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]]>Now we can talk about some advanced trigonometry formulas. In this video, I will discuss a few advanced formulas from trigonometry. As a general rule, you don’t need to have anything in this video memorized. I’m going to show you some complicated trigonometry formulas. You simply need to be able to work with the formulas if and when the test presents you with it.

Having some familiarity with these formulas ahead of time will make that much easier.

First of all, what I’ll call the sign rules, this first category has the rules for when the positive and negative signs of x changes. The sin (-x) = -sin (x), but the cos(-x) = just cos (x).

In the unit circle, starting from 0, if we move some angle clockwise and then the same angle counterclockwise, we’ll arrive at points that have opposite y-coordinates and the same x-coordinates. So in other words, we’re going to move this way, and this way. And those two points have the exact same x-coordinate, but they have opposite y-coordinates.

And so that’s why the sines, the sin of x are negatives of each other, but the cosines of x are identical.

These formulas have implications for the shape of the graph, the standard cosine graph is a reflection of itself over the y-axis. The sine graph is an image of itself under 180 degrees rotational symmetry around the origin.

Now, if you’re familiar with the ideas of an even function and an odd function, cosine is an even function and sine is an odd function. That’s a convenient thing to know, but the ACT does not ask about those kinds of symmetries.

Next, we’ll talk about the angle addition and subtraction rules. We can derive the exact values of sine and cosine for three acute angles.

Pi over 6, pi over 4, and pi over 3, those are our angles in the special triangles. If we add and subtract these angles in various combinations, we can get a few more angles. And mathematicians have derived the formulas for the sine or cosine of the sum or difference of two known angles. Assume that alpha and beta are angles for which we know the values of sine and cosine.

These four formulas are the sine of alpha plus beta, the sine of alpha minus beta. The cosine of alpha plus beta, and the cosine of alpha minus beta. So, again, you do not have to have these four complicated formulas memorized. The test will give you one of these if you’re expected to know it. But it’s useful to practice with them. So if you are given one in a problem, it is familiar.

Here’s a practice problem, pause the video and then we’ll talk about this.

Okay. For Q1 angles, Quadrant 1 angles alpha and beta, the sin of alpha is three-fifths, and the cosine of beta is twelve-thirteenths. Given that, find the cosine of alpha + beta. Well, the first thing we need to do is we need to recognize that we’re dealing with some very important Pythagorean Triplets.

And if the idea of Pythagorean Triplets are not familiar to you. I’d suggest go back and watch the video Right Triangle in the section on Geometry. We’re dealing with these Pythagorean Triplets, 3, 4, 5, and 5, 12, 13. So notice that alpha has a sine of three-fifths so we see that the adjacent leg is 4. Beta has a cosine of twelve-thirteenths, so the opposite leg is 5.

And this means that we can find the cosine of alpha, that’s four-fifths, and the sine of beta, that’s five-thirteenths. So now that we have these four values we can plug into the formula. They give us the formula, cosine of alpha + beta = cosine alpha cosine beta minus sin alpha sin beta. So we plug all these values in and multiply, we get forty eight-sixty fifths minus fifteen-sixty fifths, which is thirty three-sixty fifths. And answer choice A is the answer.

Next, we’ll talk about formulas for non-right triangles. The SOHCAHTOA relationships are wonderful for solving the sides of right triangles. But most triangles in geometry and many triangles in real life are not right triangles. For this, we will follow the conventions that vertices are denoted by capital letters, which also serve as the angle names.

And each side is the lower case of the same letter as its opposite vertex. So, for sample here, we see we have the three vertices, A, B and C. And opposite from the angle is the side indicated by the lower case letter of the same letter. So for any triangle, A,B,C, there are two important rules for these non-right triangles.

One of them is The Law of Cosines, which is kind of a generalized version of the Pythagorean Theorem and then The Law of Sines. So given the numerical values in a combination, SAS, that is to say, side, side, and an included angle. Or ASA, angle, angle, and an included side; or AAS, two angles and a non included side, or just all three sides, SSS.

Now, notice those are the four combinations that determine a triangle, they are good enough for triangle congruence. So if we’re given numerical values in any of those combinations, we could find all the other angles and sides of the triangle. Here is a practice problem. Pause the video and then we will talk about this.

Okay, so here we are given side, side, side. We are given the three side lengths, and we want to find the cosine of angle C, cosine of that larger angle. It kind of appears from the diagram that that angle is an obtuse angle, an angle greater than 90 degrees. So we’re actually expecting that the cosine of it will be negative. So that’s just a prediction, let’s see how this is borne out by the numbers.

Plugging in, we get 2 squared, which is 4 + 3 squared, which is 9 minus 12*cos(C) = 16. Subtract the 9 and the 4, so we get -12*cos(C) = 16 minus 9 minus 4, which is 3. Divided by -12, we get cosine of C = 3 divided by -12 or -3 over 12, which is negative one-quarter.

So, indeed, the cosine is negative and the answer is D. You do not need to have the rules discussed here memorized. But it’s good to do enough practice problems with them so you are familiar with them and are comfortable with using them. So that way if you have a problem and the test hands you the formula and says, use this, you’ll already be comfortable with it.

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]]>Inverse trig functions. All the trig functions have an input that is an angle and they give an output that is a ratio. Sometimes we know the ratio and we need to find the angle measure. For this, we need the inverse trig functions, which undo the direction of the original trig functions.

They switch around what’s the input and what’s the output. Each trig function has its associated inverse function. One way to denote this inverse is by writing the prefix “Arc” in front of the function. So Arc sine is the inverse of sine. Arc cosine is the inverse of cosine.

And Arc tangent is the inverse of tangent. Another way of denoting this inverse is to write in power of -1 immediately following the function. So sine to the -1 (x) is the inverse of sine x. Cosign to the -1 of (x) is the inverse of cosine (x), and tangent of the -1 of x is the inverse of tangent x.

Notice that we find the inverses only of the three major trig functions, not the others. We don’t have to worry about the inverses of co-tangent, secant, and co-secant. The test doesn’t ask about those. You need to be comfortable with both notations for the inverse trig functions. Both the negative one notation as well as the arc notation.

For the purpose of the test, we only have to concern ourselves with quadrant one angles and positive ratios. The rules get more complicated when we consider angles of other quadrants, but the test doesn’t explore those issues.

**The basic idea is that the inverse function reverses the input-output pair of the original function. **

In other words, if the cosine of K equals 0.375, then the arc cosine of .375 would equal the angle K. When we are working with angles in triangles in this context, we use degrees, not radians. So we always use degrees for triangles in this context.

In the right triangle shown, which of the following is equal to the measure of angle A? So pause the video and then we’ll talk about this.

Okay, well certainly it’s true that the cos(A) = 15/17. And so it can’t be that A would equal the cosine of something other than 15 over 17. It also is true that that ratio of 15 over 17, since that’s the cosine, it would not be the sine or the tangent, so neither of those work.

It’s also true that the sine of A is 8 over 17. Opposite over hypotenuse, 8 over 17. And so that means that the sine inverse of 8 over 17 would have to equal A. So B is the answer here.

Here’s another problem, a slightly harder problem. Pause the video, and then we’ll talk about this.

Okay, so let’s think about this carefully. The arcsin of d over f is just angle D because the sine of angle D is d over f, so the arcsin of d over f is just D. The arccos(d/f), well what angle has a cosine of d/f? Well the angle E. So the arccos(d/f) = E.

So that expression arcsin(d/f) + arccos(d/f), that’s just angle D + angle E. And those two angles have to add up to 90 degrees. So if C is the answer. The inverse trig functions can be written with either of two different notations, either the arc notation Arcsine, Arccosine and Arctangent.

Or the power-of-negative-one notation. Sine to the negative 1, cosine to the negative 1, tangent to the negative 1. You have to recognize both of those. Inverse trig functions have inputs of ratios and outputs of angles.

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]]>Trigonometric functions. A whole other side of trigonometry opens up when we begin to consider sine and cosine not merely as functions of angle, but as functions of x that we can graph against y. We have to be very careful to distinguish between x and y. The graphing variables on the one hand, from the point x, y where the radius at an angle theta intersects the unit circle.

So we have to be really careful about this distinction, in order to understand this.

