It's about relationships between the trig functions

Much of what you do in higher math will involve trigonometric functions and the relationships between them. Their unique properties and relationships sometimes even mean that we can substitute one or more trig functions for parts of other functions in order to make them solvable.

The Pythagorean Identity

The most useful relationship in trigonometry

We'll begin with the most important of all relationships between the trigonometric functions, the Pythagorean identity. It's very simple to derive.

We begin with a right triangle labeled with an angle, θ, its opposite and adjacent sides, and the hypotenuse:

Now express the sine function in terms of o and h, and square both sides:

$$sin{\theta} = \frac{o}{h}\: \color{#E90F89}{\longrightarrow} \: [sin(\theta)]^2 = \frac{o^2}{h^2}$$

Do the same with cosine:

$$cos{\theta} = \frac{a}{h}\: \color{#E90F89}{\longrightarrow} \: [cos(\theta)]^2 = \frac{a^2}{h^2}$$

Other versions of the Pythagorean identity

We can derive a couple of other Pythagorean-like identities from the first. Here's how: Start with the Pythagorean identity:

$$sin^2(\theta) + cos^2(\theta) = 1$$

Divide each term by sin²(θ):

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

Then simplify and use our definitions of cotangent and cosecant to get the second Pythagorean identity:

$$1 + cot^2(\theta) = csc^2(\theta)$$

Now let's go back to the Pythagorean identity and divide by cos²(θ) this time:

$$sin^2(\theta) + cos^2(\theta) = 1$$

One trig. function in terms of another / others

Many textbooks and curricula approach analytic trigonometry by asking you to memorize a bunch of trigonometric formulae. I'm not a huge fan of that. In fact, there is almost no trig. equation that can't be converted from one form to another by remembering the basic definitions of the trig functions and the Pythagorean identity.

The sum-angle identities: $sin(a + b)$ and $cos(a + b)$

There are many occasions to come where knowing how to calculate the sine and cosine of a sum of angles, a and b, will be a big advantage. It's not to difficult to figure out those formulas. We start with two right triangles, with angles a and b, stacked on top of one-another. Notice that the hypotenuse of the green triangle has a length of 1. That makes the other sides easy to calculate; they're just sin(b) and cos(b), as shown:

Now let's enclose these triangles inside a rectangle so that they're inscribed inside it. Inscribed means that all vertices and sides are as close to the rectangle as possible without going outside of it.

Then by recognizing that our hypotenuse of length 1 is a transversal of the parallel top and bottom sides of the rectangle, we can label that upper angle a+b (alternate interior angles are congruent). And we can label the two sides of the upper-left triangle as cos(a+b) and sin(a+b). Looks like we're headed in the right direction.

Now we need to label all of the other segments of our square. The goal is to show that sin(a+b) on the left is equal to the sum of two smaller segments on the right (of the rectangle), and that cos(a+b) is the difference between the length of the bottom side of the rectangle and the short segment on the top.

First the pink triangle. If we label the unknown sides as x and y (just temporary labels) we have

Now the cosine of angle a gets us side x:

$$ \begin{align} cos(a) &= \frac{x}{cos(b)} \\[4pt] x &= cos(a) \, cos(b) \end{align}$$

The double-angle identities: sin(2a) and cos(2a)

It's also convenient to have formulas that convert sines and cosines of double angles into functions of the measure of the single angle. We want to express, for example, sin(2a) as a function only of the angle a, not 2a.

Well, now that we have the sum formulas (above), that's pretty easy. Just make sin(a + b) be sin(a + a) and cos(a + b) = cos(a + a). The resulting formulae are easy to find by just making the substitution and simplifying:

The power-reduction formulas

Sometimes, and particularly when doing some integrals in calculus, it's good to have a relationship between the square of a sine, cosine or tangent function and sin(x), cos(x) or tan(x). Here's a set of those. We begin with the double-angle formula:

$$cos(2a) = cos^2(a) - sin^2(a)$$

Now substitute cos²(a) = 1 - sin²(a) from the Pythagorean identity:

$$cos(2a) = [1 - sin^2(a)] - sin^2(a)$$

Simplify that to

$$cos(2a) = 1 - 2 sin^2 (a) \; \; \color{#E90F89}{\text{*}}$$

and rearrange to solve for sin²(a):

$$sin^2(a) = \frac{cos(2a) - 1}{2}$$

The half-angle identities: sin(a/2) and cos(a/2)

Well, why not find a formula for a half-angle. It turns out that formulae like these are very convenient to have around, too. For example, when we work with waves or in a field called Fourier transform (you couldn't get an MRI without it!), we use identities like the half-angle formulae.

We begin with the *starred formula above:

$$cos(2a) = 1 - 2 sin^2 (a) \; \; \color{#E90F89}{\text{*}}$$

Now we'll let α (the Greek letter "alpha") stand for 2a, then α = 2a and a = α/2:

$$cos(\alpha) = 1 - 2 \, sin^2 \left( \frac{a}{2} \right)$$

Download a trig. identity sheet

Learning should be more about remembering than memorizing facts. The two are different. I think it's important to know (from memory) the definitions of the trigonometric functions, that tan(x) = sin(x)/cos(x), and the main Pythagorean identity, sin²(x) + cos²(x) = 1. After that, just knowing that these many other identiites exist is enough. You can look them up in a number of places, including a sheet you can download right here:

Practice problems

Using basic trig. identities, prove that each of the following statements is true: