Can I prove Pythagoras' Theorem using that $\sin^2(\theta)+\cos^2(\theta)=1$?

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.

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Please see here for a guide to writing math with MathJax, and see here for a guide to formatting posts with Markdown. – Zev Chonoles Jul 21 '13 at 17:16
The identity $\cos^2\theta + \sin^2\theta = 1$ is based on the Pythagorean Theorem. So I would consider this inappropriate. – Ted Shifrin Jul 21 '13 at 17:17
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 – Atish Dipongkor Jul 21 '13 at 17:23
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. – goblin May 8 at 14:41

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.

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That means I can't prove it this way – Atish Dipongkor Jul 21 '13 at 17:25
@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). – Zev Chonoles Jul 21 '13 at 17:27
Is that complete circular reasoning?? – Atish Dipongkor Jul 21 '13 at 17:31
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. – Mark Bennet Jul 21 '13 at 17:33
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 ;-) – 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.

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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): $$(v-w)^2 = v^2 + w^2 \qquad \text{if}\qquad (v,w) = 0.$$

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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.

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