So far we've looked at trig functions as ratios of sides, and where those are known, that's easy to calculate. The other way we've used them is as "black boxes" — we put a number into a calculator and we get one back out. Here's a triangle for which we know the lengths of all sides:
$$ \begin{align} sin(\theta) &= \frac{o}{h} = \frac{5}{9.434} = 0.530 \\[5pt] cos(\theta) &= \frac{a}{h} = \frac{8}{9.434} = 0.056 \\[5pt] tan(\theta) &= \frac{o}{a} = \frac{5}{8} = 0.625 \end{align}$$
Later we'll learn how to recover angles when the side lengths are shown, but let's move on for now.
The other way we've used trigonometry is to enter angle measures (in degrees or radians) into a calculator or computer, and it just gives us the value of the trig function for that angle — magically.
Here's an example of how the trig-function keys will look on your scientific calculator:
It's far from magic, as we will see below, and you'll learn more about that if you study infinite series as part of a calculus course.
In trigonometry and many subfields of math, we use the unit radians for angle measure. The radian arises more or less naturally from the geometry of the circle. There are 2π (yes, 6.283 ..., but we usually just say 2π) radians in a circle, so 180˚ = π radians. The figure on the left shows the most frequently-used radian measures. They are generally multiples of π/6 (30˚) and π/4 (45˚).
You should memorize this unit circle, including the radian measurements. It will grease the wheels for what's ahead. The tool below might help. It reduces the task to simple counting by multiples of 30˚ and 45˚. If you come back to it once in a while, a little more will sink in each time and soon you'll have it.
Note: Unit circle angles begin with zero (degrees or radians) on the far right, and increase in the counterclockwise direction.
You can learn more about angle measures here.
There are $2 \pi = 6.28$ radians and 360˚ in a circle, $\pi$ radians and 180˚ in a half circle.
Usually, we don't multiply the $\pi$ when reporting radian measurements. We just say 1.1$\pi$ radians, and so forth.
Press the forward arrow as many times as you want to advance the arrow and find the angle, in degrees and radians, on the unit circle.
The circle is "counted around" in 180˚ increments, then in 90˚, 45˚ and 30˚ increments. Go around the circle enough times that you can correctly anticipate the next angle, both in degrees and radians.
Press the
Pro tip: I can't emphasize enough how important it is that you know your way around the unit circle. Now that that's (hopefully) done, we can take a deeper look at the trigonometric functions.
These two essential right triangles, their angle measures and the ratios of their side lengths will pop up time and again throughout your studies and work in math and science. Knowing how to jot them down and use them, and from where the measurements came, will be invaluable to you. Study them carefully and memorize their proportions.
The 45-45-90 triangle is just what it sounds like, a right triangle that has two 45˚ angles, and is thus an isosceles triangle. We'll work with just symbols for the lengths of the sides, like this:
Now the Pythagorean theorem has to hold for this right triangle, so
$$^2 + x^2 = r^2$$
Now combine the like terms on the left,
$$2x^2 = r^2$$
and let's solve for $x$ (you'll see why in a minute), by first dividing both sides by 2.
$$x^2 = \frac{r^2}{2}$$
Now to solve for $x$, take the square root of both sides:
$$x = \sqrt{\frac{r^2}{2}}$$
Taking the square root of the numerator gives r, with $\sqrt{2}$ in the denominator:
$$x = \frac{r}{\sqrt{2}}$$
Finally, just for completeness, not that it really matters mathematically, the denominator of such an equation is usually "rationalized" by multiplying by $\sqrt{2}/\sqrt{2}:$
$$x = \frac{r \sqrt{2}}{2}$$
Now imagine if we let r, the length of the hypotenuse of our triangle, be 1. Then we get this triangle, complete with side measures:
What's great about such a triangle, is that it will give us the lengths of the sides of a 45-45-90 triangle with any hypotenuse length because we have the ratios of the side lengths, $1:\sqrt{2}/2.$
The 30-60-90 (degrees) right triangle is another frequently-encountered triangle that's worth knowing a lot about.
It isn't quite as easy to find an expression for the side lengths x and y, unless we combine two of these triangles to form an equilateral triangle like this:
Notice here that I've replaced x from the original figure with r/2. We know that the gray vertical line both bisects the top 60˚ angle (by construction) and bisects the base of the triangle. For a proof the the angle bisector of an isosceles triangle also bisects the opposite side, click here.