First of all, let’s think about how the values of sine and cosine change once around the unit circle. So here we are in the unit circle. In the first quadrant, as theta increases from 0 to 2 pi, so of course as we are going up and around the circle–the x is decreasing, the y is increasing. This means that sine increases from 0 to 1 and cosine decreases from 1 to 0.

In the second quadrant, as theta increases from pi over 2 to pi, well now. Notice that if you imagined a tangent vector, it would be pointed down to the third quadrant. And so both x and y are decreasing. Sine decreases from 1 to 0 and cosine decreases from 0 to -1. So this is the only instance in which both the sine and the cosine are decreasing at once.

In the third quadrant, of course, they are both negative in the third quadrant. But what happens in the third quadrant from pi to 3 pi over 2, is that sine continues to decrease. It decreases from 0 to negative 1, we actually go down to the bottom of the unit circle but cosine now starts to increase. Cosine had already reached it’s minimum value at pi, so now it’s increasing from -1 to 0. It’s still negative but increasing.

And in the fourth quadrant, sin increases from -1 to 0 and cosine increases from 0 to positive 1. So both of them are increasing in the fourth quadrant. When we graph y = sin(x) and y=cos(x), the angle becomes the horizontal axis, and the output becomes the vertical axis. And so these are the graphs. The graph of sin(x) which starts at 0 increases as we move towards pi over 2, decreases and equals 0 again at pi, then becomes negative as we move from pi to 3pi over 2, and then comes back up to 0 at 2pi.

Cosine starts at its maximum, starts at 1, decreases to 0 at pi over 2, goes down to negative 1 at pi. Curves back up to zero at 3pi over 2, and then increases to its maximum again at 2pi, very beautiful, elegant curves. Both graphs are really the same shape, simply shifted from each other. This shape is called the **sinusoidal curve**, and it is the shape of many waves.

It’s a very important shape in mathematics. Notice that both graphs repeat the same pattern like wallpaper. This is because when the radius turns 2pi, all the way around the circle, the values go through all the same cycles again. So you can go around and around the circle. You’ll just repeat those same values, which is why the pattern repeats like wallpaper.

This distance on the x-axis pi over 2, is known as the period, the distance at which all the values repeat. So in other words, the sin(x) = sin(x) + pi over 2 for any value of x, similarly for cos(x). The standard function y = sin(x) and y = cos(x) each have a period of 2pi in a range given by y is greater than or equal to negative 1, and less than or equal to 1.

Now the test may ask about these standard functions, but it is much more interested in transformations of these functions. So in other words, when we start multiplying and adding a bunch of different values in the equation. Both sine and cosine will respond to the transformations in the same way. So here I will just talk about the transformations of y = sin(x).

Everything I say is going to be true for cosine(x) as well. First, I will distinguish between operations on the outside and operations on the inside. On the outside, so that would be for example, multiplying on the outside. When we multiply the entire sine function on the outside, or outside the parenthesis, we’ll more adding d.

Those are the outside operations. These affect the graph vertically.

By contrast, the inside operations, these are the things inside of the parentheses that were feeding into the sine function. The b times the x, and then adding something into the x before it goes into the sine, these affect the graph horizontally.

Very important outside operations affect the graph vertically, inside operations affect the graph horizontally. Also, when we multiply or divide, that tends to scale the graph. In other words, we either stretch it and make it bigger in one direction or we shrink it and make it smaller in one direction.

So for example, in all four of these drawings, the green M is the same size.

In the first one, the purple M is a vertically stretched version. Now in the second one, it’s a vertically shrunk one. We’ve made it smaller in the vertical direction, even though in both of those it stays in the same place horizontally. In the third one, we stretch it horizontally, everything stays in the same place vertically, but it’s gotten wider.

And then the fourth one we’ve contracted it or shrunk it horizontally, whereas again it remains unchanged in the vertical direction. When we add or subtract, that tends to move the unchanged shape in one direction or another. So again here, we have the original green M, and then we just see it moved in four different directions.

First of all, let’s talk about multiplication outside.

So this is going to be a vertical change, because it’s an outside change and it’s going to scale. It’s not going to shift it, it’s going to scale it and change the size. If a is greater then 1, then the graph expands vertically, it vertically stretches.

And so, here’s an example, in all of these, the green graph will be the ordinary y = sin(x) and the purple graph will be the expanded graph, will be the transformed graph.

So here, y = 3sin(x) is much taller, goes up much higher goes down much lower. Notice that it intersects the x-axis at exactly the same place. Horizontally, it hasn’t changed, it’s only changed vertically. If a is a number between 0 and then 1, a fraction, then the graph contracts vertically, it shrinks vertically.

So again, the unchanged graph is the green, and here we have a purple graph which goes up to a much lower peak and then down to a much lower trough. So it’s much closer to being flat than the original graph. If we add on the outside, that just shifts it. If we add a positive number, the graph unchanged in shape, moves up. So again, the green is the unchanged and the purple, we’ve just shift the graph up one unit, but it’s exactly the same shape.

And everything horizontally stays in the same place. If what we add is a negative number, in other words, we subtract on the outside, then the graph unchanged in shape moves down. And so here we have the original graph and then Y equals sin minus 2, the purple graph which has been shifted into the negative region below the x-axis.

Now, we look at multiplication inside, this is a horizontal change.

If b is greater than 1, then the graph horizontally contracts. And this could be a little bit anti-intuitive because you think that multiplying by a larger number would make things larger. You’ll find that everything that happens on the inside is anti-intuitive, it’s the opposite of the way you think it would work. So here, we have the graph of y = sin(x), and then the graph of sin(2x).

And 2x, we see, it’s been horizontally shrunk, it has the same heights, the same peaks and trough heights. But now it has been shrunk, so repeats much more frequently, it becomes a higher frequency wave. In the period of this one is just pi. Now, if we give a fraction we multiply by a fraction inside then the graph is horizontally expands.

So here we have the graph y=sin(x), and the graph y=sin(x/2). And so this one has horizontally expanded, it stretched out, and this actually has a period of 4pi. We don’t even show the full period of it, it would go off the screen because it is a very very wide period.

Changing the value of b changes the period. An ordinary y = sin x has a period of just 2pi, if we have y = sin(bx), this is a period of 2pi divided by the absolute value of b. That’s a handy formula to know. If b is greater than 1, the period gets smaller, and if b is a fraction between 0 and 1, then the period gets bigger.

And again, this is anti-intuitive, it’s very good to be careful about this, because the trap would be to think well b is bigger period gets bigger. Don’t think that way, you have to be very careful when we’re doing operations on the inside. Addition on the inside. When we add on the inside, then the graph unchanged in shape, moves to the left.

In other words when we add moves in the negative direction, again anti-intuitive. So here we have the original sine curve, and then we’ve just shifted it to the left in that graph of y=sin(x+pi/6). If we have a negative inside, if we subtract inside, then the shape unchanged moves to the right. So adding a positive number moves you to the left.

And subtracting a positive number moves you to the right. So here, we have the original graph again and the graph of y = sin(x)- 2pi/3). And so whole graph has shifted over to the right. In the big world of mathematics, and for example, even in your own math class, you might be expected to deal with a function in which a, b, c, and d all change.

All four of those change at once. So we’re doing a vertical shift, a vertical stretch, a horizontal shift,and a horizontal stretch all at once. As a general rule, that’s a little more than the a,c,d is gonna ask you. The test tends not to give more than one transformation in each direction. At most, it tends to have just one operation on the inside, and just one on the outside. That enormously simplifies things.

Here is the practice problem, pause the video and then we’ll talk about this.

Okay, trigonometric functions with the equation y = a*cos(bx). Where a and b are real numbers, is graphed in the standard x, y plane. Which of the following is the period of the function, which one is it? Well, remember period can be defined in a few different ways. We can define where it intersects the x-axis, and then it would have to go through an upper loop and a lower loop, or a lower loop and a upper loop.

So for example, from here down and then up to here, that would be a period. Of course, those are in between, it’s a bit hard to tell exactly what the values are there. Another way to find period is peak to peak, another way to find it is trough to trough. Well with the cosine, peak to peak is very convenient because one peak is right here at zero.

The next peak is right here, at 5pi, and so that has to be the period right there, 5pi. The answer is B.

