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

Proof

Prerequisites

  • 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

$ \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"

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

Prerequisites

Proof

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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