Now we can set up the Pythagorean theorem for one of the two 30-60-90 triangles shown:
$$r^2 = y^2 + \left( \frac{r}{2} \right)^2$$
Squaring the last term gives
$$r^2 = y^2 + \frac{r^2}{4}$$
Let's solve for y in terms of r by separating terms across the = sign:
$$y^2 = r^2 - \frac{r^2}{4}$$
The difference on the right is easy to calculate; just use 4 as a common denominator:
$$y^2 = \frac{3r^2}{4}$$
Now taking the square root of both sides gives
$$y = r \frac{\sqrt{3}}{2}$$
Now if we take r = 1, we arrive at the prototype 30-60-90 triangle:
So for a 30-60-90 triangle, the shorter leg (recall that the sides that form the right angle of a right triangle are called "legs") has a length that is half the length of the hypotenuse, and the longer leg is $\frac{\sqrt{3}}{2}$ of the hypotenuse.
$$ \require{cancel} \begin{align} sin(30^{\circ}) &= \frac{1}{2} \\[8pt] cos(30^{\circ}) &= \frac{\sqrt{3}}{2} \\[5pt] tan(30^{\circ}) &= \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\cancel{2}} \frac{\cancel{2}}{\sqrt{3}} = \frac{\sqrt{3}}{3} \end{align}$$
$$ \begin{align} sin(45^{\circ}) &= \frac{\sqrt{2}}{2} \\[8pt] cos(45^{\circ}) &= \frac{\sqrt{2}}{2} \\[5pt] tan(45^{\circ}) &= \frac{\cancel{\frac{\sqrt{2}}{2}}}{\cancel{\frac{\sqrt{2}}{2}}} = 1 \\[5pt] \end{align}$$
$$ \begin{align} sin(60^{\circ}) &= \frac{\sqrt{3}}{2} \\[8pt] cos(60^{\circ}) &= \frac{1}{2} \\[5pt] tan(60^{\circ}) &= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\sqrt{3}}{\cancel{2}} \frac{\cancel{2}}{1} = \sqrt{3} \\[5pt] \end{align}$$
Now we want to use the unit circle to create graphs of the three major trigonometric functions,
$$ \begin{align} f(x) &= sin(x) \\[5pt] f(x) &= cos(x) \\[5pt] f(x) &= tan(x) \end{align}$$
We'll do that by drawing a series of triangles, both 45-45-90 and 30-60-90, inside of the unit circle, and calculating the trig function values.
Let's start at 30˚ = θ/6. Notice that we can use that triangle to calculate all three trig functions. We'll set those aside as we work around the circle with triangles, and look at all of the values together at the end of the trip.
Now we really should work our way around the circle in increments of 30˚, but I'm going to skip around a bit and hope you'll get the idea. Next is a 135˚ angle, or 45˚ past 90˚, working counterclockwise (as usual) around the unit circle.
One quirk of the unit circle is that the length of the radius (the hypotenuses of all of our triangles) is always 1, but the other triangle sides can have negative or positive values that match with the x,y axis directions. That's how certain sine, cosine or tangent values end up negative.
You might have noticed that the 90˚ (which we passed up) angle doesn't really define a triangle. The key to understanding what happens there is to consider an 89˚ angle, in which the opposite side is very small and the adjacent side is nearly equal to the hypotenuse, making the sine function small (close to zero) and the cosine function close to one. Well, at 90˚, those values are zero and one, respectively, and we make similar arguments at all of the multiples of 90˚ around the circle.
Now let's do a 240˚ = 4θ/3 angle. Notice that because the x-and y-coordinates of the tip of the radius are negative, we treat the sides of our triangle as negative, and that affects the values of the trig functions.
Finally, let's look at an angle in the 4^{th} quadrant, 315˚ = 7θ/4, which is seven increments of 45˚ around the unit circle, and produces a 45-45-90 triangle, just with some "negative" sides.
Notice that all of this would just repeat as we moved our radius vector around the unit circle again and again. Next we'll gather all of this data in a table and determine some patterns in the sine, cosine and tangent functions.
Here's a table of trig function values at key angles in one round trip of the unit circle.
The trigonometric functions are repetitive, or cyclic, and we usually refer to them as periodic, meaning that they repeat the same basic pattern predictably.
