Mixing
Can I prove Pythagoras' Theorem using that $sin^2(theta)+cos^2(theta)=1$?
$begingroup$
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.
geometry
edited Jul 21 '13 at 17:16
Zev Chonoles
110k16228425
asked Jul 21 '13 at 17:11
Atish DipongkorAtish Dipongkor
1434
$endgroup$
|
show 2 more comments
$begingroup$
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.
geometry
edited Jul 21 '13 at 17:16
Zev Chonoles
110k16228425
asked Jul 21 '13 at 17:11
Atish DipongkorAtish Dipongkor
1434
$endgroup$
1
$begingroup$
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
1
$begingroup$
The identity $cos^2theta + sin^2theta = 1$ 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
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: $$|x-y|^2 = (x-y) bullet (x-y) = x bullet x- xbullet y-y bullet x+y bullet y = |x|^2+|y|^2$$ 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
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
|
show 2 more comments
$begingroup$
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.
geometry
edited Jul 21 '13 at 17:16
Zev Chonoles
110k16228425
asked Jul 21 '13 at 17:11
Atish DipongkorAtish Dipongkor
1434
$endgroup$
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.
geometry
geometry
edited Jul 21 '13 at 17:16
Zev Chonoles
110k16228425
asked Jul 21 '13 at 17:11
Atish DipongkorAtish Dipongkor
1434
edited Jul 21 '13 at 17:16
Zev Chonoles
110k16228425
asked Jul 21 '13 at 17:11
Atish DipongkorAtish Dipongkor
1434
edited Jul 21 '13 at 17:16
Zev Chonoles
110k16228425
edited Jul 21 '13 at 17:16
Zev Chonoles
110k16228425
edited Jul 21 '13 at 17:16
Zev Chonoles
110k16228425
110k16228425
asked Jul 21 '13 at 17:11
Atish DipongkorAtish Dipongkor
1434
asked Jul 21 '13 at 17:11
Atish DipongkorAtish Dipongkor
1434
asked Jul 21 '13 at 17:11
Atish DipongkorAtish Dipongkor
1434
1434
1
$begingroup$
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
1
$begingroup$
The identity $cos^2theta + sin^2theta = 1$ 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
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: $$|x-y|^2 = (x-y) bullet (x-y) = x bullet x- xbullet y-y bullet x+y bullet y = |x|^2+|y|^2$$ 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
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
|
show 2 more comments
1
$begingroup$
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
1
$begingroup$
The identity $cos^2theta + sin^2theta = 1$ 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
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: $$|x-y|^2 = (x-y) bullet (x-y) = x bullet x- xbullet y-y bullet x+y bullet y = |x|^2+|y|^2$$ 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
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
1
1
$begingroup$
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
$begingroup$
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
1
1
$begingroup$
The identity $cos^2theta + sin^2theta = 1$ is based on the Pythagorean Theorem. So I would consider this inappropriate.
$endgroup$
– Ted Shifrin
Jul 21 '13 at 17:17
$begingroup$
The identity $cos^2theta + sin^2theta = 1$ 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
Jul 21 '13 at 17:23
$begingroup$
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
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: $$|x-y|^2 = (x-y) bullet (x-y) = x bullet x- xbullet y-y bullet x+y bullet y = |x|^2+|y|^2$$ 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
May 8 '16 at 14:41
$begingroup$
Its probably worth adding that in the framework of inner product spaces, Pythagoras is remarkably easy to prove: $$|x-y|^2 = (x-y) bullet (x-y) = x bullet x- xbullet y-y bullet x+y bullet y = |x|^2+|y|^2$$ 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
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
$begingroup$
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
|
show 2 more comments
5 Answers
5
active
oldest
votes
$begingroup$
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.
edited May 6 '16 at 14:39
Patrick Abraham
17912
answered Jul 21 '13 at 17:22
André NicolasAndré Nicolas
452k36423808
$endgroup$
$begingroup$
That means I can't prove it this way
$endgroup$
– Atish Dipongkor
Jul 21 '13 at 17:25
$begingroup$
@Atish: Yes, it does mean that. Using $cos^2theta+sin^2theta=1$ 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
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
|
show 2 more comments
$begingroup$
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.
answered Jul 21 '13 at 17:31
Mark BennetMark Bennet
80.9k981179
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add a comment |
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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):
$$
(v-w)^2 = v^2 + w^2 qquad text{if}qquad (v,w) = 0.
