### Sine and Cosine Functions

**Main definitions and formulas**:

- When you draw an angle θ
(measured in radians) in standard position (i.e. starting on the positive
*x*-axis), the coordinates of its terminal side on the unit circle are the*cosine*and*sine*of θ . - Master formula for right triangles: SOHCAHTOA!
sinθ = opposite side hypotenusecosθ = adjacent side hypotenusetanθ = opposite side adjacent side - A function
*f*is*odd*if*f*(−*x*) = −*f*(*x*), or equivalently, its graph has rotational symmetry around the origin. - A function
*f*is*odd*if*f*(−*x*) =*f*(*x*), or equivalently, its graph has mirror symmetry across the*y*-axis.

**Example 1**:

**Example 2**:

**Example 3**:

**Example 4**:

*f*(

*x*) = sin(x + (π /2)) and

*g*(

*x*) = cos(x − (π/2)). For each one, determine if the function is odd, even, or neither.

**Example 5**:

**Example 6**:

*f*(

*x*) = sin(x −(π /2)) and

*g*(

*x*) = cos(x + (π /2)). For each one, determine if the function is odd, even, or neither.

## Sine and Cosine Functions

## A right triangle has short sides of lengths 7 and 8. Find the sine, cosine, and tangent of all the angles in the triangle.

- Recall: SOHCAHTOA sinθ = [Opposite/Hypotenuse], cosθ = [Adjacent/Hypotenuse], tanθ = [Opposite/Adjacent]
- First find the length of the longest side by using the Pythagorean Theorem: a
^{2}+ b^{2}= c^{2} - 7
^{2}+ 8^{2}= x^{2}⇒ 49 + 64 = x^{2}⇒ 113 = x^{2}⇒ √{113} = x - Set up a triangle to help you determine which side will be the opposite and which side will be the adjacent according to your reference angle.

### sinθ = [7/(√{113} )] = [(7√{113} )/113], sinϕ = [8/(√{113} )] = [(8√{113} )/113], sinα = sin[(π)/2] = 1

cosθ = [8/(√{113} )] = [(8√{113} )/113], cosϕ = [7/(√{113} )] = [(7√{113} )/113], cosα = cos[(π)/2] = 0

tanθ = [7/8], tanϕ = [8/7], tanα = tan[(π)/2] = undefined

## A right triangle has one leg of length 9 and hypotenuse of length 12. Find the sine, cosine, and tangent of all angles in the triangle.

- Recall: SOHCAHTOA sinθ = [Opposite/Hypotenuse], cosθ = [Adjacent/Hypotenuse], tanθ = [Opposite/Adjacent]
- First find the length of the longest side by using the Pythagorean Theorem: a
^{2}+ b^{2}= c^{2} - x
^{2}+ 9^{2}= 12^{2}⇒ x^{2}+ 81 = 144 ⇒ x^{2}= 63 ⇒ x = √{63} - Set up a triangle to help you determine which side will be the opposite and which side will be the adjacent according to your reference angle.

### sinθ = [(√{63} )/12], sinϕ = [9/12], sinα = sin[(π)/2] = 1

cosθ = [9/12], cosϕ = [(√{63} )/12], cosα = cos[(π)/2] = 0

tanθ = [(√{63} )/9], tanϕ = [9/(√{63} )] = [(9√{63} )/63], tanα = tan[(π)/2] = undefined

## Use the following right triangle to find the sine, cosine, and tangent of all the angles in the triangle:

- Recall: SOHCAHTOA sinθ = [Opposite/Hypotenuse], cosθ = [Adjacent/Hypotenuse], tanθ = [Opposite/Adjacent]
- First find the length of the longest side by using the Pythagorean Theorem: a
^{2}+ b^{2}= c^{2} - x
^{2}+ 4^{2}= 11^{2}⇒ x^{2}+ 16 = 121 ⇒ x^{2}= 105 ⇒ x = √{105}