$\theta$ (deg.) |
$\theta$ (rad) |
sin $(\theta)$ |
cos $(\theta)$ |
tan $(\theta)$ |
||
0˚ | 0 | 0 | $\frac{\sqrt{0}}{2}$ | 1 | $\frac{\sqrt{4}}{2}$ | 0 |
30˚ | $\frac{\pi}{6}$ | $\frac{1}{2}$ | $\frac{\sqrt{1}}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{3}$ |
45˚ | $\frac{\pi}{4}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $1$ |
60˚ | $\frac{\pi}{3}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{2}$ | $\frac{1}{2}$ | $\frac{\sqrt{1}}{2}$ | $\sqrt{3}$ |
90˚ | $\frac{\pi}{2}$ | $1$ | $\frac{\sqrt{4}}{2}$ | $0$ | $\frac{\sqrt{0}}{2}$ | $\infty$ |
120˚ | $\frac{2\pi}{3}$ | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{3}}{2}$ | $-\frac{1}{2}$ | $-\frac{\sqrt{1}}{2}$ | $-\sqrt{3}$ |
135˚ | $\frac{3\pi}{4}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $-\frac{\sqrt{2}}{2}$ | $-\frac{\sqrt{2}}{2}$ | $-1$ |
150˚ | $\frac{5\pi}{6}$ | $\frac{1}{2}$ | $\frac{\sqrt{1}}{2}$ | $-\frac{\sqrt{3}}{2}$ | $-\frac{\sqrt{3}}{2}$ | $-\frac{\sqrt{3}}{3}$ |
180˚ | $\pi$ | $0$ | $\frac{\sqrt{0}}{2}$ | $-1$ | $-\frac{\sqrt{4}}{2}$ | $0$ |
Forget about the gray columns for now. You can pick up a bunch of patterns as we walk around the circle and look at sin(θ), cos(θ) and tan(θ). Notice that the tangent function is a different beast, and we'll get to that later. Notice also that both the sine and cosine functions oscillate between ±1 and pass periodically through y = 0.
The gray columns are just re-expressions of the sine and cosine values to the left. Each is just re-stated with a square root in the numerator so that the increasing-decreasing pattern is more obvious.
The signs of the trig functions change according to which quadrant the tip of radius vector is in (UL = upper left, LR = lower right, and so on). This table shows how the sign of each function depends on the quadrant of the angle:
Quadrant | sin(θ) | cos(θ) | tan(θ) |
---|---|---|---|
UR (I) |
+ |
+ |
+ |
UL (II) |
+ |
— |
— |
LL (III) |
— |
— |
+ |
LR (IV) |
— |
+ |
— |
Graphs of the sine (black) and cosine (
All of the points in the table above are plotted on this graph. Be sure to think about these graphs for a while.
Notice a few patterns: They oscillate between ±1; they cross or are farthest apart (vertically) at multiples of $\pi/4$ and $\text{sin}(\theta)$ is the same curve as $\text{cos}(\theta)$, but shifted to the right by $\pi/2$ radians or 90˚. Notice also that $\text{sin}(x) = \text{cos}(x)$ when $x$ is $\frac{\pi}{4}$ and then every $\pi$ radians from there on.
These curves are also referred to as "sine waves". We usually don't say "cosine waves" because the cosine function is the same as the sine, just shifted by $\pi/2$. It is common to represent many periodic natural phenomena using sine and cosine curves: Waves, the motion of a pendulum, and the motion of a bouncing spring, for example.
In order to model periodic behavior using sines and cosines, we need, as usual, some transformations to translate and scale our function at will. In our typical way (see functions section), the transformations of $f(x) = sin(x)$ are shown on the right.
When we use trig. functions to model real data later, we'll make a slight modification of this prototype equation in order to keep the units straight.
The transformations of the other trig. functions are analogous.
Move the slider to change $c$ in the function $f(x) = sin(f/c)$ to get an idea of the effect of horizontal stretching on a periodic function. The function is written both as $f(x) = sin(x/c)$ and as $f(x) = sin[(1/c)·x]$.
If this function represented a wave (e.g. a sound, light or water wave) then the horizontal stretching parameter, $c$, would change the wavelength, the domain distance between successive peaks or troughs. The parameter $f = 1/c$ is called the frequency factor.
Notice that for small c, the frequency (number of cycles on the graph) is large, and for small $c$, it's large. Because the sine function is odd, when $c$ is negative, the sine curve is inverted across the x-axis.
(Notice also that the graph disappears when we try to divide by zero!)