$$
answered May 21 '15 at 13:55
Emanuele PaoliniEmanuele Paolini
17.8k22052
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add a comment |
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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.
answered May 6 '16 at 14:52
user247327user247327
10.6k1515
$endgroup$
$begingroup$
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
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@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 at 9:48
add a comment |
$begingroup$
Yes you can. None of the other answers give a rigorous proof of the Pythagorean theorem but this one does. I'll show that $forall x in mathbb{R}, (sin(x))^2 + (cos(x))^2 = 1$ then will decide that as a result of that, I will define the distance from any points in $mathbb{R}^2$, $(x, y)$ and $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$. Then I will show that according to this definition of distance, the Pythagorean theorem holds for all right angle triangles in $mathbb{R}^2$, not just those whose legs are parallel to one of the axes.
Sin and cos are defined by the following differential equations
- $cos(0) = 1$
- $sin(0) = 0$
- $forall x in mathbb{R} sin'(x) = cos(x)$
- $forall x in mathbb{R} cos'{x} = -sin(x)$
Then $forall x in mathbb{R} frac{d}{dx}((sin(x))^2 + (cos(x))^2) =frac{d}{dx}((sin(x))^2) + frac{d}{dx}((cos(x))^2) = 2sin(x)sin'(x) + 2cos(x)cos'(x) = 2sin(x)cos(x) -2cos(x)sin(x) = 0$ so $(sin(x))^2 + (cos(x))^2$ is constant. Also $(sin(0))^2 + (cos(0))^2 = 1$ so $forall x in mathbb{R} (sin(x))^2 + (cos(x))^2 = 1$
As a result of this, I will define the distance from $(x, y)$ to $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$. For convenience, I will also define the distance of any point $(x, y)$, $d(x, y)$ as the distance from $(0, 0)$ to $(x, y)$. It can be shown that using this definition of distance, the Pythagorean theorem is equivalent to the statement that for any two points $(x, y)$ and $(z, w)$, $d(xz - yw, xw + yz) = d(x, y)d(z, w)$ This is true because $d(xz - yw, xw + yz) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d(x, y)d(z, w)$
answered Jan 6 at 1:07
TimothyTimothy
303212
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5 Answers
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5 Answers
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$begingroup$
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.
edited May 6 '16 at 14:39
Patrick Abraham
17912
answered Jul 21 '13 at 17:22
André NicolasAndré Nicolas
452k36423808
$endgroup$
$begingroup$
That means I can't prove it this way
$endgroup$
– Atish Dipongkor
Jul 21 '13 at 17:25
$begingroup$
@Atish: Yes, it does mean that. Using $cos^2theta+sin^2theta=1$ 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
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
|
show 2 more comments
$begingroup$
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.
edited May 6 '16 at 14:39
Patrick Abraham
17912
answered Jul 21 '13 at 17:22
André NicolasAndré Nicolas
452k36423808
$endgroup$
$begingroup$
That means I can't prove it this way
$endgroup$
– Atish Dipongkor
Jul 21 '13 at 17:25
$begingroup$
@Atish: Yes, it does mean that. Using $cos^2theta+sin^2theta=1$ 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
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
|
show 2 more comments
$begingroup$
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.
edited May 6 '16 at 14:39
Patrick Abraham
17912
answered Jul 21 '13 at 17:22
André NicolasAndré Nicolas
452k36423808
$endgroup$
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.
edited May 6 '16 at 14:39
Patrick Abraham
17912
answered Jul 21 '13 at 17:22
André NicolasAndré Nicolas
452k36423808
edited May 6 '16 at 14:39
Patrick Abraham
17912
edited May 6 '16 at 14:39
Patrick Abraham
17912
edited May 6 '16 at 14:39
Patrick Abraham
17912
17912
answered Jul 21 '13 at 17:22
André NicolasAndré Nicolas
452k36423808
answered Jul 21 '13 at 17:22
André NicolasAndré Nicolas
452k36423808
answered Jul 21 '13 at 17:22
André NicolasAndré Nicolas
452k36423808
452k36423808
$begingroup$
That means I can't prove it this way
$endgroup$
– Atish Dipongkor
Jul 21 '13 at 17:25
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@Atish: Yes, it does mean that. Using $cos^2theta+sin^2theta=1$ within a proof of Pythagoras' theorem would be circular reasoning (which is not valid).
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– Zev Chonoles
Jul 21 '13 at 17:27
$begingroup$
Is that complete circular reasoning??
