Transcription
This video is intended to provide some trigonometry background from my video series on visualizing the six trigonometric functions.
I'm Dennis Davis. If you found this video some other way, welcome! I hope you find it interesting and informative. If you haven't taken trigonometry yet, this could provide a big boost to your preparation, but it's not a comprehensive trigonometry course.
I'll cover five important trigonometry concepts, and the first is the longest, so don't worry, the pace will pick up. The word trigonometry, which I'll often abbreviate as trig, means triangle measurement, but it's really about angles. We use triangles quite a bit, actually, to define trigonometry concepts, but trig starts and ends with angles.
So we will start with an angle, any angle out in space. Angles are usually, but not always, denoted by a Greek letter, and the Greek letter theta is by convention a common choice. So we'll label our angle theta, which looks like the letter O with a horizontal line through the middle.
Now we're going to create an imaginary triangle using this angle, and we're going to use that imaginary triangle to describe the trigonometric functions of the angle theta. I'll show how this is done using a memory aid I used in high school. The sillier it is, the easier it will be to remember.
Okay, a fair warning: imagine a paper cutter and put a blue horizontal guide on it. We'll use colors for memory aids later, especially in the next video series, so please just trust me, the horizontal guide is blue.
We take the angle theta and we line up one of its arms to the blue guide, putting the angle's arms out over the right edge. Then we imagine cutting a red line with the paper cutter. This creates a triangle using the blue side, the red side, and the other arm of the angle, the one we didn't line up with the blue guide. I'll color it yellow.
It's a right triangle, of course, because the blue side and the red side are perpendicular. The square symbol in the corner means the angle is a right angle. Each side of the triangle has a special name from the perspective of angle theta. The side that's opposite, that is the red one across the triangle from theta, is called the opposite side, usually abbreviated OPP.
The longest side, which will always be the side across from the right angle, is called the hypotenuse, abbreviated hyp. The blue side that we lined up with the blue guide is called the adjacent side, adj, because it's right next to theta.
The six trig functions are simply the various ratios between the lengths of these three sides. There are only six possible ratios between pairs, and here they are: opposite divided by the other two, adjacent divided by the other two, and hypotenuse divided by the other two.
These six ratios correspond to the six trigonometric functions, and here are their names: sine, cosine, tangent, cotangent, secant, cosecant. The trig functions all have three-letter abbreviations. Abbreviations for five of them, the first five in the order I've listed them, are just the first three letters of the spelled-out function name.
But the abbreviation for cosecant can't be cos because that's already taken by cosine, so by convention, the abbreviation for cosecant is CSC. We'll start with these first three: sine, cosine, and tangent. A popular mnemonic to remember these is sohcahtoa.
Soh CAH Toa. Each three-letter component of sohcahtoa is intended to denote a trig function and its ratio. So Soh: sine is opposite over hypotenuse. Soh CAH: cosine is adjacent over hypotenuse. Toa: tangent is opposite over adjacent.
Let's do a quick example. I consider an angle theta, put it on our paper cutter, measure, and get the lengths shown in the triangle. The hypotenuse is seven point four meters. I guess I imagined a very large paper cutter, and the adjacent and opposite sides are six point one and four point two meters, respectively.
Let's sohcahtoa to find the sine, cosine, and tangent of angle theta. So Soh: sine equals opposite over hypotenuse, four point two meters divided by seven point four meters is 0.568.
CAH: cosine equals adjacent over hypotenuse, six point one meters divided by seven point four meters. Toa: tangent equals opposite over adjacent, four point two meters divided by six point one meters.
Please chant sohcahtoa to yourself and know the corresponding formulas for sine, cosine, and tangent. Here you might consider making six flashcards. You should know these frontwards and backwards, and fortunately, sohcahtoa makes it easy.
The last three trig functions are just reciprocals of the first three. Remember, it's easy to take the reciprocal of a fraction: the top and bottom switch places. So the reciprocal of a over b is b over a.
The cosecant is one over sine, so hypotenuse over opposite. The secant is one over cosine, so hypotenuse over adjacent. The cotangent is one over tangent, so adjacent over opposite.
You don't need to memorize something if you can figure it out quickly, and these last three trig functions are good examples. Don't memorize cosecant, secant, or cotangent. I haven't just memorized sine, cosine, and tangent, which by the way, sohcahtoa makes so easy it hardly counts as memorization.
We just have to know which of these other trig functions, cotangent, secant, and cosecant, is paired with each of our first three sohcahtoa functions: sine, cosine, and tangent.