Here’s another practice problem, pause the video and then we’ll talk about this.

Okay, the functions y = sin(x), and y = sin(ax) + b, for constants a and b, are graphs in the standard (x, y) coordinate plane below.

Which of the following statements about a and be is true, which statement is it? Well, we see the purple one, the purple graph, that’s the ordinary y = sin(x) graph. So we have to figure out that blue one. Well, one thing that’s true about that blue one, is that it’s been shifted down.

It’s no longer centered on a mid-line that’s up the x axis, the mid-line of it is a horizontal line at x equals negative 2. So it’s been shifted down, so the b is definitely negative. On the outside, we have added a negative number. So we can eliminate A and C right away. That means that C or D, A and B are eliminated, C or D could possibly be the answer.

Now, we are to think about what’s going on horizontally. Well, clearly it’s been stretched out. Here we have to be very careful, we’re stretching it out, which means that we’re not multiplying by a larger number. Multiplying by a larger number would actually shrink the period, to stretch out the period, we have to multiply by a fraction less than 1.

And so in fact, C is wrong and D is the correct answer.

In summary, know and recognize the shapes of the basic sine and cosine graphs. Be able to read the period of a transformed sine function from the graph, and understand the rules of transformation. How to change, how changes to the algebraic formula for the function affect the position and the appearance on the graph.

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]]>The post Trigonometry: Degrees and Radians appeared first on Magoosh Math.

]]>Now we can introduce the very important idea of radians. And I’ll start by saying, have you ever wondered why a right angle is defined as 90 degrees? Why that number 90? Where did the idea of degrees come from? The system goes back to the ancient Babylonian religion, over 3000 years ago.

So here’s a picture of the Babylonian God, Marduk, and his pet dragon. These ancient Babylonians believed that 60 and 360 were sacred numbers. They believed all kinds of things, they believed a host of strange Gods, they sacrificed barnyard animals for various things. It’s a very old religion that died out a long time ago. But we still hang on to these two numbers.

Turns out that the Babylonians thought all-the-way around a circle should divided into 360 parts, so that’s the origin of the degree system. These Babylonian values are also the reason we have 60 minutes in an hour and 60 seconds in a minute. So it turns out that even though this religion has been dead for millennium, turns out their values are still influencing a lot about how we look at the world.

The fact that Babylonians decided to measure angles this way, based on the values of their ancient pagan religion, that does not necessarily mean that, mathematically, it’s the best way to measure angles. Degrees are familiar and commonly used, but they are not mathematically the best way to measure angles. So suppose we started from scratch, and we wanted to come up with the best way to measure angles.

The best way to measure angles, the most natural way to measure angles, derives not from some arbitrary choice based on a pagan religion. But it would derive from qualities of the circle itself. That’s the way the mathematician thinks, what are the qualities of the circle itself? Think about a circle.

What is the most fundamental length associated with the circle, from which everything about the circle could be calculated? Obviously the radius. We can calculate the area from the radius, we can calculate the circumference from the radius. We can calculate everything we need from the radius.

If we could use the radius of a circle to define an angular measurement, this could be the basis of a new system for measuring angles. And this is precisely what we’re gonna do. This system is known as radians. So remember that arcs of circles have length. Suppose we made an arc that had exactly the same length as the radius.

So here, we have a little sector. And of course there are two radiuses going out from the center of the circle. And then the arc length also, even though it’s curved, it has the same length. So all three sides of this little sector have the same length. This would define an angle of one radian. And we could use this as a unit of angle.

So we could measure radians around the circle, and also that would be the same thing as measuring the number of radiuses around the circle. In fact, we already measure all the way around the circle using radius lengths. We know that the circumference equals 2 pi r. And so that means there are 2 pi radius lengths around the circumference. 2 pi radius lengths all the way around the circle.

This means that there are 2 pi radians all the way around the circle. So all the way around the circle, the thing that we used to call 360 degrees, we’re now gonna call that 2 pi radians. So if 360 degree = 2 pi radians, notice that for degrees we have to use the degree symbol. Radians are so special that we don’t need a special symbol for them.

We don’t need to write r, or radians, or anything like that. We can just write 2 pi and that implies that it is 2 pi radians. And this allows us to figure out how to relate the entire degree system to the entire radian system. So divide that equation by 2, we get 180 = pi, that’s very important. 180 degree angle is the same as a pi radian angle.

Divide that again, a right angle which we use to know as 90 degrees, we can now call that an angle of pi/2. So that is the measure over right angle in radians, pi/2. Now start with that right angle, if we divide that by 2, then we get a 45 degree angle and that’s pi/4. Again go back to the pi/2, instead divide it by 3.

We get 30 degrees = pi/6, if we multiply both sides by 2, we get 60 degrees = pi/3. These are the values of all the special triangles angles in radians. So the two special triangles one of them is a pi/4, pi/4, pi/2 triangle. The other is a pi/6, pi/3, pi/2 triangle. The equation, 180 degrees = pi is often a good place to start because 180 has so many factors.

Suppose we needed to find, say, 20 degrees in radians. Well of course, 180 = 20 x 9, and of course that equals pi. Just divide both sides by 9 and we get 20 degrees = pi/9. This can be a very quick way if the number of degrees is an obvious factor of 180. We can also change the equation 180 degrees = pi into a fraction equal to 1.

So 1 = pi/180, we can multiply any degree measurement by this fraction to change from degrees to radians. For example, 270 degrees multiplied by pi/180, we get 270 pi/180, which is 27/18. Divide top and bottom by 9, we get 3 pi/2, so 3 pi/2 is 270 degrees.

In fact, we can define the four quadrants in terms or radians. The first quadrant is 0 to pi /2, 0 to a right angle. The second quadrant is pi/2 to pi. A right angle to a full semicricle. Third quadrant is pi to 3pi/2. The first quadrant is 3pi/2 to 2pi.

This is often how the test will specify a particular quadrant. They won’t say third quadrant, they will say theta is between pi and 3pi/2. Which of the following is the value of 240 degrees in radians? Pause the video and then we’ll talk about this.

Okay, so method one, this is a more visual method. Notice that 240 degrees is twice of 120 degrees, and 120 degrees is a third of the circle.

This means that 240 degrees is 2/3 of the circle. A whole circle is 2 pi, so 2/3 of 2 pi would be 4/3 pi. So that’s what 240 degrees is in radians. That’s method one.

Method two would be the more methodical method to take that number and multiply by pi/180.

When we multiply, we cancel the 0’s, then we cancel a factor of 6, and we get 4 pi/3. And so it turns out that the answer is answer choice D. Let me just point out, it’s very methodical to just multiply by pi/180. I’m going to discourage you from making your 100% of the time go to method for converting to radians.

It’s much more efficient to think in terms of proportions, and to think visually about the problem. You will understand radians much more deeply if you’re not simply reliant on multiplying by that ratio every time. The radian values of pi are easy to relate to a visual understanding. A 3pi/4 angle is 3/4 of pi, that means it’s 3/4 of a semicircle.

So it’s very easy to imagine. Take a semicircle, divide it into quarters, 3 of those quarters is a 3 pi/4 angle. To understand radians deeply when you can think about them visually as well as numerically. The full advantages of radians are not necessarily obvious at this level.

They become far more evident in calculus, when we start doing calculus with trigonometric functions. That’s where radians really prove their full value. Obviously, that puts us well beyond anything on the ACT, so we’re not gonna have to worry about that for the ACT. That’s just a preview for why radians are so important.

In summary, degrees are the common way to measure angles, but radians are a mathematically preferable way to do so. The number of radians in the central angle of the circle equals the number of radius-lengths in its arc. This means that 360 degrees all the way around the circle is 2 pi radians and a 180 degrees, a semicircle, is pi radians.

To change from degrees to radians, we can think about simple proportions, or we can just multiply the number of degrees by the fraction pi/180.

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]]>Now we can continue our discussion of unit circle trigonometry. So in the previous video, we discussed the basic definitions of the elementary trig functions sign and cosign from the unit circle. And just a brief reminder here, what we did was of course. We put our socket to a triangle inside the unit circle so that the hypotenuse to the unit circle is the radius and of course the radius of the unit circle is 1.