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– Atish Dipongkor
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
|
show 2 more comments
$begingroup$
That means I can't prove it this way
$endgroup$
– Atish Dipongkor
Jul 21 '13 at 17:25
$begingroup$
@Atish: Yes, it does mean that. Using $cos^2theta+sin^2theta=1$ 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
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
$begingroup$
That means I can't prove it this way
$endgroup$
– Atish Dipongkor
Jul 21 '13 at 17:25
$begingroup$
That means I can't prove it this way
$endgroup$
– Atish Dipongkor
Jul 21 '13 at 17:25
$begingroup$
@Atish: Yes, it does mean that. Using $cos^2theta+sin^2theta=1$ within a proof of Pythagoras' theorem would be circular reasoning (which is not valid).
$endgroup$
– Zev Chonoles
Jul 21 '13 at 17:27
$begingroup$
@Atish: Yes, it does mean that. Using $cos^2theta+sin^2theta=1$ 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
Jul 21 '13 at 17:31
$begingroup$
Is that complete circular reasoning??
$endgroup$
– Atish Dipongkor
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
$begingroup$
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
$begingroup$
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
|
show 2 more comments
$begingroup$
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.
answered Jul 21 '13 at 17:31
Mark BennetMark Bennet
80.9k981179
$endgroup$
add a comment |
$begingroup$
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.
answered Jul 21 '13 at 17:31
Mark BennetMark Bennet
80.9k981179
$endgroup$
add a comment |
$begingroup$
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.
answered Jul 21 '13 at 17:31
Mark BennetMark Bennet
80.9k981179
$endgroup$
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.
answered Jul 21 '13 at 17:31
Mark BennetMark Bennet
80.9k981179
answered Jul 21 '13 at 17:31
Mark BennetMark Bennet
80.9k981179
answered Jul 21 '13 at 17:31
Mark BennetMark Bennet
80.9k981179
answered Jul 21 '13 at 17:31
Mark BennetMark Bennet
80.9k981179
80.9k981179
add a comment |
add a comment |
$begingroup$
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):
$$
(v-w)^2 = v^2 + w^2 qquad text{if}qquad (v,w) = 0.
$$
answered May 21 '15 at 13:55
Emanuele PaoliniEmanuele Paolini
17.8k22052
$endgroup$
add a comment |
$begingroup$
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):
$$
(v-w)^2 = v^2 + w^2 qquad text{if}qquad (v,w) = 0.
$$
answered May 21 '15 at 13:55
Emanuele PaoliniEmanuele Paolini
17.8k22052
$endgroup$
add a comment |
$begingroup$
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):
$$
(v-w)^2 = v^2 + w^2 qquad text{if}qquad (v,w) = 0.
$$
answered May 21 '15 at 13:55
Emanuele PaoliniEmanuele Paolini
17.8k22052
$endgroup$
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):
$$
(v-w)^2 = v^2 + w^2 qquad text{if}qquad (v,w) = 0.
$$
answered May 21 '15 at 13:55
Emanuele PaoliniEmanuele Paolini
17.8k22052
answered May 21 '15 at 13:55
Emanuele PaoliniEmanuele Paolini
17.8k22052
answered May 21 '15 at 13:55
Emanuele PaoliniEmanuele Paolini
17.8k22052
answered May 21 '15 at 13:55
Emanuele PaoliniEmanuele Paolini
17.8k22052
17.8k22052
add a comment |
add a comment |
$begingroup$
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.
answered May 6 '16 at 14:52
user247327user247327
10.6k1515
$endgroup$
$begingroup$
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 at 9:48
add a comment |
$begingroup$
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.
answered May 6 '16 at 14:52
user247327user247327
10.6k1515
$endgroup$
$begingroup$
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 at 9:48
add a comment |
$begingroup$
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.
answered May 6 '16 at 14:52
user247327user247327
10.6k1515
$endgroup$
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.
answered May 6 '16 at 14:52
user247327user247327
10.6k1515
answered May 6 '16 at 14:52
user247327user247327
10.6k1515
answered May 6 '16 at 14:52
user247327user247327
10.6k1515
answered May 6 '16 at 14:52
user247327user247327
10.6k1515
10.6k1515
$begingroup$
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 at 9:48
add a comment |
$begingroup$
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 at 9:48
$begingroup$
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
$begingroup$
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 at 9:48
$begingroup$
@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 at 9:48
add a comment |
$begingroup$
Yes you can. None of the other answers give a rigorous proof of the Pythagorean theorem but this one does. I'll show that $forall x in mathbb{R}, (sin(x))^2 + (cos(x))^2 = 1$ then will decide that as a result of that, I will define the distance from any points in $mathbb{R}^2$, $(x, y)$ and $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$. Then I will show that according to this definition of distance, the Pythagorean theorem holds for all right angle triangles in $mathbb{R}^2$, not just those whose legs are parallel to one of the axes.