And here's how to do that: each reciprocal pair includes exactly one function whose name starts with "co." We start by pairing tangent and cotangent because their names are similar. Sohcahtoa reminds us that tangent is opposite over adjacent, so cotangent is flipped: adjacent over opposite.
But sine and cosine don't sound like secant or cosecant. Well, since each pair has exactly one "co," sine can't be the reciprocal of secant, so it must go with cosecant, and cosine must be the reciprocal of secant. There's one "co" per reciprocal pair.
Here are six more flashcards. These will probably take a few extra seconds to answer, but that's better than memorizing. Whatever you can memorize, you can easily forget, but what you know how to figure out, that you can do forever.
Let me make four important points about these six functions. First, the trig functions are properties of angles, not triangles. An angle out in space or an angle drawn in a homework or test problem has a sine and cosine, even if there isn't a triangle drawn.
We just imagined the triangle using a paper cutter. If you like to help visualize the meaning of the ratios, triangles don't have sines, cosines, or tangents; angles do.
The trig functions are unitless since they're the ratio between two lengths. The unit of measure cancels out, so the sine of theta would never be 0.568 meters or 0.568 feet, but merely 0.568.
Thirdly, trig functions are mathematical functions, but unlike traditional functions that include their argument in parentheses, trig functions are usually written without any parentheses. When reading an equation out loud, we say sine theta or cosine gamma. We don't say "of," like we would for other functions like f of x.
There's a common verbal and written shorthand regarding trig functions, and you should be familiar with it. Finally, when we square trig functions, which we'll do a lot when we use the Pythagorean theorem, we write the exponent two between the function abbreviation and the angle variable.
This would be read out loud as sine squared theta, not sine of theta squared. It's a little different, but you'll get used to it.
Well, that was part one of this introductory video, and it was the longest. Mercifully, part two will be the shortest. We imagine a plane with perpendicular x and y-axes. By convention, the x-axis is usually drawn horizontally with a vertical y-axis.
By convention, the x-axis has a positive direction to the right and a negative direction to the left. Similarly, the y-axis has a positive direction up and a negative direction down. The point where the x and y axes intersect is called the origin.
Any of the infinitely many points on the plane can be identified by two numbers: the x-coordinate, which represents its distance, positive or negative, along the x-axis, and the y-coordinate, which represents its distance, positive or negative, along the y-axis.
So every point in a plane is identified by an ordered pair of XY coordinates. The axes divide the plane into four quadrants labeled with Roman numerals like this. All of the points in quadrant one have a positive x-coordinate and a positive y-coordinate.
In quadrant two, the x-coordinate is negative, but the y-coordinate is still positive. For points in quadrant three, both x and y coordinates are negative, and in quadrant four, the x-coordinate is positive and the y-coordinate is negative.
You shouldn't need to memorize these; just picture this Cartesian coordinate system in your head and decide which direction from the origin to go to get to each quadrant: horizontal first, right or left; vertical second, up or down.
Here are eight more flashcards. This might seem meaningless or trivial, but quadrants are a handy way to evaluate trig functions, and they'll be important later, especially in the next video series.
Expressing and measuring angles in degrees is probably familiar to most people. Here's a circle showing some common or interesting angles. Each label corresponds to the size of the angle at the origin from the positive x-axis swept around counterclockwise to the labeled point.
In this example, the arm passes through the 30-degree point on the circle, which means the arm makes a 30-degree angle with the x-axis, as shown. Of course, there are an infinite number of angles, not just the 12 shown by these labels, but these are common ones you should know in any trig class.
But don't memorize them; just know the quadrant boundaries: 0, 90, 180, and 270. And you don't have to memorize these if you can start at 0 and add 90 as you imagine each point around the circle counterclockwise, then just add or subtract 30 or 60 degrees, whichever happens to be easiest.
For example, what angle does this point represent? Well, it's 180 degrees plus 30 degrees, so 210 degrees. The 30-60-90 angles are common, and they proceed through all four quadrants all the way back around to 360 degrees, which is the same angle as zero degrees.
Also important are the 45-degree angles that bisect each quadrant. You should know all of these using the same method of adding or subtracting 45 degrees from one of the quadrant boundary angles: 0, 90, 180, 270, or 360.
Scientists, mathematicians, and engineers don't use degrees very often to specify angle sizes; they use a unit of measure called a radian. Imagine a circle with radius R. We take that radius length and line it up with the circle on the positive x-axis.
Then we wrap that length around the circle as far as it goes. The ray from the origin through the endpoint forms an angle with the x-axis whose size is 1 radian. You could remember one radian is one radius worth of angle.