### sinθ = [4/11], sinϕ = [9/12], sinα = sin[(π)/2] = 1

cosθ = [(√{105} )/11], cosϕ = [(√{63} )/12], cosα = cos[(π)/2] = 0

tanθ = [4/(√{105} )] = [(4√{105} )/105], tanϕ = [(√{105} )/4], tanα = tan[(π)/2] = undefined

## Use the following right triangle to find the sine, cosine, and tangent of all the angles in the triangle:

- First find the length of the longest side by using the Pythagorean Theorem: a
^{2}+ b^{2}= c^{2} - 3
^{2}+ 7^{2}= x^{2}⇒ 9 + 49 = x^{2}⇒ 58 = x^{2}⇒ x = √{58}

### sinθ = [7/(√{58} )] = [(7√{58} )/58], sinϕ = [3/(√{58} )] = [(3√{58} )/58], sinα = sin[(π)/2] = 1

cosθ = [3/(√{58} )] = [(3√{58} )/58], cosϕ = [7/(√{58} )] = [(7√{58} )/58], cosα = cos[(π)/2] = 0

tanθ = [7/3], tanϕ = [3/7], tanα = tan[(π)/2] = undefined

## Graph the functions f(x) = sin (x − [(π)/4]) and g(x) = cos (x + [(π)/4]). Identify the zeros, maxima, and minima.

- First start off by graphing f(x) = sin(x)
- Notice that for this problem f(x) = sin(x) shifts [(π)/4] units to the right in f(x) = sin (x − [(π)/4])
- The zeroes are the points in which the function crosses the x - axis, the maxima are the highest points, the minima are the lowest points
- Now, graph g(x) = cos(x)
- Notice that g(x) = cos(x) shifts [(π)/4] units to the left in f(x) = cos(x + [(π)/4])

### f(x) = sin (x − [(π)/4]) zeroes are ([(π)/4],0), ([(5π)/4],0), ([(9π)/4],0), maxima is ([(3π)/4],1) minima is ([(7π)/4], − 1)

g(x) = cos(x + [(π)/4]) zeroes are ([(π)/4],0), ([(5π)/4],0), maxima is ( − [(π)/4],1), ([(7π)/4],1) minima is ([(3π)/4], − 1)

## Graph the functions f(x) = sin (x + π) and g(x) = cos (x − π). Identify the zeros, maxima, and minima.

- First start off by graphing f(x) = sin(x)
- Notice that for this problem f(x) = sin(x) shifts π units to the left in f(x) = sin (x + p)
- The zeroes are the points in which the function crosses the x - axis, the maxima are the highest points, the minima are the lowest points
- Now, graph g(x) = cos(x)
- Notice that g(x) = cos(x) shifts π units to the right in f(x) = cos(x − π)

### f(x) = sin (x + π) zeroes are ( − π,0), (0,0), (π,0), maxima is ( − [(π)/2],1) minima is ([(π)/2], − 1)

It is an odd function

g(x) = cos(x − π) zeroes are ( − [(π)/2],0), ([(π)/2],0), ([(3π)/2],0), maxima is (π,1), ( − π,1) minima is (0, − 1), (2π, − 1)

## Use the following graph to identify the zeroes, maxima, and minima

- The zeroes are the points in which the function crosses the x - axis, the maxima are the highest points, the minima are the lowest points

### Zeroes ( − [(3π)/2],0), ( − [(π)/2],0), ([(π)/2],0), ([(3π)/2],0)

Maxima ( − 2π, 3), (0, 3), (2π, 3)

Minima ( − π, − 3), (π, − 3)

## Graph the functions f(x) = sin (x − [(π)/3]) and g(x) = cos (x + [(π)/3]). Identify the zeros, maxima, and minima.

- First start off by graphing f(x) = sin(x)
- Notice that for this problem f(x) = sin(x) shifts [(π)/3] units to the right in f(x) = sin (x − [(π)/3])
- Now, graph g(x) = cos(x)
- Notice that g(x) = cos(x) shifts [(π)/3] units to the left in f(x) = cos(x + [(π)/3])

