The best proof in the world for the Pythagoras theorem

Snotzenfartz

[latex]cos^2\theta+sin^2\theta
= (cos\theta+isin\theta)(cos\theta-isin\theta)
= e^{i\theta}.e^{-i\theta}=1
[/latex]

Right?

Reek_The_Meek

Pythagoras' Theorem states that a^2+b^2=c^2 where c is the hypotenuse of a right angled triangle and a and b are the other two sides.

What you have there is a generalised version.

Snotzenfartz

wat?

Reek_The_Meek

This is Pythagoras' Theorem: http://mathworld.wolfram.com/PythagoreanTheorem.html

If you were to take the unit circle (i.e. the radius = 1) and let the radius be your hypotenuse, the opposite and adjacent sides are (sinx)^2 and (cosx)^2.
That hasn't been explained very well so here's a picture: https://wiki.eee.uci.edu/images/6/6e/Triangle_on_unit_circle.jpg

Jan Hus

Sin and cos are based on the ratios present in a right angled triangle, which must be proven using Pythagoras's theorem. That is like me performing this proof. a^2=b^2+c^2-2abcosA. cos90=0 therefore a^2=b^2+c^2. The exception would exist when you can prove something independently and then using that to prove whatever. The major example would be using Stoke's Theorem to prove the fundamental theory of calculus.

Pole Monkey

It's possible to derive Euler's formula without using the Pythagoras theorem.

Jan Hus

Pole Monkey wrote: »

It's possible to derive Euler's formula without using the Pythagoras theorem.

To the best of my knowledge it is not. I am aware that you can use the equation of a circle to derive it, but that itself is based on Pythagoras.
If your proof differs from this, explain how. ( If you are using inner space products be aware that the dot product is not independent either, though this may not apply for all inner products)

Pole Monkey

Sure it's possible. You can derive it from the sine addition formula,

[latex]sin(a + b) = sinacosb + cosasinb[/latex]

You don't need the pythagoras theorem to derive that.

Jan Hus

Pole Monkey wrote: »

Sure it's possible. You can derive it from the sine addition formula,

[latex]sin(a + b) = sinacosb + cosasinb[/latex]

You don't need the pythagoras theorem to derive that.

But both sin and cos (and of course the relationship between them) comes from the ratios of angles in a right angled triangle... at least that is my understanding.