We know there are 360 degrees all the way around a circle, so how many radians make a complete circle? The length of a circle's circumference is pi times the diameter of the circle. Since the diameter is equal to two times the radius, this is the same as 2 pi R.
So there are two pi radiuses or radii around a circle, which means there are two pi radians in a circle. Let's count them out. In the circle on the left, we already have one, two, three, four, five, six, plus a little extra. Six point two eight three is two pi rounded to three decimal places, so there are indeed two pi radians around a circle.
So if all the way around the circle is 2 pi radians, then halfway around the circle must be pi radians. It follows then that 90 degrees is pi over 2 radians, and various fractions of pi correspond to the angles we've already covered in degrees.
This can look intimidating at first, but you don't have to memorize these if you can figure them out quickly. Here's a table that should help, but don't memorize this either. This table will show you the pattern, which you should know because using it, you can figure out any of the angles in radians.
Let's start with the multiples of 30, the 30-60-90 angles and their equivalent in radians. Since 30 degrees is 1/6 of pi, and all these angles are multiples of 30 degrees, each 30-60-90 angle will be a multiple of one-sixth of pi.
Let's step through them all quickly. 30 degrees is one sixth of pi. When reading and writing angles in radians, we usually put pi in the numerator rather than to the right of the fraction. So instead of writing or saying 1/6 pi, we write and say pi over 6 radians.
60 degrees is 2/6 of pi, twice what 30 degrees is. Two sixths reduces to 1/3, so pi over 3 radians. 90 degrees is 3 sixths of pi, which reduces to 1/2, so pi over 2 radians.
120 degrees is 4/6 pi, which reduces to 2/3, so 2 pi over 3 radians. 150 degrees is 5/6 pi, which can't be reduced, so 5 pi over 6 radians. 180 degrees is 6/6 pi, which reduces to 1, so 180 degrees is simply pi radians.
210 degrees is 7/6 pi, which can't be reduced, so 7 pi over 6. It's okay to use improper fractions when expressing angles in radians. 240 degrees is 8/6 of pi, which reduces to 4/3, so 4 pi over 3 radians.
270 degrees is 9/6 pi, which reduces to 3 halves, so 3 pi over 2 radians. 300 degrees is 10/6 pi, which reduces to 5/3, so 5 pi over 3 radians. 330 degrees is 11/6 pi, which can't be reduced, so 11 pi over 6 radians.
360 degrees is 12/6 pi, which reduces to 2, so 360 degrees is 2 pi radians. Each additional multiple of 30 degrees is another pi over 6 radians.
Let's do the 45-degree multiples; that will be faster since there aren't as many. 45 degrees is 1/4 of 180 degrees, which is pi radians, so 45 degrees is pi over 4 radians. Each multiple of 45 degrees is another pi over 4 radians.
So 90 degrees must be two fourths of pi, which reduces to 1/2, so pi over 2 radians. Of course, this is the same answer we got when we considered 90 degrees to be 3 times 30 degrees, so 3/6 pi radians.
135 degrees is 3/4 of pi, which cannot be reduced, so 3 pi over 4 radians. 180 degrees is 4/4 of pi, which reduces to 1, so 180 degrees is simply pi radians, just like before.
225 degrees is 5/4 of pi, which can't be reduced, so 5 pi over 4 radians. 270 degrees is six fourths of pi, which reduces to 3 halves, so 3 pi over 2 radians.
315 degrees is 7/4 of pi, which cannot be reduced, so 7 pi over 4 radians. And 360 degrees is 8/4 of pi, which reduces to 2, so 360 degrees is indeed 2 pi radians.
Practice. Go slowly, build up your speed. What angle is this in degrees and radians? You should be able to state the angle measurement for all of these points. Note that I didn't say memorize. When have I ever said that?
As usual, with a little understanding of the patterns, you can figure them out instead of memorizing them. Instead of flashcards, draw a circle with these 16 points and skip around randomly, saying the size of the corresponding angle in degrees and radians.
Remember, the angle we mean is the one formed by the positive x-axis and aligned from the origin through the point. Also, you should be able to convert between degrees and reduced radians, as shown on the right chart.
I've shaded out the helper columns in the middle as a reminder that they aren't part of your drills. You could make 34 flashcards if you wanted: 17 covering all the angles from 0 to 360 degrees to be converted to radians, and 17 covering all the angles from 0 to 2 pi radians to be converted to degrees.
And if given one of these angles in degrees or radians, you should be able to point to its corresponding dot on the circle. Sometimes it seems like 90% of trigonometry is based on the Pythagorean theorem, and it might even be closer to a hundred percent, so it's definitely something you should be familiar with.
