θ+sin2θ=1 PRESUMES the Pythagorean Theorem. Is that mean I can't use it to prove Pythagorean theorem @TedShifrin $\endgroup$ – Atish Dipongkor - MVP Jul 21 '13 at 17:23
  • $\begingroup$ Its probably worth adding that in the framework of inner product spaces, Pythagoras is remarkably easy to prove: $$\|x-y\|^2 = (x-y) \bullet (x-y) = x \bullet x- x\bullet y-y \bullet x+y \bullet y = \|x\|^2+\|y\|^2$$ So for any version of Pythagoras you want to prove, its probably worth translating the whole thing into inner product language; if you can do this (and it isn't always easy to do), but if you can, then the proof of Pythagoras becomes trivial. $\endgroup$ – goblin GONE May 8 '16 at 14:41
  • $\begingroup$ What are you trying to ask? Do you mean if you define the distance formula as $\sqrt{x^2 + y^2}$ without justification, can you prove that that definition satisfies the Pythagorean theorem a slightly different way than I do in my Quora answer at quora.com/… and my page it links? $\endgroup$ – Timothy Dec 3 '18 at 5:21

The usual proof of the identity $\cos^2 t+\sin^2 t=1$ uses the Pythagorean Theorem. So a proof of the Pythagorean Theorem by using the identity is not correct.

True, we can define cosine and sine purely "analytically," by power series, or as the solutions of a certain differential equation. Then we can prove $\cos^2 t+\sin^2 t=1$ without any appeal to geometry.

But we still need geometry to link these "analytically" defined functions to sides of right-angled triangles.

Remark: The question is very reasonable. The logical interdependencies between various branches of mathematics are usually not clearly described. This is not necessarily always a bad thing. The underlying ideas of the calculus were for a while quite fuzzy, but calculus was still used effectively to solve problems, Similar remarks can be made about complex numbers.

  • $\begingroup$ That means I can't prove it this way $\endgroup$ – Atish Dipongkor - MVP Jul 21 '13 at 17:25
  • $\begingroup$ @Atish: Yes, it does mean that. Using $\cos^2\theta+\sin^2\theta=1$ within a proof of Pythagoras' theorem would be circular reasoning (which is not valid). $\endgroup$ – Zev Chonoles Jul 21 '13 at 17:27
  • $\begingroup$ Is that complete circular reasoning?? $\endgroup$ – Atish Dipongkor - MVP Jul 21 '13 at 17:31
  • $\begingroup$ You can prove Pythagoras for a triangle of hypotenuse $1$ and then generalise using similar triangles. It is just as easy to prove Pythagoras for a general right-angled triangle directly. $\endgroup$ – Mark Bennet Jul 21 '13 at 17:33
  • $\begingroup$ That does not mean, that any proof of Pythagoras' theorem from $\cos^2 t + \sin^2 t = 1$ is wrong. It would be complicated, boring and I would say pointless, but it doesn't mean that you can't prove it this way. You certainly can! The connection between analytic and geometric definitions of a triangle is surely there ;-) $\endgroup$ – dtldarek Jul 21 '13 at 19:05

My take on this is that in Euclidean space the Pythagorean theorem is equivalent to $\sin^2(\theta)+\cos^2(\theta)=1$. One simply uses similar triangles - every right-angled triangle is similar to a triangle with hypotenuse $1$. The sin and cos functions make sense in the Euclidean plane because similarity preserves the ratios between lengths and the angles between lines.

There are some quite deep geometrical ideas here. In non-euclidean geometry we don't have the same simple scale invariance (similarity) to work with. So the parallel postulate is essential to the proof.


Euclidean geometry relates to entities such as lines, points, angles which satisfy a set of axioms. In this setting it is quite difficult to define what is the amplitude of an angle and what is the sinus and co-sinus of an angle. Instead Pythagoras Theorem is relatively simple to prove starting by Euclide's axioms. In such a setting the relation $\sin^2 \theta + \cos^2 \theta = 1$ would be a consequence of Pythagoras Theorem.

Nowadays trigonometric functions are defined by means of purely analytical tools (such has Taylor series) which have no dependency on Euclide axioms but rely on the axioms of real numbers. In this setting one can define the Euclidean Plane as a 2-dimensional real affine space with a scalar product. In this case Pythagoras Theorem could be proven as you suggest, but actually it would be anyway over-complicated because Pythagoras Theorem is then a simple consequence of the purely algebraic fact (ensured by the properties of a scalar product): $$ (v-w)^2 = v^2 + w^2 \qquad \text{if}\qquad (v,w) = 0. $$


Without any assumptions about what properties the distance formula satisfies, you can't prove anything about what the distance formula is. If you take on blind faith that it has been proven that $\forall x \in \mathbb{R}\sin^2(x) + \cos^2(x) = 1$ and you assume that distance is a binary function from $\mathbb{R}^2$ to $\mathbb{R}$, in other words, a function from $(\mathbb{R}^2)^2$ to $\mathbb{R}$ satisfying the following properties:

  1. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
  2. $\forall x \in \mathbb{R}\forall \text{ nonnegative }y \in \mathbb{R}d((0, 0), (y \cos(x), y \sin(x))) = y$

Then you can use them to prove that the distance formula is $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$. This just shows that the Pythagorean theorem holds for all right angle triangles whose legs are parallel to the axes. If you make the additional assumption $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$, then you can show that the Pythagoren theorem holds for all right angle triangles, not just ones whose legs are parallel to the axes.