### Zeroes: ([(π)/6],0),([(7π)/6],0), Maxima: ( − [(π)/3],1),([(5π)/3],1), Minima: ([(2π)/3], − 1)

## Graph the functions f(x) = sin (x − π) and g(x) = cos (x + π). Identify the zeros, maxima, and minima.

- First start off by graphing f(x) = sin(x)
- Notice that for this problem f(x) = sin(x) shifts p units to the right in f(x) = sin (x − π)
- Now, graph g(x) = cos(x)
- Notice that g(x) = cos(x) shifts π units to the left in f(x) = cos(x + π)

### Zeroes: ( − [(π)/2],0), ([(π)/2],0), ([(3π)/2],0), Maxima: ( − π, 1), (π, 1), Minima: (0, − 1), (2π, − 1)

## Use the following graph to identify the zeroes, maxima, and minima

### Zeroes ( − 2π,0), ( − [(3π)/2],0), ( − π,0), ( − [(π)/2],0), (0,0), ([(π)/2],0), ([(3π)/2],0), (2π,0)

Maxima ( − [(7π)/4], 2), ( − [(3π)/4], 2), ([(3π)/4], 2), ([(7π)/4],2)

Minima ( − [(5π)/4], − 2), ( − [(π)/4],2), ([(π)/4],2), ([(5π)/4], − 2)

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

#### Answer

## Sine and Cosine Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Sine and Cosine
- Unit Circle
- Coordinates on Unit Circle
- Right Triangles
- Adjacent, Opposite, Hypotenuse
- Master Right Triangle Formula: SOHCAHTOA
- Odd Functions, Even Functions
- Example 1: Sine and Cosine
- Example 2: Graphing Sine and Cosine Functions
- Example 3: Right Triangle
- Example 4: Odd, Even, or Neither
- Extra Example 1: Right Triangle
- Extra Example 2: Graphing Sine and Cosine Functions

## Mathematics: Trigonometry

## Transcription: Sine and Cosine Functions

*Ok we are going to work on some extra examples with you, I hope you have time to practice a little bit on your own.*0000

*Here we are given a right triangle that has 1 leg of length 5, and a hypotenuse of length 13.*0006

*What we want to do is find the sin, cos, and tan of all the angles in the triangle.*0024

*I hope you know that the master formula here is going to be SOHCAHTOA.*0029

*That is the one we are going to use but in order to use SOHCAHTOA you have to know the lengths of all the sides of the triangle.*0035

*Let me set a variable for this length, we need to solve x ^{2} + 5^{2} = 13^{2}.*0044

*That is x ^{2} + 25 =169, x^{2} = 169-25, that is 144, so x=12.*0057

*That tells us our third side length, it looks like my triangle is not really drawn on the scale because 12 is a lot bigger than 5.*0073

*But we can still do the calculations here even if we did not draw the triangle perfectly to scale, you can still figure out the calculations.*0079

*Let me label the angles (theta) and (phi) and we will look at (theta) first.*0088

*Sin(theta) using SOHCAHTO is opposite/hypotenuse, so that is 12 is the opposite angle of (theta) and the hypotenuse is 13.*0094

*Cos(theta)=adjacent of the side is 5/13, the tan(theta) =opposite of adjacent that is 12/5.*0112

*Sin(phi) let us do that one first, is equal to the opposite/hypotenuse.*0134

*The opposite side to (phi) is 5 and the hypotenuse is still 13.*0146

*The cos(phi) is equal to the adjacent side, the side adjacent to (phi) is 12/13.*0155

*The tan(phi) is equal to the opposite of adjacent, the opposite side of (phi) is 5, and the adjacent side is 12.*0165