That means that the adjacent side winds up equaling the x, the opposite side winds up equaling y. And we can define cosine as sine. Cosine of theta equals x and sine of theta equals y.

In other words sine and cosine are defined in terms of. If we have a radius at an angle theta where does that radius intersect the unit circle? And that the coordinates of that point x, y where the radius equals the unit circle provide the definition for sine and cosine.

And so that’s the basic unit circle definition of sine and cosine. In the last video we talked about how to find the values of sine, and cosine, simply on the axes. From the positive and negative x-axis, and the positive and negative y-axis. In this video, we can talk about getting into the quadrants. When we start looking at slanting angles that aren’t on the axis.

So first let’s talk about the quadrants. The quadrant of theta tells us the positive or negative sign of sine and cosine, precisely because we know the positive and negative signs of x and y in the different quadrants. Let’s think about this. In the first quadrant, x and y are both positive. And so that means that sin and cos are both positive there.

That makes sense. All the outputs are first quadrant outputs, all of those are positive. Now, when we move to the second quadrant In the second quadrant, remember that y is still positive but now cosine is negative. And so that means sine is positive and cosine is negative in the second quadrant. When we move to the third quadrant both y and x are negative so both sine and cosine are, and are negative.

When we move to the fourth quadrant, that now we have that the y is negative but the x is positive. And so that means that the sign is negative and the cosine is positive there. It’s very easy to tell the positive or negative signs of these two functions a different quadrant purely from the quadrant. The quadrant itself gives us the information that’s really important.

We can also look at tangent so tangent is sine over cosine of that is y over x. And so that’s gonna be positive wherever y and x have the same sign. Well, y and x are both positive in the first quadrant. And they’re both negative in the third quadrant.

So positive over positive is positive, but negative over negative is also positive. And so tangent is positive when sine and cosine have the same sign, either they’re both positive or both negative. And this happens in the first and the third quadrant.

In the other two quadrants, x and y have opposite signs. One is positive, and one is negative. So you have a negative over a positive or a positive over a negative that produces a negative fraction. And so tangent is negative, when sine and cosine have opposite signs in the second and fourth quadrants. Notice that sine, cosine, and tangent, all three are positive in quadrant one.

That’s the only quadrant in which all three are positive, and then in each other quadrant. Two of them are negative and only one of them is positive. And so as we go around all of our positive sin is positive, tangent is positive, cosine is positive. Some people like to remember that as all students take calculus.

If that helps you remember it that’s great. Now how does this help us? Well now we’re on our way to figuring out the values of sine, cosine and tangent for angles that are not on the axis, angles that are in the quadrants. And it’s very important to realize that we have the information of a positive or a negative purely from the quadrants.

So now suppose we need the values of sin of 150 degrees and cosine of 150 degrees. This is a quadrant two angle. So let’s think about what we know first. From the quadrant we know right away that of course we’re in the second quadrant, Y is positive, X is negative. So that means that the sine of under 50° will be positive, the cosine will be negative.

We easily have the positive and negative sines, now we need to get the numerical values. So let’s talk about that. Notice that 150°, Equals 180 minus 30. And so that means that one way to get to the angle of 150 degrees would be to go a full 180, and then back up 30. Let’s think about that. And so that means we could draw a little triangle in Q2.

And we see what we get that little triangle in QII there with the 30 degree angle. That is congruent to the first quadrant triangle that would have a 30 degree angle. What that means is, the lengths of those sides are gonna be exactly the same as the lengths of the sides of the 30 degree angle. And of course the lengths of the side determine the values to sin and cosine.

So it means that at least the absolute value of sin will be the same for 150 and 30. The absolute value of cosine will be the same for 150 and 30. So what are the values of sin and cosine of 30 degrees? Well we talked about this a little bit a couple videos back. And so you may remember that we figured out there that the sin(30°) = 1/2, the cos(30°) = root 3/2.

Well now we’re golden, because sine of sine and cosine of 150 degrees will have those same absolute values. So now we know the absolute values, all we need to know is the positive and negative signs. But we already have that figured out we get that from the quadrants. And so the sine is still gonna be positive and the cosine is gonna be negative in the second quadrant and so that tells us right away. Just take these absolute values and adjust the positive and negative signs.

So, sine of 150 degrees is gonna be one-half cosine of 150 degrees is gonna be negative root 3 over 2. And that’s how we find those values.

To solve that we used a reference angle. This is a really important idea. This is one of the most important ideas in trigonometry. The idea of a reference angle for any angle greater than 90 degrees.

So we wrap around into the second quadrant, the third quadrant, and the fourth quadrant. What we wanna do is we imagine a segment. That goes from where that point ends up, dropped to the x-axis. You’ll be drawing to either the positive or negative x-axis, and you’d make a little triangle and that little triangle would have an acute angle in it. That acute angle is the reference angle.

And it turns out that that triangle you make is always gonna become congruent to the triangle that you would have in the first quadrant with the reference angle. And so all the values of the trig function In any quadrant, will be equal to, they’ll have the same absolute value as the same trig functions of the reference angle. Of course the reference angle will have positive outputs. So the only thing you’ll have to figure out is what their sine or cosine should be negative in another quadrant.

And again, we can figure that out from the quadrant itself. So any trig function of any large angle has the same absolute value as that trig function of the reference angle. And then we just find the positive or negative from the quadrant and we’re done. So we get the positive or the negative from the quadrant. We get the numerical value from the reference setting.

So let’s stick with the 30 degrees and let’s think about what would have a reference angle of 30 degrees. We’ve already done it in the second quadrant. What about in the third quadrant? So in the second quadrant it’s 150 degree angle that has the reference of 30. In the third quadrant we have to go all the way to 180 degrees and then we have to go 30 beyond that so would be at 210 degrees.

And so 210 degrees has a reference angle 30. So it has exactly the same absolute values as sine and cosine of 30 degrees, but now both of them have to be negative. Now we could say that the sine of 210 degrees is negative one-half. The cosine of 210 degrees is negative root 3 over 2. So again we’re using the same absolute values. And just putting a negative sign in front of both of them well now let’s go to QIV.

Well in QIV, we have to wrap all the way around to 360 degrees, and then back up 30 degrees so that would put us at 330 degrees. And so that’s the angle in QIV that has a reference angle of 30 degrees. So again, same absolutely value as the sine of 30 degrees and the cosine of 30 degrees. In QIV we know that x is positive and y is negative.

So sine is negative and cosine is positive. Those same values with the negative sine sin(330°) = 1/2, cos(330°) = positive root 3/2. And so notice that we get to recycle the SOHCAHTOA values in the first quadrant. We get to recycle those in each quadrant just by flipping around the positive and negative signs. That’s the beauty of reference angle.

So the test tends not to ask us directly to evaluate sine and cosine for angles such as this. Nevertheless, knowing and understanding reference angle is an extremely powerful problem solving tool. And again, I’ll put in another advertisement here. If you’re moving on to calculus, you will have to do evaluations like this once you’re in calculus.

And so this is really an important idea to understand if you’re thinking about taking calculus later on.

Here’s a practice problem, test-like practice problem. Pause the video and then we’ll talk about this. Okay. If the tangent of y, some angle y, equals 2 root 2, and we know that y is between 180 and 270 degrees. So in other words, it is a third quadrant angle which of the following is the value of cosine?

And so I’ll just point out here third quadrant, we know both x and y are negative, so right away we know the cosine is going to be negative. And in fact all five answer choices are negative, so we’re gonna have to pick a negative anyway. Let’s think about how we’d find that numerical value. So we’re gonna use the reference angle.

Let’s just suppose we build an acute triangle where angle has a tangent of two root two. And so we’re gonna let R be the reference angle of Y. So R and Y, they’re two, R is an acute angle, Y is a third quadrant angle, but R is the reference angle of Y. And we know that if the tangent is two over two, we could just make the opposite equal to two over two, and the adjacent equal to one.

This is actually a triangle that would not fit inside the unit circle, it’s a little bigger. That doesn’t matter, the ratios will still be the same. And so we need to find d. And so we can do that just with the Pythagorean Theorem. So ordinary Pythagorean Theorem, d squared = 1 squared + (2 root 2) squared.