Sin and cos are defined by the following differential equations
- $cos(0) = 1$
- $sin(0) = 0$
- $forall x in mathbb{R} sin'(x) = cos(x)$
- $forall x in mathbb{R} cos'{x} = -sin(x)$
Then $forall x in mathbb{R} frac{d}{dx}((sin(x))^2 + (cos(x))^2) =frac{d}{dx}((sin(x))^2) + frac{d}{dx}((cos(x))^2) = 2sin(x)sin'(x) + 2cos(x)cos'(x) = 2sin(x)cos(x) -2cos(x)sin(x) = 0$ so $(sin(x))^2 + (cos(x))^2$ is constant. Also $(sin(0))^2 + (cos(0))^2 = 1$ so $forall x in mathbb{R} (sin(x))^2 + (cos(x))^2 = 1$
As a result of this, I will define the distance from $(x, y)$ to $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$. For convenience, I will also define the distance of any point $(x, y)$, $d(x, y)$ as the distance from $(0, 0)$ to $(x, y)$. It can be shown that using this definition of distance, the Pythagorean theorem is equivalent to the statement that for any two points $(x, y)$ and $(z, w)$, $d(xz - yw, xw + yz) = d(x, y)d(z, w)$ This is true because $d(xz - yw, xw + yz) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d(x, y)d(z, w)$
answered Jan 6 at 1:07
TimothyTimothy
303212
$endgroup$
add a comment |
$begingroup$
Yes you can. None of the other answers give a rigorous proof of the Pythagorean theorem but this one does. I'll show that $forall x in mathbb{R}, (sin(x))^2 + (cos(x))^2 = 1$ then will decide that as a result of that, I will define the distance from any points in $mathbb{R}^2$, $(x, y)$ and $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$. Then I will show that according to this definition of distance, the Pythagorean theorem holds for all right angle triangles in $mathbb{R}^2$, not just those whose legs are parallel to one of the axes.
Sin and cos are defined by the following differential equations
- $cos(0) = 1$
- $sin(0) = 0$
- $forall x in mathbb{R} sin'(x) = cos(x)$
- $forall x in mathbb{R} cos'{x} = -sin(x)$
Then $forall x in mathbb{R} frac{d}{dx}((sin(x))^2 + (cos(x))^2) =frac{d}{dx}((sin(x))^2) + frac{d}{dx}((cos(x))^2) = 2sin(x)sin'(x) + 2cos(x)cos'(x) = 2sin(x)cos(x) -2cos(x)sin(x) = 0$ so $(sin(x))^2 + (cos(x))^2$ is constant. Also $(sin(0))^2 + (cos(0))^2 = 1$ so $forall x in mathbb{R} (sin(x))^2 + (cos(x))^2 = 1$
As a result of this, I will define the distance from $(x, y)$ to $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$. For convenience, I will also define the distance of any point $(x, y)$, $d(x, y)$ as the distance from $(0, 0)$ to $(x, y)$. It can be shown that using this definition of distance, the Pythagorean theorem is equivalent to the statement that for any two points $(x, y)$ and $(z, w)$, $d(xz - yw, xw + yz) = d(x, y)d(z, w)$ This is true because $d(xz - yw, xw + yz) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d(x, y)d(z, w)$
answered Jan 6 at 1:07
TimothyTimothy
303212
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Yes you can. None of the other answers give a rigorous proof of the Pythagorean theorem but this one does. I'll show that $forall x in mathbb{R}, (sin(x))^2 + (cos(x))^2 = 1$ then will decide that as a result of that, I will define the distance from any points in $mathbb{R}^2$, $(x, y)$ and $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$. Then I will show that according to this definition of distance, the Pythagorean theorem holds for all right angle triangles in $mathbb{R}^2$, not just those whose legs are parallel to one of the axes.