How can you prove the Pythagoren identity? If you take on blind faith that the Pythagoren theorem is a well known theorem and therefore must be true, then you can deduce the Pythagorean identity. Then how do you prove the Pythagorean theorem in the first place? If you take for granted that distance satisfies the following properties:

  1. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
  2. $\forall x \in \mathbb{R}\forall \text{ nonnegative }y \in \mathbb{R}d((0, 0), (y \cos(x), y \sin(x))) = y$
  3. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$

Then you can prove the Pythagoren theorem as follows. $\cos$ and $\sin$ are defined by the following differential equations

  • $\cos(0) = 1$
  • $\sin(0) = 0$
  • $\cos' = -\sin$
  • $\sin' = \cos$

so $\frac{d}{dx}\cos^2(x) + \sin^2(x) = \frac{d}{dx}\cos^2(x) + \frac{d}{dx}\sin^2(x) = 2\cos(x)(-\sin(x)) + 2\sin(x)\cos(x) = 0$ so $\cos^2(x) + \sin^2(x)$ is constant. Also $\cos^2(0) + \sin^2(0) = 1$ so $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$.

From this, we can show that the distance formula is $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$. Since the distance formula also satisfies property 3, the Pythagorean theorem is true.

How do we know there even exists a way of defining distance that satisfies those properties? Since the proof that $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$ didn't even use the concept of distance, we know that it's true. From the statement $\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$, it's easy to show that the function $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ satisfies the first 2 properties. It can also be shown to satisfy the third property as follows. $d((0, 0), (xz - yw, xw + yz)) = \sqrt{(xz - yw)^2 + (xw + yz)^2} = \sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = \sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = \sqrt{(x^2 + y^2)(z^2 + w^2)} = \sqrt{x^2 + y^2}\sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$

Some people make other assumptions about what properties the distance formula satisfies. How do we know there exists a way of defining distance that satisfies all of them? Because it has been proven in this answer that $d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$ is the unique function satisfying the following properties

  1. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (x + z, y + w)) = d((0, 0), (z, w))$
  2. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((x, y), (z, w))$ is nonnegative
  3. $\forall \text{ nonnegative } x \in \mathbb{R}d((0, 0), (x, 0)) = x$
  4. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}d((0, 0), (x, -y)) = d((0, 0), (x, y))$
  5. $\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R}d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$

and it also satisfies the additional properties

  1. The area of any square is the square of the length of its edges
  2. $\forall x \in \mathbb{R}d((0, 0), (\cos(x), \sin(x))) = 1$

The second of the two assumptions I made about distance don't appear as any of there 7 assumptions. That's because that assumption can be deduced from properties 3, 5, and 7 of these 7 properties.


It is not true that you must use the Pythagorean theorem to prove that $sin^2(x)+ cos^2(x)= 1$. It depends upon how you have defined sine and cosine. It is, for example, perfectly proper to define $sin(x)= \sum_{n= 0}^\infty \frac{(-1)^n}{(2n+ 1)!}x^{2n+ 1}$ and $cos(x)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}$ or, equivalently, to define $sin(x)$ as the function satisfying the differential equation y''+ y= 0 with initial conditions y(0)= 0, y'(0)= 1 and $cos(x)$ as the function, y, satisfying the differential equation y''+ y= 0 with initial conditions y(0)= 1, y'(0)= 0. In either case, you can then prove that $sin^2(x)+ cos^2(x)= 1$ without reference to the Pythagorean theorem.

  • $\begingroup$ This doesn't appear to answer the question of how to prove the Pythagorean theorem using that identity. It just appears to be a reply to somebody else's answer to this question that says the proof of that identity uses the Pythagorean theorem. You didn't explain how after you prove that identity, you can prove the Pythagorean theorem from it. I'm not sure what the author is trying to ask but if he or she is asking what I guessed he or she was asking as described in my comment on the question, this answer would not answer that question. $\endgroup$ – Timothy Dec 3 '18 at 16:18
  • $\begingroup$ @user247327 I agree with Timothy's criticism of your answer, but I suspect that your approach is still workable. Starting with the conventional (opposite/hypotenuse), (adjacent/hypotenuse) definitions, leads (through the theory behind taylor polynomials) to the taylor series for sine and cosine. I suspect that using the taylor series as the starting def's (which presumably implies that $sin^2x + cos^2x = 1$) one would then have to use the theory behind the taylor series to derive that these formulas imply that sine and cosine "represent" opposite/hypotenuse and adjacent/hypotenuse. $\endgroup$ – user2661923 Jan 10 '19 at 9:48
  • $\begingroup$ @user2661923 I think to give a satisfactory answer, you need to use the assumption $d((0, 0), (\cos(x), \sin(x)) = 1$. Then using the assumptions $d((0, 0), (\cos(x), \sin(x))) = 1$ and $\sin^2(x) + \cos^2(x) = 1$ and some other assumptions, you can prove the Pythagorean theorem. I just recently edited my answer and think it is now pretty much a fixed up version of this answer. $\endgroup$ – Timothy May 27 '19 at 5:48
  • $\begingroup$ @user2661923 I now feel like this answer would not have been worth a down vote if André Nicolas' answer didn't exist and is worth one for no other reason than it doesn't add anything to it. Otherwise, it would not have been a totally useless answer because it can be figured out from the answer that the Pythagorean identity can be proven without reference to the Pythagorean theorem and then using the additional assumption $\forall x \in \mathbb{R}\forall \text{ nonnegative }y \in \mathbb{R}d((0, 0), (y \cos(x),y \sin(x))) = 1$ and some other intuitive properties of distance, you can prove the $\endgroup$ – Timothy Jun 27 '19 at 2:45
  • $\begingroup$ Pythagorean theorem. $\endgroup$ – Timothy Jun 27 '19 at 2:45

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