Well that’s 1 squared, and then 2 root 2, that’s 2 squared, which is 4 times root two squared which is two and so that’s four root two which is eight and then eight plus one is nine, so that’s d squared. D squared has to be positive because it’s a length so it’s a square root of nine, which is three, so now we know d. So now inside that little reference triangle, we can find the cosine.

The cosine(R) is one-third. Well, now we’re golden. We know that the cosine(Y) will have exactly the same absolute value and we just have to adjust the positive for a negative sign. And so of course again third quadrant we need something negative. So we take this same absolute value and we make it negative and so the cosine of y would be -1/3.

So we go back to our answers and we choose -1/3.

In summary, we can figure out the positive or negative sign from the quadrant of the angle. For any angle greater than 90 degrees we know that the sine and cosine of that angle will have the same absolute values as the sine of the reference angle and the cosine of the reference angle.

And so we could figure out the little SOHCAHTOA of the reference angle. And you could figure this out either from the properties of special triangles, you might have to use the Pythagorean theorem, as we used in the practice problem. But we can figure out everything in that acute triangle and then we can adjust the positive or negative signs from the quadrant.

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]]>This is where trigonometry really gets interesting. So, first I’ll say that one of the far ranging patterns in mathematics, this is big picture. Is that once mathematicians had figured out how something works in a limited context, the next step is always to consider how the idea can be expanded to a broader context.

And that’s exactly what we’re gonna do in this video. One of the most elegant examples of this in all of mathematics, it’s really actually incredible in the greater view of mathematics–is simply expanding trigonometry from the limited SOHCAHTOA context, to the much more inclusive unit circle context. So that’s what we’re gonna talk about in this lesson.

And so I’m assuming at this point you’re comfortable with SOHCAHTOA, and that’s going to be our jumping off point. So, far we have defined everything. In the previous videos we’ve just talked about SOHCAHTOA in right angles. That’s all we’ve talked about, SOHCAHTOA and right triangles. And of course, in a right triangle, the acute angles must be bigger than 0 degrees, and smaller than 90 degrees.

That’s the range for the angle. And that’s a quite limited range if you think about it, because angles can be much much bigger than that. In the real world there many angles that are outside of that range, so consider the following. First of all, there are certainly architectural features that can have obtuse angles.

There’s are all kinds of geometric shapes. And we can see in the real world obtuse angles, we could have an obtuse angle between three points on a map, that sort of thing.

Now think about it. I could turn around once, or if I’m feeling really ambitious, I could turn around three times.

Well if I turn around three times, how much angle have I turned through? I’ve turned through 1,080 degrees of angle, okay? That just something simple I can do with my body. Now, let’s think about something mechanical, say the tire of a car. So let’s say the tire of a car, every time it turns is another 360 degrees. And say I drive that car from Boston to San Francisco, how much angle is that tire go through, probably millions of degrees.

So in the real world, there’s all kinds of angles that are much, much bigger than 90 degrees, and that’s why we need a larger context than SOHCAHTOA. To allow for the expanded range of angles, mathematicians move everything about trigonometry to the unit circle. So let’s briefly review. What is the unit circle?

The unit circle is in the xy plane. And it is a circle with a radius of 1, and a center at the origin. And so this is a picture of the unit circle. There it is in the xy plane, in the Cartesian plane. The equation of the unit circle is x squared + y squared = 1. This is something we talked about in the unit on coordinate geometry.

So that’s the equation of the unit circle. Notice that because it has a radius of 1, it has a circumference of 2 pi and has an area of pi. Those are the basic geometric facts about the unit circle. We will move our SOHCAHTOA triangle inside the unit circle, so that the vertex of the angle that we care about, the vertex of that angle is at the origin and the hypotenuse is the radius.

So here’s the SOHCAHTOA triangle situated inside the unit circle. Let’s think about this. So the radius has a length of 1 and so that’s the hypotenuse, okay? The point where the radius intersects the unit circle, we’re gonna call that point x, y.

And that’s gonna be a very special point, the point where that radius intersects the unit circle. And notice that x would be the horizontal distance from the y-axis to the point. In other words, this x distance here from the y axis over to that point, that’s the distance x and that equals the adjacent. Similarly, the y coordinate is the height above the x axis.

And so that’s equal to the opposite. And so something really important is happening here, let’s take a closer look. So again that horizontal distance from the origin along that horizontal leg, that is the distance x. And the vertical distance along the vertical leg from the x axis up to the point, that’s y.

Well this is really interesting. Because now it’s very clear that sine, that’s opposite over adjacent, y over 1, opposite over hypotenuse, so that’s y over 1. So that is just y, because of course the hypotenuse is 1. Similarly, the cosine, adjacent over hypotenuse, x over 1 is x. Sine equals y, cosine equals x.

Sine and cosine equal, respectively, the x and y coordinates of the point where the radius intersects the unit circle. And this in fact is the unit circle definition of sine and cosine. So in other words, sine and cosine in this system are no longer defined merely in terms of opposite and adjacent, they’re defined in terms of the point where that radius intersects the unit circle.

And the x coordinate is the cosine, and the y coordinate is the sine. The brilliance of this new definition is that, while it is perfectly consistent with the SOHCAHTOA understanding within that range, it’s 100% equal to our SOHCAHTOA understanding if the angle happened to be between 0 and 90, this new definition allows for unlimited angle. For example, when the angle is 0, that makes no sense in a triangle.

We can’t have a triangle with an angle of 0, then it would be a flat thing. It wouldn’t really be a triangle. But we can talk about that in the unicircle definition, because the radius; think about a radius at angle 0. Where does it intersect the unicircle? It intersects it along the x-axis, the positive x-axis, at (1,0).

So that’s the point of intersection of the radius with the unit circle,(1, 0). Well what that means is that sine of 0 equals the y coordinate, 0. And cosine of 0 equals the x coordinate, 1. In accordance with the uni circle definition we can say sine of 0 is 0, cosine of 0 is 1. And so right there, that is the value we get from a new definition that would not be possible.

That’s totally impossible in the SOHCAHTOA system, but our new system allows us to assign values to those. Let’s continue around the unit circle. Let’s go to 90 degrees, and again in a right triangle we only have one 90 degree angle, we can’t have two 90 degree angles in a triangle. But let’s just look at this, at 90 degrees, we’d be along the positive y-axis, we’d be at the point (0,1).

And so of course what that means is sine of 90 degrees is the wide coordinate 1, cosine of 90 degrees, is 0. All right, now we’ll go all the way around to 180. So now we’ll be entirely outside a triangle. Now we’re getting into angles that would never be in a triangle, but that’s fine because we’re in a new definition.

If we started the positive x-axis and go on 180 degrees, then we’re at the negative x-axis. And we’re at the point (-1, 0). Of course what this means is that the sine of 180 degrees is 0. And the cosine of 180 degrees is -1, so that’s interesting. So now we’re getting 0 outputs, but we’re also getting negative outputs, from cosine.

Keep on going around the unit circle. We get to 270 degrees. And so that’s three-quarters of the way around the circle, starting at the positive x axis going across the first quadrant, second quadrant, third quadrant, and then we wind up at the negative y axis. We’re at the point (0,-1).

And so here, sine of 270 degrees equals -1, cosine of 270 degrees equals 0. So it turns out both sine and cosine can have negative outputs as we move into other quadrants. We’ll talk more about that in the next video. I just wanna point out here, that the angle could be 360 degrees or much greater.

We just wrap around the unit circle multiple times, and then see where we land. We could go negative, and negative angle would just mean that we’re going clockwise, instead of counterclockwise. Again, we go around and see where we land. So the angle could be anything on the number line, so now in terms of function, so let’s think about this in terms of functions for a minute.

Sine and cosine. When we’re in the SOHCAHTOA region, they had a range between 0 and 90 degrees, and they had a domain between 0 and 90 degrees. That was a very limited domain in terms of the unit circle. Now these two functions have domains of all real numbers, and that is a gigantic change. Now, we’re talking about functions, that are defined every point on the real axis.

So here’s a practice problem. Pause the video, and then we’ll talk about this.

Okay, so this is a problem that’s just good practice, with the values on the axes, all these angles are values either on the x axis or the y axis. And so it’s just a matter of figuring out that point, and then figuring the sine and the cosine.

