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.
In this section we'll start to derive some of those relationships. We'll begin with how trig. functions relate to one another, then move on to solve trigonometric equations. Because the trigonometric functions are periodic, trig. equations like cos(x) = 1 have an infinite number of periodically related solutions (cos(x) = 1 for x = even multiples of π).
You might want to refresh your memory about basic trigonometry before working through this section. Other pages related to trigonometry are
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:
$$sin{\theta} = \frac{o}{h}\: \longrightarrow \: [sin(\theta)]^2 = \frac{o^2}{h^2}$$
Do the same with cosine:
$$cos{\theta} = \frac{a}{h}\: \longrightarrow \: [cos(\theta)]^2 = \frac{a^2}{h^2}$$
Now add the squares of the sin and cos functions. Remember that it's customary to write cos^{2}(x) instead of the more cumbersome [cos(x)]^{2}:
$$sin^2(\theta) + cos^2(\theta) = \frac{o^2}{h^2} + \frac{a^2}{h^2}$$
Notice that we can add the two fractions on the right because they have a common denominator:
$$= \frac{o^2 + a^2}{h^2}$$
We recognize that o^{2} + a^{2} is the sum of the squares of the two sides of our right triangle, and that it must equal h^{2} by the Pythagorean theorem.
$$= \frac{h^2}{h^2} = 1$$
That leads us to one of the most important and useful relationships between trigonometric functions, the Pythagorean identity.
The Pythagorean identity is the most important relationship among trig. functions. You should remember it.
We can derive a couple of other Pythagoreanlike identities from the first. Here's how: Start with the Pythagorean identity:
$$sin^2(\theta) + cos^2(\theta) = 1$$
Divide each term by sin^{2}(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 cos^{2}(θ) this time:
$$sin^2(\theta) + cos^2(\theta) = 1$$
$$\frac{sin^2(\theta)}{cos^2(\theta)} + \frac{cos^2(\theta)}{cos^2(\theta)} = \frac{1}{cos^2(\theta)}$$
Our definitions of tangent and secant give us a third Pythagorean identity
$$tan^2(\theta) + 1 = sec^2(\theta)$$
$$ \begin{align} sin^2(\theta) + cos^2(\theta) &= 1 \\[3pt] csc^2(\theta)  cot^2(\theta) &= 1 \\[3pt] sec^2(\theta)  tan^2(\theta) &= 1 \end{align}$$
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 table below was borrowed from a popular mathematics book. It shows how to express any trig function (left column) in terms of any other single trig function (top row). Each of these identities can be proven fairly simply. You should try some.
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 oneanother. 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 upperleft 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:
$$ \begin{align} cos(a) &= \frac{x}{cos(b)} \\[4pt] x &= cos(a) \, cos(b) \end{align}
and the sine gets us side y:
$$ \begin{align} sin(a) &= \frac{y}{cos(b)} \\[4pt] y &= sin(a) \, cos(b) \end{align}
Likewise, we need the side z (temporary variable) of that little triangle on the upperright:
$$ \begin{align} sin(a) &= \frac{z}{sin(b)} \\[4pt] z &= sin(a) \, sin(b) \end{align}$$
So here's the fully labeled square:
I'll remove the triangles to get rid of the noise. Now you can easily see that sin(a+b) = sin(a)cos(b) + cos(a)sin(b), and that cos(a+b) = cos(a)cos(b)  sin(a)sin(b).
What we've done here is found a formula for the sine and cosine of a sum of angles that only depends on the measure of each angle separately. That will be very useful later. Here are the trig. sum formulas:
$$ \begin{align} sin(a + b) &= sin(a) cos(b) + cos(a) sin(b) \\[3pt] cos(a + b) &= cos(a) cos(b)  sin(a) sin(b) \end{align}$$
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:
$$ \begin{align} sin(2a) &= 2sin(a) cos(b) \\[3pt] cos(2a) &= cos^2(a)  sin^2(a) \end{align}$$
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 doubleangle formula:
$$cos(2a) = cos^2(a)  sin^2(a)$$
Now substitute cos^{2}(a) = 1  sin^{2}(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 sin^{2}(a):
$$sin^2(a) = \frac{cos(2a)  1}{2}$$
We can do a similar derivation for the cos^{2}(a) power reduction formula:
$$cos^2(a) = \frac{cos(2a) + 1}{2}$$
Well, why not find a formula for a halfangle. 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 halfangle formulae.
We begin with the
$$cos(2a) = 1  2 sin^2 (a) \; \; \color{#E90F89}{\text{*}}$$
Now we'll let α (the Greek letter "alpha") stand for 2a, then α = 2a and a = α/2:
$$cos(\alpha) = 1  2 \, sin^2 \left( \frac{a}{2} \right)$$
Now we just solve for the sine of α/2, which involves taking a root of both sides to get the sine halfangle formula:
$$sin \left( \frac{1}{2} \right) = \sqrt{\frac{cos(\alpha)  1}{2}}$$
In a similar way, we can derive the cosine halfangle formula:
$$cos \left( \right) = \sqrt{\frac{cos(\alpha) + 1}{2}}$$
These are the derivations of some of the more useful trig identities, but there are others you might want to use from time to time. Hit the button below to download a more complete list of identities to keep around when you're working on trig. problems.
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, sin^{2}(x) + cos^{2}(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:
Using basic trig. identities, prove that each of the following statements is true:
1. 
$sec(x) cot(x) = csc(x)$ 

2. 
$sin(x) tan(x) = \frac{1  cos^2(x)}{cos(x)}$ 

3. 
$\frac{cos(x)}{1 + sin(x)} + \frac{1 + sin(x)}{cos(x)} = 2 sec(x)$ 

4. 
$\frac{tan(x)  sin(x)}{1 + cos(x)} = tan(x)$ 

5. 
$\frac{sin(x)}{tan(x)} + \frac{cos(x)}{cot(x)} = sin(x) + cos(x)$ 

6. 
$sin^2(x) + cos^2(x) = 1$ 

7. 
$csc(x)  csc(x) cos^2(x) = sin(x)$ 

8. 
$cos^4(x)  sin^4(x) = 1  2 sin^2(x)$ 

9. 
$\frac{1 + tan(x)}{1 + cot(x)} = tan(x)$ 

10. 
$(3 cos(x)  4 sin(x))^2 + (4 cos(x) + 3 sin(x))^2 = 25$ 

11. 
$\frac{tan(x) + cot(x)}{sec(x) csc(x)} = 1$ 

12. 
$\frac{sin(x) + cos(x)}{sin(x)}  \frac{cos(x)  sin(x)}{cos(x)} = sec(x) csc(x)$ 

13. 
$\frac{1  sin(x)}{cos(x)} = \frac{cos(x)}{1 + sin(x)}$ 

14. 
$tan^2(2x) + sin^2(2x) + cos^2(2x) = sec^2(2x)$ 
alpha  Α  α 
beta  Β  β 
gamma  Γ  γ 
delta  Δ  δ 
epsilon  Ε  ε 
zeta  Ζ  ζ 
eta  Η  η 
theta  Θ  θ 
iota  Ι  ι 
kappa  Κ  κ 
lambda  Λ  λ 
mu  Μ  μ 
nu  Ν  ν 
xi  Ξ  ξ 
omicron  Ο  ο 
pi  Π  π 
rho  Ρ  ρ 
sigma  Σ  σ 
tau  Τ  τ 
upsilon  Υ  υ 
phi  Φ  φ 
chi  Χ  χ 
psi  Ψ  ψ 
omega  Ω  ω 
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