*Finally, let me label the right angle (alpha), the sin(alpha), I’m not going to use SOHCAHTOA on this but the sin (alpha) is just the sin (pi/2) is 1.*0180

*The cos(alpha) is the cos (pi/2) which is 0 and then the tan(alpha).*0194

*Again we have not really learned about tan yet, but it turns out that the tan of a 90 degree angle or the tan of pi/2 is undefined.*0202

*I learned why that is true when we get to studying the tan function later on.*0211

*To recap here, everything came out of the master formula SOHCAHTOA, memorize that word SOHCAHTOA or memorize the phrase Some Old Horse Caught Another Horse Taking Oats Away.*0225

*That helps you to remember how to figure out all these sin, cos, and tan, once you know the lengths of the sides of the right angle.*0237

*Our last example here is two functions that we need to graph, two modified sin waves and remember that the way to do that is to start out with sin and cos graphs that you know.*0000

*Let me start graphing sin(x), there is pi, 2pi, 1, -1, pi/2, 3pi/2.*0012

*Remember that sin(x) starts at 0, goes up to 1, comes back to 0 down to -1.*0035

*What I graphed there in black is sin(x), now sin(x)-pi/2 I will do that in blue.*0052

*Sin(x)-pi/2 that takes the graph and it shifts it over to the right by pi/2 units.*0060

*I’m going to take this graph in black and shift it over to the right by pi/2 units.*0067

*Let me extend that a little bit so I will know how to shift it, now I’m going to shift it over to the right by pi/2 units.*0077

*Now it starts at 0 to pi/2, comes back to 0 to 3pi/2 down to -1, back to 0.*0082

*What I have drawn there in blue is sin(x) – pi/2, it is just a basic sin graph but moved over by pi/2 units.*0104

*We have to figure out if its odd, even, or neither*0113

*Remember that odd has rotational symmetry, even has mirror symmetry across the y axis.*0116

*Clearly this blue graph that I have drawn here has mirror symmetry across the y axis but not rotational symmetry.*0124

*This is an even function because it has mirror symmetry across the y axis.*0135

*Let us move on to g(x), that is cos, again I will start with the cos graph that I have memorized and hopefully you have two.*0151

*There is pi, 2pi, 3pi/2, pi/2, 0, -pi, -pi/2, 1 and -1, that is a little low, let me draw that a little higher, -1.*0163

*Remember the cos graph starts at 1 goes down to 0, down to -1 at pi and comes back to 0 and back 1 at 2pi.*0189

*What I graphed there in black was cos(x), the basic cos curve, cos(x + pi/2), I will do this in red.*0210

*You want to think about that as cos(x - -pi/2) and I do that to create that negative sign because when you have a shift you always want to have a negative sign there.*0222

*It helps you figure out which way it is shifting, that means shifting –pi/2 units to the right, which means pi/2 units to the left.*0235

*I take this graph and I shift it p/2i units to the left, so that means it will start here go down to 0, down to -1, back to 0 up to 1 and back to 0.*0244

*That was a little bit of too high there, that curve in red is cos(x + pi/2) and if we look at that curve it has rotational symmetry around the origin.*0264

*If you rotate it back graph, 180 degrees around the origin it would look the same.*0286

*It definitely does not have a mirror symmetry around the y axis, so this is an odd function.*0292

*Odd functions have rotational symmetry around the origin, that is an odd function.*0303

*Remember the way to check whether they are odd or even is to check which kind of symmetry they have or maybe they do not have either kind of symmetry.*0310

*That is the end of our lessons on sin and cos, this is www.educator.com.*0320

*Hi, this is educator.com and we are here to talk about sine and cosine functions.*0000

*I'll start out by giving you the definitions and kind of the master formulas.*0007

*Then we'll go through and work on a bunch of examples.*0010

*The definition of sine and cosine of an angle is, you start out with the axis and the unit circle it's important to know that.*0015