Move the slider to see the effect of the vertical-stretching parameter, $A$, on our sine function.
If this function represented a wave like a sound or light wave, the vertical distance between $f(x) = 0$ and the maximum (or minimum) value of the sine function would be the amplitude, which would be proportional to volume (loudness) of sound or brightness (intensity) of light.
Notice that when $A \lt 0$, the function is reflected across the x-axis, as we would expect for any function.
Vertical translation of a periodic function is performed just like on any other function: Simply add a constant number (k) to the value of the function for each point in the domain. Vertical translation is used in modeling to move the average value of the function up or down. For example, to model a tide that fluctuates by ± 3 feet from an average depth of 10 feet, we'd want a sine or cosine function (turns out it doesn't matter which), that oscillates between 7 and 13 feet. The function would read f(some stuff) + 10.
Horizontal translation is very important when we model waves because it represents a property called phase. In our tide example, we might need to shift our model wave over to the right a bit to get the timing of the tides just right (and we'd also have to adjust the wavelength, which would represent the time between high or low tides). Phase is extremely important in many areas of science like x-ray crystallography (used to study molecular structure) and medical imaging.
Now let's think about a class of trig functions called the reciprocal trig functions. The tangent is the most prominent of these, but there are three more. We can look at the tangent using SOH-CAH-TOA; we know that
$$sin(\theta) = \frac{o}{h}, \phantom{00} cos(\theta) = \frac{a}{h}, \phantom{00} tan(\theta) = \frac{o}{a}.$$
Now notice that
$$\frac{sin(\theta)}{cos(\theta)} = \frac{\frac{o}{h}}{\frac{a}{h}} = \frac{o}{\cancel{h}} \frac{\cancel{h}}{a} = \frac{o}{a}$$
This means that the tangent function is the ratio of the sine and cosine functions:
$$tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$$
Now because $cos(\theta)$ — the denominator of the tangent expression — can be zero, the graph of the tangent function will have vertical asymptotes, as shown in the graph. The tangent of odd multiples of $\pi/2$ is infinite because $cos(\pi/2)$ (and odd multiples of $\pi/2$) is zero.
This has an interesting consequence if you are a rock climber or a lineperson (person who strings cable, e.g. electric wires, from pole to pole. You'll find an explanation here.
There are three other trigonometric functions left to explore, and like the tangent, all are ratios of other trig. functions, so we can expect asymptotic behavior there, too.
$$tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$$
Finally, let's define the three other trig. functions. The reciprocal of the sine function (not to be confused with the inverse) is called the cosecant (csc) function. The reciprocal of the cosine is (paradoxically, I know) the secant (sec) and the reciprocal of the tangent is the cotangent (cot). csc(θ), sec(θ) and cot(θ) are defined below.
$$ \begin{align} \text{sine} &\phantom{000} sin(\theta) = \frac{o}{h} \\[5pt] \text{cosine} &\phantom{000} cos(\theta) = \frac{a}{h} \\[5pt] \text{tangent} &\phantom{000} tan(\theta) = \frac{o}{a} \tag{*} \end{align}$$
$$ \begin{align} \text{cosecant} &\phantom{000} \text{csc} \phantom{00} csc(\theta) = \frac{1}{sin(\theta)} = \frac{h}{o} \\[5pt] \text{secant} &\phantom{000} \text{sec} \phantom{00} sec(\theta) = \frac{1}{cos(\theta)} = \frac{h}{a} \\[5pt] \text{cotangent} &\phantom{000} \text{cot} \phantom{00} cot(\theta) = \frac{cos(\theta)}{sim(\theta)} = \frac{a}{o} \end{align}$$
(*) Strictly speaking, the tangent is a reciprocal trig function, too, but we'll consider it here to be one of the "big three" most-frequently-used trig functions.
It should be easier to remember which of the main trig. functions are connected to sec, csc and tan, but due to an accident of history, it's not. Sine should go with secant, and cosine should go with cosecant. Unfortunately, those are reversed, and you'll have to remember that. Tangent goes with cotangent, though!
