Trigonometric functions with the help of a circle (trigonometric ratios of any angle)

Trigonometric functions with the help of a circle

Recovering the right-triangle ratios geometrically. Take a right triangle with the right angle at the foot of a perpendicular, and an acute angle θ. Writing the sides as before, $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$ and $\sin\theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$ — this is just the familiar definition, set up so it can be generalised.

The general definition. Let θ be any angle, placed in standard position: its vertex at the origin O, and its initial ray OA along the positive x-axis. Consider a circle of radius r centred at O, and let OB be the terminal ray of θ. Let P(x, y) be the point where OB meets the circle. Drop a perpendicular from P to OA, meeting it at M. Then in the right triangle PMO, $OM = x$, $PM = y$, and $OP = r$, so

This matches the right-triangle definition exactly when θ is acute — but now nothing in the construction requires θ to be acute, or even positive. So for any θ ∈ ℝ, we simply define

Because every angle θ determines a unique point P on the circle (and every point on the circle determines a unique angle up to full revolutions), these six ratios really are functions of θ — hence the name trigonometric functions.

Two important consequences.

1. Independent of r. If we chose a bigger or smaller circle, P's coordinates would scale by the same factor as r, so the ratios x/r and y/r are unchanged. The trigonometric functions depend only on the angle θ, never on which circle we drew.
2. Coterminal angles agree. Angles that differ by a whole number of revolutions (360°, 720°, −360°, ...) have the same terminal ray, hence the same point P, hence identical trigonometric function values.

Why $r^2 = x^2+y^2$. Since P(x, y) lies on the circle of radius r centred at O, Pythagoras' theorem on triangle PMO gives $x^2+y^2 = r^2$. As θ varies, the pair $(x,y) = (r\cos\theta, r\sin\theta)$ traces out the whole circle.

The unit-circle special case. If we choose $r=1$ (the unit circle), the formulas collapse to their simplest form: $\cos\theta = x$ and $\sin\theta = y$ directly, i.e. $P(x,y) \equiv P(\cos\theta,\sin\theta)$. This is the form used throughout the rest of the chapter, since it is the cleanest to work with. …