Sin and cos are defined by the following differential equations
- $cos(0) = 1$
- $sin(0) = 0$
- $forall x in mathbb{R} sin'(x) = cos(x)$
- $forall x in mathbb{R} cos'{x} = -sin(x)$
Then $forall x in mathbb{R} frac{d}{dx}((sin(x))^2 + (cos(x))^2) =frac{d}{dx}((sin(x))^2) + frac{d}{dx}((cos(x))^2) = 2sin(x)sin'(x) + 2cos(x)cos'(x) = 2sin(x)cos(x) -2cos(x)sin(x) = 0$ so $(sin(x))^2 + (cos(x))^2$ is constant. Also $(sin(0))^2 + (cos(0))^2 = 1$ so $forall x in mathbb{R} (sin(x))^2 + (cos(x))^2 = 1$
As a result of this, I will define the distance from $(x, y)$ to $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$. For convenience, I will also define the distance of any point $(x, y)$, $d(x, y)$ as the distance from $(0, 0)$ to $(x, y)$. It can be shown that using this definition of distance, the Pythagorean theorem is equivalent to the statement that for any two points $(x, y)$ and $(z, w)$, $d(xz - yw, xw + yz) = d(x, y)d(z, w)$ This is true because $d(xz - yw, xw + yz) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d(x, y)d(z, w)$
answered Jan 6 at 1:07
TimothyTimothy
303212
$endgroup$
Yes you can. None of the other answers give a rigorous proof of the Pythagorean theorem but this one does. I'll show that $forall x in mathbb{R}, (sin(x))^2 + (cos(x))^2 = 1$ then will decide that as a result of that, I will define the distance from any points in $mathbb{R}^2$, $(x, y)$ and $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$. Then I will show that according to this definition of distance, the Pythagorean theorem holds for all right angle triangles in $mathbb{R}^2$, not just those whose legs are parallel to one of the axes.
Sin and cos are defined by the following differential equations
- $cos(0) = 1$
- $sin(0) = 0$
- $forall x in mathbb{R} sin'(x) = cos(x)$
- $forall x in mathbb{R} cos'{x} = -sin(x)$
Then $forall x in mathbb{R} frac{d}{dx}((sin(x))^2 + (cos(x))^2) =frac{d}{dx}((sin(x))^2) + frac{d}{dx}((cos(x))^2) = 2sin(x)sin'(x) + 2cos(x)cos'(x) = 2sin(x)cos(x) -2cos(x)sin(x) = 0$ so $(sin(x))^2 + (cos(x))^2$ is constant. Also $(sin(0))^2 + (cos(0))^2 = 1$ so $forall x in mathbb{R} (sin(x))^2 + (cos(x))^2 = 1$
As a result of this, I will define the distance from $(x, y)$ to $(z, w)$ as $sqrt{(z - x)^2 + (w - y)^2}$. For convenience, I will also define the distance of any point $(x, y)$, $d(x, y)$ as the distance from $(0, 0)$ to $(x, y)$. It can be shown that using this definition of distance, the Pythagorean theorem is equivalent to the statement that for any two points $(x, y)$ and $(z, w)$, $d(xz - yw, xw + yz) = d(x, y)d(z, w)$ This is true because $d(xz - yw, xw + yz) = sqrt{(xz - yw)^2 + (xw + yz)^2} = sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = sqrt{(x^2 + y^2)(z^2 + w^2)} = sqrt{x^2 + y^2}sqrt{z^2 + w^2} = d(x, y)d(z, w)$
answered Jan 6 at 1:07
TimothyTimothy
303212
edited 2 days ago
answered Jan 6 at 1:07
TimothyTimothy
303212
answered Jan 6 at 1:07
TimothyTimothy
303212
answered Jan 6 at 1:07
TimothyTimothy
303212
303212
add a comment |
add a comment |
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– Zev Chonoles
Jul 21 '13 at 17:16
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The identity $cos^2theta + sin^2theta = 1$ is based on the Pythagorean Theorem. So I would consider this inappropriate.
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– Ted Shifrin
Jul 21 '13 at 17:17
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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
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– Atish Dipongkor
Jul 21 '13 at 17:23
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Its probably worth adding that in the framework of inner product spaces, Pythagoras is remarkably easy to prove: $$|x-y|^2 = (x-y) bullet (x-y) = x bullet x- xbullet y-y bullet x+y bullet y = |x|^2+|y|^2$$ 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.
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– goblin
May 8 '16 at 14:41
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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?
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– Timothy
Dec 3 '18 at 5:21