The Pythagorean theorem applies to right triangles, which is fortunate for us because the trig functions are based on ratios between the lengths of the sides of right triangles, and our imaginary paper-cutter procedure creates imaginary right triangles.
The theorem states that the squared length of the hypotenuse of a right triangle is equal to the sum of the squared lengths of the other two sides, or algebraically, a squared plus b squared equals c squared, as long as c is the hypotenuse of the triangle.
Or more specifically to us, adjacent squared plus opposite squared equals hypotenuse squared. An easy and popular example is a 3-4-5 triangle. If the short sides of a right triangle are 3 and 4, then the hypotenuse will be 5 because 3 squared plus 4 squared equals 5 squared, and that's the Pythagorean theorem.
Very, very simple, very, very important. Lastly, perhaps most importantly, is the concept of the unit circle, which is a circle whose radius is one. We don't specify the units because in trigonometry they'll cancel out since we're dealing with ratios between lengths.
So the radius could be one inch, one mile, one millimeter; it doesn't matter. We just say one or one unit without being specific. We draw out the unit circle with its center at the origin of an XY plane.
Every point on the circumference of the unit circle is exactly one unit away from the origin, the center of the circle. So to be clear, its rightmost point, corresponding to angle zero, is at coordinates 1, 0. The topmost point, corresponding to angle pi over 2 radians or 90 degrees, is at coordinates 0, 1.
The leftmost point is at negative 1, 0, and the bottommost is at 0, negative 1. Using our paper cutter illustration, we take our angle theta and line it up so the angle's point is at the origin and one of its arms is lined up with the blue x-axis.
The other arm will intersect the circle somewhere on its circumference. Since the radius of a unit circle is 1, this point will always be exactly one unit away from the center. It's through this point that we drop our vertical red line, thus ensuring that the length of the hypotenuse of our right triangle is 1.
The unit circle simplifies sohcahtoa since H, the hypotenuse, is 1. Sine theta equals opposite over hypotenuse, and cosine theta equals adjacent over hypotenuse. But when we consider that the hypotenuse is 1, these ratios simplify for unit circles only.
The sine of theta is simply the length of the opposite side, and the cosine of theta is the length of the adjacent side. This is a good time to point out another definition for tangent. Sohcahtoa reminds us that tangent equals opposite over adjacent.
Toa: since sine theta equals opposite and cosine theta equals adjacent, opposite over adjacent equals sine over cosine, so tangent theta is equal to sine theta over cosine theta. This is algebraically the same as opposite over adjacent anyway.
This means we can read the cosine and sine directly from the unit circle since they are the x and y coordinates of the yellow point on the circle corresponding to angle theta. Don't panic, but you should know the cosine and sine of each of these angles.
Of course, I'll show you an easy way to know these without memorizing them. I don't know if I was a clever student or just lazy, but my advice is to be both if you can. If you'll allow me, I'll assume that the angles at the quadrant boundaries are trivial.
You should see these in your head, and know each one has a 0 and a 1 that's either positive or negative. So I'll remove these from our diagram and we'll focus on the remaining 12. To know their sine and cosine, you need to see this circle in your head, and you need to know three numbers.
I think of the numbers as large, medium, and small, and remembering them is as simple as 3, 2, 1. The large number is square root of 3 over 2, the medium number is square root of 2 over 2, the small number is square root of 1 over 2.
Well, the square root of 1 is just 1, so square root of 1 over 2 simplifies to 1 over 2 or 1/2. Here are their approximate decimal values. I don't think you need to memorize these; just know the fractions corresponding to large, medium, and small.
Let me show you where these fractions come from because if I leave it out, someone will complain. We'll start with 1/2, which is the sine of 30 degrees. Let's prove it. Let's prove that this red line segment is 1/2 units long.
The dot represents pi over 6 radians or 30 degrees, so we draw a right triangle with a 30-degree angle. The hypotenuse is one since we're on a unit circle. The interior angles of a triangle add up to pi radians or 180 degrees.
We know there are already 90 and 30-degree angles in the triangle, so this last angle must be 60 degrees. Next, we draw a mirror image triangle below the x-axis, so now we have a large triangle with all three interior angles of 60 degrees.
This makes it an equilateral triangle, meaning that all three sides have the same length. So the total length of the red side must be one, like the other two. So the original red line, the sine of 30 degrees, must be half of this or 1/2 the length of the blue line.
Representing the cosine of theta is the size large number. I've told you it's square root of 3 over 2, but let's prove it. For this proof, we just need the Pythagorean theorem. We know the lengths of two sides: 1 and 1/2, and the third side is the cosine of theta.