Several cycles of each of the six trig functions are plotted below. Note that the sin(x) and cos(x) graphs are vertically expanded by a factor of ten compared to the other graphs. This means that the sine graph would fit in between the U-shaped graphs (they are not parabolas!) in the csc(x) graph, same for the cos(x) and sec(x) graphs. Except for sine and cosine,
Determine the values (to the thousandths place) of all six trigonometric functions of the angle $\theta$.
$$ \begin{align} o &= \sqrt{21^2 - 17^2} \\[5pt] &= 12.329 \text{ cm} \end{align}$$
Now that all of the sides are known, we can calculate all of the ratios:
$$ \begin{align} sin(\theta) &= \frac{12.329}{} = 0.587 \\[5pt] cos(\theta) &= \frac{17}{21} = 0.810 \end{align}$$
$$ \begin{align} tan(\theta) &= \frac{12.329}{17} = 0.725 \\[5pt] csc(\theta) &= \frac{21}{12.329} = 1.703 \\[5pt] sec(\theta) &= \frac{21}{17} = 1.235 \\[5pt] cot(\theta) &= \frac{17}{12.329} = 1.379 \\[5pt] \end{align}$$
There is much more to trigonometry than can fit into one section like this. You'll need to know how to use inverse trig. functions, how to relate trig functions to one another (analytic trig), and how to use trigonometry on non-right triangles. Here are links to other trigonometry-related pages:
For problems 1 and 2, calculate the measure of any missing angles and the lengths of any missing sides of the triangle.
1.
First, the length of the hypotenuse is
$$h = \sqrt{63^2 + 14^2} = \sqrt{4165} = 64.537$$
Now the trig functions are:
$$ \begin{align} sin(\theta) &= \frac{14}{64.537} = 0.217 \\[5pt] cos(\theta) &= \frac{63}{64.537} = 0.976 \\[5pt] tan(\theta) &= \frac{14}{63} = 0.222 \\[5pt] csc(\theta) &= \frac{64.537}{14} = 4.61 \\[5pt] sec(\theta) &= \frac{64.537}{63} = 1.024 \\[5pt] cot(\theta) &= \frac{63}{14} = 4.50 \end{align}$$
2.
First, the length of the adjacent side is
$$h = \sqrt{21^2 + 8.5^2} = \sqrt{368.75} = 19.203$$
Now the trig functions are:
$$ \begin{align} sin(\theta) &= \frac{8.5}{21} = 0.405 \\[5pt] cos(\theta) &= \frac{19.203}{21} = 0.914 \\[5pt] tan(\theta) &= \frac{8.5}{19.203} = 0.443 \\[5pt] csc(\theta) &= \frac{21}{8.5} = 2.47 \\[5pt] sec(\theta) &= \frac{21}{19.203} = 1.094 \\[5pt] cot(\theta) &= \frac{19.203}{14} = 2.259 \end{align}$$
The cosine of an angle of a right triangle is $\frac{3}{11}$. Calculate the values of the other five trigonometric functions of this angle, and sketch the triangle.
Here's a diagram of the triangle, including the length of the missing side:
The missing side is $o = \sqrt{11^2 - 3^2} = 10.583$.
Now the values of the six trig functions are:
$$ \begin{align} sin(x) &= \frac{10.583}{11} = 0.962 \\[5pt] cos(x) &= \frac{3}{11} = 0.273 \\[5pt] tan(x) &= \frac{10.583}{3} = 3.528 \\[5pt] csc(x) &= \frac{11}{10.583} = 1.039 \\[5pt] sec(x) &= \frac{11}{3} = 3.667 \\[5pt] cot(x) &= \frac{3}{10.583} = 0.283 \end{align}$$
The tangent of an angle of a right triangle is $\frac{7}{5}$. Calculate the values of the other five trigonometric functions of this angle and sketch the triangle.
Here's a diagram of the triangle, including the length of the missing side:
The missing side is $o = \sqrt{11^2 - 3^2} = 10.583$.
Now the values of the six trig functions are:
$$ \begin{align} sin(x) &= \frac{10.583}{11} = 0.962 \\[5pt] cos(x) &= \frac{3}{11} = 0.273 \\[5pt] tan(x) &= \frac{10.583}{3} = 3.528 \\[5pt] csc(x) &= \frac{11}{10.583} = 1.039 \\[5pt] sec(x) &= \frac{11}{3} = 3.667 \\[5pt] cot(x) &= \frac{3}{10.583} = 0.283 \end{align}$$
A certain point on the unit circle is $\left(\frac{7}{25}, \; y \right)$ with $y \gt 0$. Determine the exact value of $y$.
