(see attached file)

My proof works like this:
1. Due to the fact that a right-angle is always created when the two points ("A" and "B") on a diameter line of a circle are joined to a third point ("C") anywhere on the circumference; then
2. All right angle triangles can be inscribed within a circle (work with me here).
3. Because sin2theta + cos2theta = 1, if we measure the angle (theta) from the centre of the circle to point "C", regardless of its angle, sin2theta + cos2theta = 1
4. The measurement of the angle (omega) across the diameter through the centre (180 degrees) also yeilds sin2(180)+cos2(180)=1
5. Therefore 1=1;
6. Therefore the sum of the squares sin(theta) + Cos(theta) = Square of the hypotenuse (omega).

Now: I need to know why this is a crock. I'm sure there are a thousand reasons why this is NOT a proof for Pythagoras, but I'm 43 years old and not mentally nimble enough to know why. Where have I gone wrong?

>1. Due to the fact that a right-angle is always created
>when the two points ("A" and "B") on a diameter line of a
>circle are joined to a third point ("C") anywhere on the
>circumference; then
Fine so far.

>2. All right angle triangles can be inscribed within a
>circle (work with me here).
I will work with you by explaining one way to make this step clearer. Let ABC be a triangle with a right angle at C, and let S be its circumcircle, O the centre of S. Then AOB = 2*ACB = 180, so O lies on AB. Thus AB is a diameter of S.

>3. Because sin2theta + cos2theta = 1, if we measure the
>angle (theta) from the centre of the circle to point "C",
>regardless of its angle, sin2theta + cos2theta = 1
While the result sin²è + cos²è = 1 is valid, the usual proof uses pythagoras' theorem, so unless you have another proof of it you should not use it to help you to prove pythagoras' theorem (that would be circular reasoning).

>4. The measurement of the angle (omega) across the diameter
>through the centre (180 degrees) also yeilds
>sin2(180)+cos2(180)=1
This is OK. The objection I gave in section 3 does not apply here as we know cos(180) = -1 and sin(180) = 0, so that cos²(180) + sin²(180) = 0² + (-1)² = 1.

>5. Therefore 1=1;
This does not make sense as a step in the argument; I suspect that instead the sentiment you wish to express here is 'now, by transitivity of equality this gives sin²è + cos²è = sin²(180) + cos²(180).

>6. Therefore the sum of the squares sin(theta) + Cos(theta)
>= Square of the hypotenuse (omega).
Pythagoras' theorem is about the sides, rather than the angles of the triangle. This statement about angles is rather different to that which you are trying to prove, as there is no simple way to relate the angles you use to the sides you are talking about without using pythagoras' theorem or something like it.

I'm afraid, therefore, that what you have given is not a proof. However, there are a lot of proofs out there if you want to see the sort of thing that does work.

Thankyou

sfwc
<><

sin²θ = cos²θ = 1.

Thankyou

sfwc
<><

However, the earlier step using sin^2 + cos^2 = 1 remains a difficulty. I know of no proof of this fact that is not already equivalent to the full pythagorean theorem, so if he did have a proof of this step, then all the other steps could be dispensed with.

--Stuart Anderson

I accept that the sine rule may be used with respect to other angles to show that Pythagoras' theorem is equivalent to sin² + cos² = 1; as you point out, this was exactly my criticism of step 3.

Thankyou

sfwc
<><

I was looking for a relationship between the angle Ø and the squares. If in-fact sin²Ø + cos²Ø is just a trigonometric way of rephrasing the original theorum, then I'm really just saying "Pythogoras = Pythagoras".

Hmmmm .... Back to the drawing board I guess.

Thank you for taking the time to look at my work.

Greg

Yes, I see. However, this is a separate flaw in his proof, since he has clearly indicated on his diagram that he intends the lengths of the sides AC and CB to be sin(θ) and cos(θ) respectively. This is of course wrong, and it is a factual error rather than a logical misstep, I think, as these sides are proportional to sin(θ/2) and cos(θ/2), not to sin(θ) and cos(θ). These are of course equal to the sines of the inscribed angles at A and B, which do mesh nicely with the law of sines. Similarly, AC can just as well be proven to equal 1 by using sin^2(ω/2)+cos^2(ω/2) = 1, and again, ω/2 is the inscribed angle at C, fitting with the law of sines. I confess I don't see why he wanted to use the angles at the circumcenter, but this looks like a quite correctable technical glitch rather than a showstopper. All he needs to do is reduce all his angles by a factor of 2 and then the law of sines reasoning will apply.

However, I agree that there is not much point in correcting step 6 when step 3 REALLY IS a showstopper. Thank you for an interesting discussion.

--Stuart Anderson