So which of the following is equal in value to sine of -270 degrees? Hm, well let’s think about that. Negative, that means it’s a clockwise angle. We’re gonna start at the positive x-axis, and then we’re gonna go a three-quarter turn in a clockwise direction, and so that’s where we’ll wind up.

We’ll start and we’ll go three-quarters the way around the circle, and we’ll actually wind up on the positive y-axis. And so that point is (1,0) where the unit circle ends, that’s where the final radius arm intersects the unit circle at (1,0). And sine is the y coordinate of that point, and so the sine of 270 degrees is the same as sine as 90, it’s 1.

Because in fact, the angle 90 and the angle -270 aren’t exactly the same place. So they both have the same sine and so that equals 1. And so really the question is, which of those following have the value of 1? Well, what’s the sine of 180 degrees, well 180 degrees is along the negative x axis, the y value there is 0, so sine of 180 degrees is 0, that doesn’t work. What’s the sine of 270 degrees?

Well, positive 270 degrees, we start at the start at the positive x-axis, we wrap three-quarters of the way around the circle to the negative y-axis. That point would be (0,-1), the y-coordinate would be -1. And so the sin(270) = -1. So that doesn’t work. The sine of 360 is the same as the sine of (0,0) and 360 are in the same place, we’re intersecting the unit circle there at (1, 0) on the positive x-axis.

The y-coordinate is 0, so this equals 0, so this doesn’t work. Now the cosine of 180, all right, well, let’s think about this. 180, we’re on the negative x-axis. The cosine is the x-coordinate, so that point on the negative x-axis Is the point (-1, 0). And so the x coordinate, cosine is the x coordinate.

The x coordinate is -1. And so cosine of 180 degrees is -1. This doesn’t work. Now cosine of 360, remember 360 is the same as 0. So if 0 and 360 at the same place, it’s the point (1,0). The x coordinate is 1, and so this equals 1, and so this has the same value, and so e is the answer.

The unit circle allows us to define sine and cosine, for all possible angles. And once again this is absolutely remarkable, because it allows us to expand our understanding of how sine and cosine work, from a very limited context of inside a triangle, opening it up to all possible angles of all possible sides. The radius at an angle theta, intersects the unit circle at the point (x,y). And that point, the point of intersection of the radius with the unit circle, the coordinates of that point provide the definition for sine and cosine.

So sine(theta) = y coordinate of that point and cos(theta) = x coordinate of that point. And we’ll talk about more implications of this definition, in the next video.

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]]>Fundamental Trig Identities. So far, we’ve talked about the three main trig functions, sine, cosine and tangent. Those three are ratios. But technically, from the three sides of the SOHCAHTOA triangle, it’s actually possible to create six ratios. And so each of the six is a separate trig function and really all six are important to know. So we know three of them already.

So let’s take a look at the SOHCAHTOA triangle. There’s our familiar SOHCAHTOA triangle, happens to have an angle of 41 degrees, there’s an opposite adjacent hypotenuse side. And so certainly three of the ratios we can create are the familiar SOHCAHTOA ratios. But there are three more ratios we can create, and here they are.

Cotangent is adjacent over opposite, secant is hypotenuse over adjacent, and cosecant is hypotenuse over opposite. And those are the six ratios all together. So, wait a second, what are those names? Let’s look at these names very carefully, here are the full names. So, we’ve already talked about sine, cosine and tangent and now we’re talking about cotangent, secant and cosecant.

And notice the way they’re listed here. If you can remember the three on the left, the three on the right is just the same name with co in front of it. So at least some of these names have their origins in geometric relationships. Let’s talk about this for a minute. So now let’s look at a circle. Could be the unit circle, has a radius of 1, center at the origin, and so AB are CD are parallel to the y-axis.

So we have two vertical segments there, AB and CD. And it looks like B is the point where that radius line intersects the circle, it continues on. And D looks like it’s tangent to the circle where it crosses the x-axis. Okay, so notice a few things. That, in triangle OAB, the triangle inside the circle, OB, the radius is 1, and of course OA is the cosine, and AB is the sine, OK?

So that’s the familiar SOHCAHTOA ratio. Now, look at triangle, slightly bigger triangle, OCD. And so this one is the one that comes starts at O passes through B all the way out to C, drops down to D and goes back along the x-axis. Well in that triangle, OD is 1. And so that would mean that opposite CD over 1 equals the tangent, so the tangent equals CD.

And it means that hypotenuse over adjacent OC over 1 is secant. So OC equals the secant. But here’s the really cool thing about this diagram. Notice that CD, the segment that has a length equal to the tangent is actually tangent to the circle. It passes the circle and touches it at one point.

That is in fact a tangent line. Notice that OC, which is the secant, actually cuts through the circle. And so this is what’s known in geometry as a secant line. That’s why those two functions have those names because one represents the length of a tangent segment and one represents the length of a secant segment. And so, if you’re a very visual person, that might help you remember these things a little bit.

Okay, sine and cosine are the most elementary trig functions and we can actually express the other four in terms of them. And these are really important formulas to know.

Tangent we can write as sine over cosine. Cotangent, we can write as cosine over sine so notice those the two are reciprocals, tangent, cotangent are reciprocals.

Secant is the reciprocal of cosine. And cosecant is the reciprocal of sine. Notice that people get confused sometimes because they think the S and the S should go together. The C and the C should go together. They don’t.

Secant is the reciprocal of cosine. Cosecant is the reciprocal of sine. So the test may give you one of those if you need it in a problem, but it may expect you to remember it as well. So it’s really good. To have those four memorized.

Now in the first lesson on trig, we mentioned the fundamental Pythagorean identity. Cosine squared + sine squared = 1. Now, that we have two more functions we can also express the other Pythagorean identities. One of them is tangent squared + 1 = secant squared, one of them is cotangent squared + 1 = cosecant squared.

So, the test quite likely would give you these equations if a problem required them. But they may serve as a shortcut or a way to confirm the answer. Another thing I’ll say is if you’re planning to take calculus,I guarantee, I absolutely guarantee that you need to know all three of these equations cold, once you’re in calculus.

So I’ll say a few things about these. Of course you can blindly memorize them, but we don’t recommend that. What we really recommend is understanding them.

And so if you start with the one at the top, cosine squared plus sine squared = 1. You could divide every thing on both sides by cosine squared you would get the top of Pythagorean identity at the bottom tangent and secant.

Or you could divide everything in cosine squared plus sine squared = 1 by sine squared. And then you’d get the bottom one, cotangent squared and cosecant. Alternately you could go back to the original SOHCAHTOA triangle with ABC and start with the Pythagorean Theorem, A squared + B squared = C squared. You may remember that we got this top Pythagorean identity, cosine squared + sine squared = 1.

We got that from taking a squared plus b squared plus c squared and dividing everything, all three terms, by c squared. Well, instead of dividing by c squared, we can divide all three terms by either a squared or b squared. And if you do that and then sub in from the ratios what the trig functions are, you’ll produce these two Pythagorean identities.

And so I strongly suggest do that on your own, show in a couple different ways that you can come up with all these equations because then you’ll really understand them.

Okay, now we can move on to a practice problem. Pause the video and we’ll talk about this.

All right, in the triangle to the right, in terms of b and c, which of the following is the value of tangent theta?

All right, well let’s think about this. We have two sides there, we’re given b and c. And of course, c is the hypotenuse, b is the opposite, and tangent is opposite over adjacent. We have the opposite, we don’t have the adjacent, so we’re gonna need that third side.

Well we can use the Pythagorean theorem. So the Pythagorean theorem tells us that b squared plus whatever the adjacent side squared is, equals c squared. And we can solve this with the adjacent side. Adjacent squared equal c squared minus b squared take a square root of both sides. Notice that taking a square root, we cannot take a square root of c and b separately.

We have to leave it as that expression, c squared minus b squared. But that is an expression for the length of the adjacent sides, c squared minus b squared. Well now, now we’re golden because tangent is opposite over adjacent. We have the opposite we have the adjacent. So opposite over adjacent and that would equal b over the square root of c squared minus b squared.

And in fact that is answer C. We go back to the problem and we chose answer C. In summary, we introduced the other three trig functions. Cotangent, secant and cosecant. We discussed how to express the other four in terms of sine and cosine. So it’s very good to understand how they fit into the SOHCAHTOA triangle.