Here's a diagram of the unit circle with our triangle roughly placed in it:
Now it's clear that our task is just to find the missing side of the triangle, the $y$ coordinate of the point on the circle. That is
$$ \begin{align} y &= \sqrt{1^2 - \left( \frac{7}{25} \right)^2} \\[5pt] &= \sqrt{\frac{625-49}{625}} \\[5pt] &= \sqrt{\frac{576}{625}} \\[5pt] &= \frac{24}{25} \end{align}$$
A certain point on the unit circle is $\left(\frac{15}{17}, \; y \right)$ with $y \lt 0$. Determine the exact value of $y$.
Here's a diagram of the unit circle with our triangle roughly placed in it:
Now it's clear that our task is just to find the missing side of the triangle, the $y$ coordinate of the point on the circle. That is
$$ \begin{align} y &= -\sqrt{1^2 - \left( \frac{15}{17} \right)^2} \\[5pt] &= -\sqrt{\frac{289-225}{289}} \\[5pt] &= -\sqrt{\frac{64}{289}} \\[5pt] &= -\frac{8}{17} \end{align}$$
Sketch a graph of $f(x) = sin(x) + cos(x)$ over the domain $[0, \, 2\pi]$.
Let's make a quick table of sums of sin(x) and cos(x) at a few select points:
$\theta$ | $sin(\theta)$ | $cos(\theta)$ | $s(\theta) + c(\theta)$ |
$0^{\circ}$ | $0$ | $1$ | $1$ |
$45^{\circ}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\sqrt{2}$ |
$90^{\circ}$ | $1$ | $0$ | $1$ |
$135^{\circ}$ | $\frac{\sqrt{2}}{2}$ | $-\frac{\sqrt{2}}{2}$ | $0$ |
$180^{\circ}$ | $0$ | $-1$ | $-1$ |
$225^{\circ}$ | $-\frac{\sqrt{2}}{2}$ | $-\frac{\sqrt{2}}{2}$ | $-\sqrt{2}$ |
$270^{\circ}$ | $\frac{\sqrt{2}}{2}$ | $-\frac{\sqrt{2}}{2}$ | $0$ |
$305^{\circ}$ | $-\frac{\sqrt{2}}{2}$ | $-\frac{\sqrt{2}}{2}$ | $-\sqrt{2}$ |
Here is what the graph looks like:
It's still a periodic function with a period of $2 \pi$ radians or 360˚, but it has been shifted by -45˚.
Sketch a graph of $g(x) = sin^2(x) + cos^2(x)$. Note that $sin^(x)$ is just a commonly-used shorthand notation for $[sin(x)]^2$, just the square of the result of $sin(x)$.
Let's make a quick table of sums of sin^{2}(x) and cos^{2}(x) at a few select points:
$\theta$ | $sin^2(\theta)$ | $cos^(\theta)$ | $s^2(\theta) + c^2(\theta)$ |
$0^{\circ}$ | $0$ | $1$ | $1$ |
$45^{\circ}$ | $\frac{1}{2}$ | $\frac{1}{2}$ | $1$ |
$90^{\circ}$ | $1$ | $0$ | $1$ |
$135^{\circ}$ | $\frac{1}{2}$ | $\frac{1}{2}$ | $1$ |
$180^{\circ}$ | $0$ | $1$ | $1$ |
$225^{\circ}$ | $\frac{1}{2}$ | $\frac{1}{2}$ | $1$ |
$270^{\circ}$ | $\frac{1}{2}$ | $\frac{1}{2}$ | $1$ |
$305^{\circ}$ | $\frac{1}{2}$ | $\frac{1}{2}$ | $1$ |
Notice that the sum $sin^2(\theta) + cos^2(\theta) = 1$ for all values of $\theta$. We haven't calculated any values in between, but it's true. This is a relationship that we'll use a lot going into analytic trigonometry and it's called the Pythagorean Identity:
$$sin^2(\theta) + cos^2(\theta) = 1$$
For any angle, $\theta$.
You should know how to label the most-frequently-used angles around the unit circle (circle of radius r = 1), both in degrees and radians. It's not that difficult if you just think of it as counting around the circle in different increments.
Minutes of your life: 3:25
A mnemonic (nee·mon'·ick) is a word or phrase designed to help a person remember something. An example would be the pseudo-word "ROYGBIV" or the phrase "Rogers of York Gave Battle in Vain." Both are designed to help us remember the colors of the visible spectrum: red, orange, yellow, green, blue, indigo & violet.
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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