So the Pythagorean equation becomes cosine squared theta plus 1/2 squared equals 1 squared. This exactly corresponds to area of blue plus area of red equals area of yellow. Our equation can be simplified to cosine squared theta plus 1/4 equals 1.
Isolating the constants, we get cosine squared theta equals 3/4. To get cosine theta, we take the square root of both sides and simplify to get cosine of theta equals square root of 3 over 2.
This means that the length of the blue side, the size large, is square root of 3 over 2. To prove the medium length is the square root of 2 over 2, we draw a right triangle through the 45-degree mark.
Again, using interior angles of a triangle, this other angle must also be 45 degrees. Since two angles are equal, their opposite sides must be equal, so the length of the red sides are equal.
We'll find this length: mental area of blue plus area of red equals area of yellow, which is one. So if blue and red add up to one and they're equal, their areas must each be one half.
We'll solve the blue square; it's the same as the red. Its area is one half, and its area is also the square of cosine theta. So cosine squared theta equals one half.
To find cosine theta, we take the square root of both sides, so cosine theta equals the square root of one half. We need to tidy up this expression, so let's distribute the radical to the numerator and denominator and simplify to one over square root of two.
Well, that would be our answer, but it's considered untidy to leave radicals in the denominator of fractions, so we'll multiply numerator and denominator by square root of two to yield the medium-sized length: square root of two over two.
Due to symmetry between the quadrants, each angle corresponding to these remaining yellow dots has a cosine and sine consisting of just these three numbers, though they can be positive or negative depending on the angle's quadrant.
Let's consider the three angles in the first quadrant. With practice, you'll know that these correspond to pi over 6, pi over 4, and 2 pi over 6, which simplifies to pi over 3 radians. These angles are also 30, 45, and 60 degrees, but try to use radians, especially if you intend on studying math, science, engineering, or electronics.
Let's consider the cosines for these angles. Remember, the cosine is the x-coordinate of the yellow point. Here they are. I'm color coding cosine and sine to match the paper-cutter colors: horizontal guide blue is cosine, vertical cut red is sine.
So which of these cosines is small, medium, and large? Yes, it's that easy. So the cosine of pi over 3 is the small value, which is 1/2. The cosine of pi over 4 is the medium value, which is the square root of 2 over 2.
And the cosine of pi over 6 is the large value, which is the square root of 3 over 2. Now let's look at the sine for each of the first quadrant angles. You might be able to pause and do it on your own.
Up here in quadrant one, both coordinates are positive, which means that the cosine and sine are both positive. Here they are. If you could envision the circle and points in your head, then assigning large, medium, and small is trivial.
The sine of pi over 3 is the large value, which is the square root of 3 over 2. The sine of pi over 4 is the medium value, which is the square root of 2 over 2. And the sine of pi over 6 is the small value, which is 1/2.
Let's try a few more. What is this angle, and what is its cosine and sine? Well, assuming this is a unit circle, the cosine and sine of the angle are the x and y coordinates of the point. The angle is close to pi radians, but it's back pi over 6 radians, so 5 pi over 6 radians, and the coordinates are negative large and positive small.
So a negative square root of 3 over 2, comma, 1/2. The cosine is always the first coordinate, and the sine is always the second coordinate when the point is on the unit circle.
How about this angle? It's 7 pi over 4. Its coordinates are positive medium and negative medium. By the way, the mediums always go together, and they correspond to pi over 4 angles that bisect each quadrant.
The pi over 6 angles get a large and a small; just be careful of the order and the positive negative sign. Anyway, for 7 pi over 4, positive medium, negative medium is square root of 2 over 2, comma, negative square root of 2 over 2.
The cosine is always the first coordinate, and the sine is always the second coordinate when the point is on a unit circle. One more try: this one on your own. Be careful of the signs; they're both negative in quadrant 3: negative 1/2, comma, negative square root of 3 over 2.
The cosine is always the first coordinate, and the sine is always the second when the point is on the unit circle. Well, we have covered a lot of material. It isn't everything you would cover in a trig course, but it's a pretty good review or a great foundation if you haven't taken trig yet.
This video is actually intended to lay the foundation for my other video series on the graphical representation of the trig functions. It continues the theme of not memorizing anything that you could otherwise figure out quickly.
As a memory aid, I assign each trig function a color, and yes, cosine is blue and sine is red. I hope you'll check it out and find it helpful. Please share this video with your friends who you think might benefit from it.
I'm Dennis Davis. Thank you for watching.