Welcome to the Calculus series.

This article's purpose is to outline all the trigonometric knowledge necessary for moving onto the next topic. This is a single long article, so it might not be as detailed as you need to fully understand the concepts. Refer to the Khan Academy trigonometry course if necessary.

Ratios

The trigonometric functions can all be defined in relation to a right triangle.

Everything is relative to the angle labeled with $\theta$. Three sides can be labeled: adjacent, opposite, and hypotenuse. "Hypotenuse" refers to the longest side of a right triangle. "Adjacent" is the other side that makes up the angle. "Opposite" is the side that doesn't touch the angle at all.

The three fundamental trigonometric functions are sine, cosine, and tangent, notated as: $\sin(\theta)$, $\cos(\theta)$, and $\tan(\theta)$.

$\sin(\theta)$ is $\frac{Opposite}{Hypotenuse}$.

$\cos(\theta)$ is $\frac{Adjacent}{Hypotenuse}$.

$\tan(\theta)$ is $\frac{\sin(\theta)}{\cos(\theta)}$.

$\frac{\sin(\theta)}{\cos(\theta)}$ can be expanded into

$\frac{Opposite}{Hypotenuse} \div \frac{Adjacent}{Hypotenuse}$

Remember that dividing a fraction by another fraction is multiplying by the reciprocal of the denominator.

$\frac{Opposite}{Hypotenuse} \cdot \frac{Hypotenuse}{Adjacent}$

Which ultimately simplifies to

$\tan(\theta) = \frac{Opposite}{Adjacent}$.

Use the mnemonic "Soh Cah Toa" to memorize these ratios:

• Sine is Opposite divided by Hypotenuse
• Cosine is Adjacent divided by Hypotenuse
• Tangent is Opposite divided by Adjacent

Let's run with this idea and look at some special triangles.

Special Triangles

Take a look at this equilateral triangle.

Each side measures 1 unit and all interior angles measure $60^\circ$. A dotted line extends from the top vertex and cuts the bottom side in half.

To find $h$, use the Pythagorean theorem.

$\frac{1}{2}^2 + h^2 = 1^2$

$\frac{1}{4} + h^2 = 1$

$h^2 = \frac{3}{4}$

$h = \frac{\sqrt3}{2}$

Using knowledge of similar triangles, you now know that all triangles with interior angles $30^\circ$, $60^\circ$, and $90^\circ$, the ratio of the side lengths is $1 : \frac{\sqrt3}{2} : \frac{1}{2}$.

Even further, since we have all the side lengths of this right triangle, we can calculate the trig ratios for the acute angles.

For $\sin(30^\circ)$, the opposite side measures $\frac{1}{2}$ units and the hypotenuse is 1 unit. Therefore, $\sin(30^\circ)$ is $\frac{1}{2}$. Try out these examples:

1. $\cos(30^\circ)$

   Solution

   Adjacent side measures $\frac{\sqrt3}{2}$ units, Hypotenuse is 1 unit.

   $\cos(30^\circ) = \boxed{\frac{\sqrt3}{2}}$

2. $\sin(60^\circ)$

   Solution

   Opposite side measures $\frac{\sqrt3}{2}$ units, Hypotenuse is 1 unit.

   $\sin(60^\circ) = \boxed{\frac{\sqrt3}{2}}$

3. $\cos(60^\circ)$

   Solution

   Adjacent side measures $\frac{1}{2}$ units, Hypotenuse is 1 unit.

   $\cos(60^\circ) = \boxed{\frac{1}{2}}$

Let's look at another right triangle we can use the Pythagorean theorem on. This time, it's isosceles.

We've set the hypotenuse to be 1 unit long. Calculating length of the legs $l$:

$l^2 + l^2 = 1^2$

$2l^2 = 1$

$l^2 = \frac{1}{2}$

$l = \frac{1}{\sqrt2}$

$l = \frac{\sqrt2}{2}$

Again, using knowledge of similar triangles, if a right triangle is isosceles, the ratio of its side lengths is $1 : \frac{\sqrt2}{2} : \frac{\sqrt2}{\sqrt2}$.