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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 $0 < \theta < \frac{\pi}{2}$

### 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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