Trigonometric Identities

Theorem: $$\sin^2 x + \cos^2 x \equiv 1\!$$

Proof:


 * Starting with $$sin x$$ and $$cos x$$,

$$\sin x = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}$$

$$\cos x = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}$$


 * Squaring both sides and adding them together gives:

$$\sin^2 x + \cos^2 x = \frac{\mathrm{opposite}^2 + \mathrm{adjacent}^2}{\mathrm{hypotenuse}^2} \equiv 1$$

This can be extended to find other trigonometric functions:


 * Starting with $$\sin^2 x + \cos^2 x \equiv 1\!$$, divide every term by $$\cos^2 x \!$$ gives:

$$\tan^2 x + 1 \equiv \sec^2 x \!$$


 * Alternatively, dividing by $$\sin^2 x \!$$ gives:

$$1 + \cot^2 x \equiv \csc^2 x \!$$

Note: See this page for a proof of Pythagoras's Theorem