It’s very good to understand how they’re related to sine and cosine. And finally we discuss the three Pythagorean Identities.

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]]>Now that we’ve introduced SOHCAHTOA, we can talk about the very important topic of SOHCAHTOA and special triangles.

So this lesson will assume, first of all, that you’re familiar with SOHCAHTOA, which we covered in the previous lesson. And it will also assume that you’re familiar with the idea of the special right triangles.

These special right triangles are the 45-45-90 triangle and the 30-60-90 triangle. So, if you don’t know much about these triangles or it’s really fuzzy, it might help to go back to that lesson in the geometry module, special right triangles, which explains the two triangles in-depth. So here, we’ll be assuming that you already understand the basics of those triangles.

First of all, the 45-45-90 triangle. As you may remember, these are the angles and proportions. Sometimes people call this the 1-1- radical 2 triangle, that denotes the sides of the triangle. Sometimes it’s also called the Isosceles Right Triangle. This is the only possible triangle that is an Isosceles Right Triangle.

Because we know all the proportions for the sides, we can find the value of the three SOHCAHTOA functions for 45 degrees. So we’ll do that right now.

Sine, of course, if opposite over hypotenuse, so starting at either 45 degree angle the opposite is 1 and the hypotenuse is radical 2. Now we have to rationalize this fraction which means that we’re gonna multiply root 2 over root 2.

And that will give us root 2 over 2. That is the rationalized fraction, the rationalized form of the reciprocal of root 2. And that is the sign of 45 degrees. Well of course, the cosine of 45 degrees is going to be exactly the same thing. It’s gonna be 1/root 2 and we’re gonna rationalize it and get root 2/2.

So the sine and the cosine are equal precisely because the opposite and the adjacent are equal. The tangent opposite over adjacent, that is going to be 1/1, which is just 1.

By the way, notice, if an angle is less than 45 degree, then it has a tangent of less than 1. If an angle is greater than 45, then it has a tangent greater than 1.

So that’s an important point. And of course, when it’s exactly 45 degrees, the tangent is exactly 1. Now we’ll talk about the 30-60-90 triangle. As you may remember, we get this from cutting an equilateral triangle in half, these are the proportions. This is sometimes known as the 1-root 3-2 triangle that denotes its sides.

Sometimes, we also call it the half equilateral triangle to help us remember that, in fact, we got it from cutting an equilateral triangle in half. We can use SOHCAHTOA ratios for both the 30 degree and 60 degree angle using this triangle, because we know all the sides. So first of all, let’s start with 30 degrees. From 30 degrees the opposite is 1 the hypotenuse is 2.

So the sine is just one half. For the cosine, the adjacent is root 3, and my hypotenuse is still 2, so that’s root 3/2. For the tangent, the opposite is 1, and the adjacent is root 3. So that’s 1/root 3, again we have to rationalize this, multiply by root 3/root 3, and we get root 3/3.

And that is the rationalized form of the tangent of 30. So a couple of things to notice. Notice that 30 degrees is an angle less than 45 degrees. So the cosine has to be bigger than the sine precisely because for an angle less than 45 degrees, the adjacent has to be bigger than the opposite. And for much the same way, the tangent has to be less than 1.

Now, for the sine of 60 degrees. So the opposite over the hypotenuse now is root 3/2. The adjacent over the hypotenuse now is one over two, or one half. And the tangent now, opposite over adjacent is root 3/1, which is just root 3. So for 60 degrees, which is an angle greater than 45 degrees, notice that the sine has to be greater than the cosine and the tangent has to be greater than 1.

Both of those precisely because for an angle greater than 45 degrees, the opposite is bigger than the adjacent. Conceivably, the test could expect you to know any of those nine values. Most often, the test supplies these values but it is still good to understand their basis.

Here’s the practice problem. Pause the video and we’ll talk about this.

Okay, so let’s think about this. We can find the value of each one of those. And so the sine of 30 degrees is one-half, the cosine is root 3/2. We’re gonna square that. And by squaring it we’re square the first one, twice the product of the terms and then the square of the second one.

Notice that we canceled the 2 and the one-half, so that means that we just have 1/4- root 3/2 + 3/4. The 1/4 plus 3/4 is just 1. So it’s 1- root 3/2. We go back to our answer choices and we select that as our answer. In summary, here are the values that are good to know.

For the sine of 45 equals to cosine of 45 that’s root 2/2. The tangent is 1, the sine of 30 is one-half. The cosine of 60 is also one-half. The cosine of 30 is root 3/2, so is the sine of 60. The tangent of 30 is root 3/3 and the tangent of 60 is root 3. These are nine very good values to know.

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]]>This lesson assumes that you are familiar with the ideas of similar triangles covered in the geometry module. If the idea of similar triangles is absolutely unfamiliar to you, it might be helpful to watch that video in the geometry module before watching the trigonometry videos.

Recall that if we know just two angles in one triangle are equal to two angles in the other triangle, then the two triangles must be similar. And that means that they have the same basic shape. One is just a scaled up or a scaled down version of the other. All three angles are the same, and it’s the same basic shape. Once we know that the two triangles are similar, we know that all their sides are proportional.

It’s very easy to show that two triangles are similar, and once we know that, we get a lot of information. All of trigonometry is based on these key pieces of information about similar triangles. Suppose we think about all the right triangles in the world with, say, a 41 degree angle.

So here are some random right triangles that have a 41 degree angle. Of course, there are many different sizes and orientations, but all have the same basic shape. All of these 41 degree right triangles are similar, because they all share a 41 degree angle, as well as a 90 degree angle. That’s two angles they share in common, so they have to be similar.

This means all the sides are proportional. In other words, I could find the ratio in any one of them, and all these same ratios with the same in all the rest of them. The 41 degree angle is between a leg and a hypotenuse. We would call that leg, the leg that touches the 41 degree angle, the leg that is adjacent to that angle.

The other leg is opposite from the 41 degree angle, so we call that the opposite. So here we have the triangle with the three sides labeled, the hypotenuse, the opposite, and the adjacent. Now, the three principle ratios here are the sine ratio, sine = sin(41 degrees) = opposite over hypotenuse.

The cosine = adjacent over hypotenuse, the tangent = opposite over adjacent. Students often remember those three ratios using the mnemonic SOHCAHTOA. What is meant by SOHCAHTOA?

Well, SOHCAHTOA, sine is opposite over hypotenuse, that’s the S-O-H.

Cosine is adjacent over hypotenuse, that’s the S-A-H.

And tangent is the opposite over adjacent.

So we have to remember that it’s SOH-CAH-TOA. Notice that all three of these are written as functions of the angle 41 degrees, because if we change the angle, all the ratios would be different. Nevertheless, as long as we have a 41 degree right triangle, no matter the size or orientation, all these ratios will be the same.

The sine and cosine and tangent of 41 degrees, and of any other possible angle, are already stored in your calculator. You just have to make sure that your calculator is in degrees mode instead of radians mode. We’ll talk more about radians in an upcoming video.

Therefore, if we are given a right triangle with one known acute angle and one known length, we can always find the other two lengths.

So suppose we have this set up. We have a right triangle, we have an angle of 10 degrees, a tiny little acute angle, and opposite that 10 degree angle, the opposite side is 3 centimeters. And we want to find the other two lengths, for example. Well, certainly we know that the sine of 10 degrees is opposite over hypotenuse. So that would be 3 over side AB.

Now, if we multiply both sides by AB, we get AB times sin(10) = 3. Divide by sin(10), sin(10) is some number. So we divide by that, and if we needed, we could compute this on a calculator, sin(10) degrees is about 0.1736. 3 divided by that number is about 17.3, that’s the length of the hypotenuse, AB.

We could also find side AC. We know that the tangent of 10 is opposite over adjacent. This would be 3 over AC. Same thing, multiply by AC, divide by tan(10). Now, we can find this in our calculator, tan(10) is about 0.1763, 3 divided by that number, is about 17.0.

And so we could find the two other lengths purely from the angle and the one given length. This is very powerful.

Here is a practice problem. Pause the video and then we’ll talk about this.

Okay, the first thing to notice is what we have here is a 3-4-5 triangle. That’s very important to notice because the test will often expect you to recognize a 3-4-5 triangle.

