Show that \(sin^2\theta + cos^2\theta = 1\).

By definition, the sine of an angle \(\theta\) is the side opposite the angle divided by the hypotenuse in a right triangle. Let the side opposite of such a triangle be equal to \(y\) units and the hypotenuse of said right triangle equal to \(h\) units.

$${sin\theta} = {y \over h}$$

Similarly, the cosine of an angle \(\theta\) is the side adjacent the angle divided by the hypotenuse in a right triangle. let the side adjacent equal \(x\) units and, as before, the hypotenuse equal \(h\) units.

$${cos\theta} = {x \over h}$$

The definition of a circle of radius \(r\) centered at the origin is all points satisfying the equation \(x^2 + y^2 = r^2\).

$${x^2} + {y^2} = {r^2}$$

Let's set the radius of such a circle to \(h\). We get:

$${x^2} + {y^2} = {h^2}$$

Next, we divide all terms by \(h^2\) to get the following:

$${x^2 \over h^2} + {y^2 \over h^2} = {h^2 \over h^2}$$

Simplified, this can be written:

$${\Bigl(} {x \over h}{\Bigr)^2} + {\Bigl(} {y \over h}{\Bigr)^2} = {1}$$

Since addition is commutative, we have:

$${sin^2\theta} + {\cos^2\theta} = {1}$$