*This is a unit circle meaning the radius is 1.*0029

*What you do is you draw that angle in standard position, meaning it has one side on the x-axis.*0033

*There is data right there.*0042

*Then you look at the coordinates of the point, the x and y coordinates on the unit circle.*0044

*The x-coordinate is, I'll go over that in red, that's the x coordinate right there.*0051

*The y-coordinate, that's the one in blue.*0060

*Those coordinates are, by definition, the...*0063

*I want to do the cosine, in red, cos(θ) and then, in blue, sin(θ).*0071

*That's the definition of what an angle is in terms of coordinates on the unit circle.*0083

*If you know what the angle is, you try to figure out what its x-coordinate and its y-coordinate are.*0090

*Let's call these cos(θ) and the sin(θ).*0097

*Now, we'll see some more examples of that later so that you'll know how to actually compute the cosine and sine, but the definition just refers to those coordinates.*0101

*The most common use of sine and cosine probably is in terms of right triangles.*0113

*Let me draw a right triangle.*0120

*Right triangle just means a triangle where one of the angles is a right angle, a 90-degree angle, or in terms of radians, π over 2.*0123

*What you do is, you let θ be one of the angles that is not the 90-degree angle, so one of the other angles.*0130

*Then you label each one of the sides in terms of its relationship to θ.*0139

*The one next to θ is called the adjacent side.*0145

*The one opposite θ is called the opposite side.*0151

*The long side is called the hypotenuse.*0161

*Then we have the master formula for right triangles, which is, the sine of θ is equal to the length of the opposite side divided by the hypotenuse.*0168

*The cosine of θ is equal to the adjacent side divided by the hypotenuse.*0181

*The tangent of θ, which is something we haven't officially defined yet, so we'll learn about tangent in a later lesson.*0185

*I just want to give you the right triangle formula now, because we're going to try to remember them all together.*0193

*The tangent of θ when we get to it will be the opposite side divided by the adjacent side.*0200

*We don't want to worry too much about tangent now because we haven't learned about it in detail yet.*0205

*I'll get to those later.*0209

*If you put all these formulas together, it's kind of hard to remember their relationships.*0211

*So people have come up with this acronym.*0217

*Sine is equal to opposite over hypotenuse.*0221

*Cosine is equal to adjacent over hypotenuse.*0225

*Tangent is equal to opposite over adjacent.*0227

*If you kind of read that quickly, people call it SOH CAH TOA.*0233

*If you talk to any trigonometry teacher in the world, or any trigonometry student in the world, they should have heard the word SOH CAH TOA,*0235

*because that's kind of the master formula that helps you remember all these relationships.*0242

*They're kind of hard to remember on their own, but if you remember SOH CAH TOA, you won't go wrong.*0247

*If you have trouble remembering that, there is a little mnemonic device that people also use.*0253

*Some old horse caught another horse taking oats away.*0262

*If you remember that sentence, if that's easier for you to remember than SOH CAH TOA, then you can remember all these formulas.*0268

*There is another definition that we need to learn which is, that a function is odd if f(-x) is equal to -f(x).*0276

*Let's figure out what that means.*0289

*We're going to talk about odd and even function.*0291

*Let me give you an example here.*0294

*F(x) is equal to x ^{3}.*0299

*Well, let's try f(-x) here.*0305

*I'm going to check this definition of odd function here.*0307

*So, f(-x), that means you put -x into the function, so that's (-x) ^{3} which is (1)3 times x^{3},*0310

*(-1) ^{3} is -1, so that's just -x^{3}, and that's negative of the original f(x).*0326

*For x ^{3}, f(-x) is equal to -f(x), which means it's an odd function.*0334

*There is a way to check this graphically.*0342

*If you graph f(x)=x ^{3}, it looks something like this.*0348

*That means that if you look at a particular value of x, and you look at f(x) there, and then you look at -x, f(-x)=-f(x).*0355

*That's what it looks like graphically.*0379

*What it means is that the graph has what I call rotational symmetry around the origin.*0383