So that missing side, XZ, has to equal 4. Now, notice that we want the tangent of angle X. From the perspective of X, 3 is the opposite side, and 4 is the adjacent side, very important.

It would be very different if we were finding the tangent from Y. But from the point of view of X, 3 is opposite and XZ equals 4, that’s the adjacent. And of course, tangent is opposite over adjacent. So the opposite is YZ, the adjacent is XZ, and that is 3 over 4. So it has to be answer choice E.

Here’s another practice problem.

Pause the video and then we’ll talk about this. Okay, so we’re given an angle, we’re given two lengths, SQ and QR, and we’re also told that the tangent of 35 degrees is approximately 0.700, and we want to know the area of the triangle. Well, we already know the base, we need the height, we need the length of PQ in order to figure out the area of the triangle.

Well, we know that the tangent of 35 degrees, that involves PQ, that’s PQ over SQ. That’s good, because we know SQ, that’s h over SQ, and we need that h. h = 5 times tangent of 35 degrees, and here we can use the approximation they give us, tangent of 35 degrees is 0.7. Well, 5 times 0.7 is 3.5, so h equals 3.5, very useful.

Now that we know h, we can find the area, of course, area of a triangle is one-half base times height. So that’s one-half 8, which is the full base from S to R, is a length at 8. One-half 8 times 3.5, one-half of 8 is 4, then for 4*3.5, we’ll use the doubling and halving trick.

Half of 4 is 2, double of 3.5 is 7, 2*7 is 14, that’s the area. So the area is 14. In general, for a general angle, mathematicians typically use the Greek letter theta. We can use this to make general statements true for any angle. So the sine of theta is opposite over hypotenuse, the cosine is adjacent over hypotenuse, and the tangent is opposite over adjacent.

This is the basic SOHCAHTOA pattern. Right now, these are true when we’re talking about angles inside triangles. So that means theta would have too be greater than 0 degrees and less than 90 degrees. We’d have to have a possible acute value inside a triangle. Right now, that’s where we’re gonna focus.

In this video, I’ll just discuss one more important relationship that you may have to know on the test. We know, of course, from the Pythagorean theorem that the adjacent squared plus the opposite squared has to equal the hypotenuse squared. That’s obviously true because of the Pythagorean theorem. We’ll divide each term by hypotenuse squared.

On the right side, we’ll get a hypotenuse squared divided by a hypotenuse squared, which is 1. We’ll get adjacent squared divided by hypotenuse squared. Well, adjacent divided by hypotenuse is cosine, and opposite divided by hypotenuse is sine. So we get cosine squared plus sine squared equals 1, and this is the Pythagorean Identity.

Notice incidentally, when we square trig function, we write the square after the name of the function and before the angle. So we write it as cosine squared theta, or sine squared theta. So this is an important trig formula, and we’ll return to this a few times, but this is a very good one to know. Here’s another practice problem, so pause the video and read this.

And here are the expressions from which to choose, take a good look at these, and see if you can solve the problem on your own. You can pause the video and when you’re ready, resume and we’ll solve it together.

Okay, let’s think about this. We’re gonna draw a right triangle with the rope as the hypotenuse, the horizontal base at the level of the tip of the prow, which is slightly above the water.

And the height up to the top of the pole, which is at P. Okay, well from the 35 degree angle at P, PR, that segment PR is the adjacent side, and that’s gonna help us with the vertical chain, so we’re gonna need that. So the cosine, we need the cosine to relate the adjacent to the hypotenuse. The cosine of 35 degrees is adjacent over hypotenuse, that’s PR over 25, so PR would equal 25 times the cosine of 35 degrees, very good.

So we have that length, the length of that entire segment PR. Well, PR, that’s not exactly the length that we’re looking for. The question asks very specifically the change in level between high tide and low tide. So at high tide, the prow of the boat was at the level of D, it was at the level of the surface of the dock.

And at low tide, the prow is at the level of R and B, that horizontal line at the bottom of the triangle. So what we need, the change in level, is DR. DR is the difference between high tide and low tide. Well, we know that PD + DR = the length PR, the two little segments together add up to the big segment.

So that means that 3 + DR = 25*cos(35 degrees), that’s the expression we got for PR. So if we want DR, we subtract 3 from both sides, and that’s the expression for the change in height. Then we go back to the answer choices and we choose this one, answer choice C.

In summary, it’s good to know SOHCAHTOA. Which means that the sine of theta is the opposite over hypotenuse, the cosine of theta is the adjacent over the hypotenuse.

And the tangent is the opposite over the adjacent. For any angle greater than 0 and less than 90 degrees, all the right angles with that acute angle are similar. And so all these ratios are the same for all of them. So if you pick any angle, say 23 degrees, a 23 degree right triangle, any 23 degree right triangles can be similar to any other.

And that’s why all these ratios are the same. And you can find the values for these three ratios on your calculator, although the test often supplies any numbers you need.

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]]>In the previous video, I talked about reflections in the x-y (coordinate) plane. Sometimes the test also asks about rotations on the coordinate plane.

In these cases, the center of rotation will almost always be the origin, and the angle will either be 90 degrees, one way or the other, or 180 degrees.

First, think about quadrants. Any point that rotates 90 degrees clockwise will go down a quadrant.

If it’s in quadrant IV, it would go to III, from III it will go to II, from II it will get to I, from I it will go to IV, that’s the clockwise direction, IV to III to II to I, back to IV. Any point that rotates 90 degrees counterclockwise will go up a quadrant. So, if it’s in I, it’s gonna get moved to II, from II to III, from III to IV, and from IV back to I.

So I, II, III, IV back to I, that’s the counterclockwise direction of rotation. Any point that rotates in 180 degrees will move to the opposite quadrant. So we swap back and forth between I and III, or swap back and forth between II and IV. Now we can be a little more precise.

Recall the discussion about perpendicular slopes a few lessons ago.

For a 90 degree rotation either way, the x-distances, and y-distances swap place. We see a triangle, the purple triangle has a horizontal leg, a long horizontal leg, and a short vertical leg. When we rotate it 90 degrees, anything horizontal becomes vertical, and anything vertical becomes horizontal. So the horizontal and vertical switch places in a 90 degree rotation.

The new x value has the same absolute value as the old y value, and vice versa. We have to think through plus and minus signs for the new coordinates based on the quadrants.

When we rotate by 180 degrees, the rule is even easier. If an original point, anywhere in the x-y plane is rotated 180 degrees. Then the x- and y-coordinates of the new point have the same absolute value and simply the opposite plus and minus signs.

So suppose we have these points here. Think about what would happen if we rotated 180 degrees around the origin. Well, the positive 5 and positive 3 would just become both negatives. The 4, -1 would become -4, positive 1, and the two negatives would become 2 positive. And so those would be the three images under 180 degree rotation. Sometimes, we have to know what happens to the coordinates of individual points when we rotate.

In other questions, we just have to use our visual reasoning abilities. If this given shape is rotated by say, 180 degrees, what will be its new position and new orientation? So there’s not really a formula for this, we have to use visual reasoning. If this is something you find challenging, practice. For example, draw, or maybe find in a magazine an asymmetrical figures, something that is clearly asymmetrical.

Look at it one way and then try to sketch or imagine what it would look like if it rotated 90 degrees clockwise or counterclockwise or rotated 180 degrees. And then actually rotate the figure, and check and see how close you were. So do the sketch first, where you imagine it, and then actually rotate the figure and see how close you were.

Here’s a practice problem.

Pause the video, and then we’ll talk about this. Okay, so that figure is in the first quadrant, we’re gonna rotate it 180 degrees. That means it’s gonna wind up in the third quadrant. So right away, we know that it’s not gonna be in the second quadrant. And, if it rotates 180 degrees, it gets flipped over. So these two long sides right now, they’re pointed down.

If it’s flipped 180 degrees over, they’re gonna be pointing up, and so the answer is D. The test will ask us to rotate things in the x-y plane, almost always by either 90 degrees, 90 degrees clockwise, 90 degrees counterclockwise, or 180 degrees, and almost always around the origin, so that’s make it easier. The test could give us coordinates of any one point and asks to find the coordinate of the new rotated point.

Or, the test may simply give us a shape or figure on one quadrant and ask us to visualize how it would appear when rotated.

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