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Proof: Pythagorean Identity

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\sin^2 \theta + \cos^2 \theta = 1, \forall \theta \in \R

Proof Edit

Prerequisites Edit

  • The antiderivative of 0 is a constant
  • Series definition of sine and cosine (in particular \sin(0) = 0 and \cos(0) = 1)
  • Differential of \sin is \cos, differential of \cos is -\sin
  • Linearity of the derivative, the Chain rule

Proof Edit

\forall \theta \in \R,

\frac{d}{d\theta}( \sin^2 \theta + \cos^2 \theta ) = \frac{d}{d\theta} ( \sin^2 \theta ) + \frac{d}{d\theta} ( \cos^2 \theta ) (linearity of the derivative)

 = \frac{d}{d\theta}(\sin\theta)\frac{d}{d\sin\theta} (\sin^2\theta) + \frac{d}{d\theta}(\cos\theta)\frac{d}{d\cos\theta} (\cos^2\theta) (chain rule)

 = 2\sin\theta\cos\theta - 2\cos\theta\sin\theta (evaluating the differentials)

 = 0

As the derivative of the expression is zero, this implies \sin^2 \theta + \cos^2 \theta = k, \forall \theta \in \R for some constant k. Evaluating at \theta = 0, \sin^2 0 + \cos^2 0 = 0 + 1 = 1, which means k=1, implying \sin^2 \theta + \cos^2 \theta = 1, \forall \theta \in \R.

Geometric "proof" Edit

It is possible to use geometry to prove the statement, however it only holds for 0 < \theta < \frac{\pi}{2}

Prerequisites Edit

Proof Edit

Given an arbitrary right triangle, the following are true:

\sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}

\cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}

Here,

\rm \sin^2\theta = \left({opposite\over hypotenuse}\right)^2

\rm \cos^2\theta = \left({adjacent\over hypotenuse} \right)^2

And therefore,

 \mathrm \sin^2\theta + \cos^2 \theta =  {opposite ^2\over hypotenuse^2} + {adjacent^2 \over hypotenuse^2} = {opposite^2 + adjacent^2 \over hypotenuse^2}

Via the Pythagorean Theorem, the legs (here: opposite and adjacent) are "a," and "b" where hypotenuse is "c".

a^2 + b^2 = c^2

\mathrm{adjacent}^2 + \mathrm{opposite}^2 = \mathrm{hypotenuse}^2

\mathrm {adjacent^2 + opposite^2 \over hypotenuse^2} = 1

So,

\mathrm sin^2 \theta + \cos^2 \theta = 1

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