Right Triangle Ratios

Consider an angle $\theta$ between 0 and $\pi/2$ ($0^\circ$ and $90^\circ$ ) in standard position, and a perpendicular segment extending from the terminal side of $\theta$ to the $x$ -axis. For now, we consider the perpendicular extending from the intersection of the terminal side of $\theta$ and the unit circle, i.e. the perpendicular extends from $(\cos{\theta},\sin{\theta})$ to the $x$ -axis.

Notice that the figure formed by the terminal side of $\theta,$ the perpendicular segment, and the $x$ -axis, is a right triangle. As a result of the circular definitions given previously, the vertical leg of this right triangle (which is the side opposite $\theta,$ ) has length $\sin{\theta}.$ The horizontal leg (which is the side adjacent $\theta,$ ) has length $\cos{\theta}.$ The hypotenuse has length 1, since it corresponds to a radius of the unit circle. By the Pythagorean Theorem, we immediately see that $\sin^2 x+\cos^2 x=1,$ which is a commonly used identity. More on identities later.

Consider the ratios of sides of this triangle:

$$\frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sin\theta}{1} = \sin\theta \frac{\text{hypotenuse}}{\text{opposite}} = \frac{1}{\sin\theta} = \csc\theta$$

$$\frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\cos\theta}{1} = \cos\theta \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{1}{\cos\theta} = \sec\theta$$

$$\frac{\text{opposite}}{\text{adjacent}} = \frac{\sin\theta}{\cos\theta} = \tan\theta \frac{\text{adjacent}}{\text{opposite}} = \frac{\cos\theta}{\sin\theta} = \cot\theta$$

Since similar triangles have corresponding sides that are in the same proportion, the values of these ratios are equal for a given acute angle $\theta$ regardless of the right triangle it appears in. In other words, $\sin\theta=\frac{\text{opposite}}{1}$ for this particular right triangle with hypotenuse 1, but in general $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ for any right triangle, and likewise for the other ratios.

So for $0<\theta<\frac{\pi}{2}$ we define the trigonometric ratios of a right triangle:

$$\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} \csc\theta = \frac{\text{hypotenuse}}{\text{opposite}}$$

$$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} \sec\theta =\frac{\text{hypotenuse}}{\text{adjacent}}$$

$$\tan\theta = \frac{\text{opposite}}{\text{adjacent}} \cot\theta =\frac{\text{adjacent}}{\text{opposite}}$$