*If you put a big dot on the origin and if you spun this graph around 180 degrees, it would look the same.*0393

*That's because f(x) and f(-x) being opposites of each other.*0400

*If you spin the graph around 180 degrees and it looks the same, if it looks symmetric with itself, that's called an odd function.*0408

*The way you remember that odd functions have that property is, just remember x ^{3}, x to the third, because 3 is an odd number and x^{3} is an odd function.*0417

*Something, kind of, has that property that x ^{3} has, then it's an odd function.*0428

*They're companion definition to that is that f is even if f(-x) equals f(x).*0435

*The difference there was that negative sign on the odd definition.*0443

*No negative sign here on the even definition.*0447

*Let me give you an example of an even one.*0450

*Let's define f(x) to be x ^{2}.*0456

*Well, f(-x) is equal to, you plugin -x into the function, so (-x) ^{2} well that's just the same as x^{2}, which is the same as the original f(x).*0462

*So, f(-x) is equal to f(x), that checks the definition, so it's even.*0478

*Of course you'll notice that x ^{2}, the 2 there is an even number, that's no coincidence.*0486

*That's why we call even functions even is because they sort of behave like x ^{2}.*0494

*If you graph those, let me graph x ^{2} for you.*0500

*That's a familiar parabola that you learned how to graph in the algebra section.*0505

*If you take a value of x, and look at f(x),*0511

*then you take f(-x), f(-x) is not -f(x), it's f(x) itself.*0517

*It's the same value as f(x) itself.*0532

*You get this f(-x) is equal to f(x).*0537

*What that means is that you have a, kind of, symmetry across the y-axis with even functions.*0542

*If a function is symmetric across the y-axis, if it looks like a mirror image of itself across the y-axis, then that's an even function.*0555

*That's why I say it has mirror symmetry across the y-axis.*0567

*That's what an even function looks like.*0572

*There is a common misconception among students.*0575

*People think, well with numbers, every number's either odd or even.*0578

*People think, well, every function is odd or even and that's not necessarily true.*0583

*Just for example, here's a line but that is not either*0590

*that does not have a rotational symmetry around the origin nor does it have mirror symmetry around the y-axis.*0600

*That function, this line, is not an add or an even function.*0607

*It's a little misleading people think every function has to be an odd or an even function.*0612

*That's not true.*0616

*It's just true that some functions are odd, some functions are even, some functions are neither one.*0618

*We'll practice some examples of that.*0624

*First, we're going to look at some common angles and we're going to figure out what the cosines and sines are.*0626

*Let me draw a big unit circle here.*0635

*That's a circle of radius 1.*0646

*Let's remember where these angles are, 0, of course is on the positive x-axis.*0648

*Here's the x-axis, here's the y-axis, there is zero, *0655

*π over 2, remember that's the same as a 90-degree angle, that's a right angle, so that's up there π over 2;*0660

*π is over here, that's a 180 degrees;*0666

*3π over 2, is down here, and 2π is right here.*0671

*We want to find the cosine and sine of each one of those angles.*0676

*Now remember, cosine and sine, by definition, are the x and y coordinates of those angles.*0680

*What are these x and y coordinates?*0689

*The 0 angle, it's x-coordinate is 1, and it's y-coordinate is 0.*0691

*That tells us that cos(0) is 1, and that sin(0) is 0.*0698

*Pi over 2 is up here, and so it's cosine is the x-coordinate,*0712

*well, the x-coordinate of that point is 0.*0726

*The y-coordinate is 1, and so that's the sin of π over 2.*0730

*Pi is over here at (-1,0), so that's the cosine and sine of π.*0739

*Cos(π) is -1, sin(π) is 0.*0748

*Finally, 3π over 2 is down here at (0,-1), so that's the cosine and sine of 3π over 2.*0760

*Cos(0), sin(π/2), is -1.*0772

*And one more, 2π is back in the same place as 0, so it has the same cosine and sine.*0783

