Sum of Squares of Sine and Cosine/Proof 2

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Theorem

$\cos^2 x + \sin^2 x = 1$


Proof

From the trigonometric definitions of sine and cosine:

\(\displaystyle \sin x\) \(=\) \(\displaystyle \frac{\text{opposite} } {\text{hypotenuse} }\)
\(\displaystyle \cos x\) \(=\) \(\displaystyle \frac{\text{adjacent} } {\text{hypotenuse} }\)

Then:

\(\displaystyle \sin^2 x + \cos^2 x\) \(=\) \(\displaystyle \frac{\text{opposite}^2 + \text{adjacent}^2} {\text{hypotenuse}^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac{\text{hypotenuse}^2} {\text{hypotenuse}^2}\) Pythagoras's Theorem
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$