*Cos(2π) is 1, sin(2π) is 0.*0790

*That's how you figure out the cosines and sines of angles.*0805

*As you graph them on the unit circle, and then you look at the x and y coordinates.*0808

*The x-coordinate is always the cosine, and the y-coordinate is always the sine.*0813

*By the way, these are very common values, 0, π/2, 3π/2, and 2π.*0819

*You should really know the sines and cosines of these angles by heart.*0826

*They come up so often in trigonometry context that it's worth memorizing these things, and being able to sort of regurgitate them very very quickly.*0830

*If you ever forget them though, if you ever can't quite remember what the cosine of π/2, or the sine of 3π/2 is,*0841

*Then, what you do is draw yourself a little unit circle, and you figure what the x and y coordinates are and you can always work them out.*0849

*It's worth memorizing them to know them quickly, but if you ever get confused, you are not quite sure, just draw yourself a unit circle and you'll figure them out quickly.*0857

*We're going to use these values, so I hope you will remember these values for the next example.*0869

*In the next example, we're being asked to draw the graphs of the cosine and the sine functions, so let's remember what those values are.*0879

*We're to label all the zeros, and the maxima and the minima of these functions.*0888

*Let me set up some axis here.*0892

*I'm going to label my x-axis in terms of multiples of π.*0898

*The reason I'm going to do that is because we're talking about cosines and sines of multiples of π.*0905

*We're really talking about radians here.*0910

*That's π, that's 2π, that's π/2, and that's 3π/2.*0915

*That's 0, of course.*0929

*The y-axis, I'm going to label as 1 and -1.*0931

*I've set up my scale here, remember that π is about 3.14 so it's a little bit bigger than 3.*0940

*I've set up my scale here so that the π is about a little bit beyond 3 units on the graph.*0945

*I'll extend it a little bit on the left here as well.*0952

*We've got -π, I'll draw that around -3 and -π/2.*0955

*I want to graph the sine and cosine function according to those values that we figured out.*0963

*Remember that the sine and cosine function are correspond to the coordinates of angles on the unit circle.*0968

*So sine and cosine,remember, are the x and y coordinates of angles on the unit circle.*0978

*Now, those coordinates will never get bigger than 1 or smaller than -1.*0986

*That's why on my y-axis, I only went up to -1 and 1, because the coordinates will never get bigger than -1 and 1.*0994

*Let me start out with the cosine function.*1004

*I'll do that one in blue, y=cos(x).*1007

*We'll remember the values that we learned in the previous question.*1014

*Cos(x), cos(0) is 1, cos(π/2) was 0, cos(π) is -1, cos(3π/2) is 0 and cos(2π) is 1.*1018

*What you get is this smooth curve.*1044

*After 2π, remember, after you circle 2π radians, then everything starts repeating.*1057

*What happens after 2π is that it repeats itself.*1065

*It repeats itself in the negative direction as well.*1071

*Now we know what the graph of cos(x) looks like.*1080

*I'll do the sine graph in red.*1085

*Remember that sin(0) is 0, sin(π/2) is 1, sin(π) was 0 again, sin(3π/2) is -1, sin(2π) is 0.*1092

*I'm doing this from memory and hopefully you remember these values as well.*1110

*But if you can't remember these values, you know you can always look back at the unit circle and figure them out again just from their coordinates.*1114

*The sine graph, I'm going to connect this up into a smooth curve.*1122

*It repeats itself after this.*1139

*It repeats in the negative direction as well.*1146

*That's what y=sin(x) looks like.*1151

*It actually has the same shape as cos(x) but it's shifted over on the graph.*1153

*Now, we're asked to label all zeros, maxima and minima.*1160

*Let me go through and label the zeros first.*1163

*This is on the cos(x), this is (π/2,0), that's the 0 right there.*1167

*There is one right there, (3π/2,0).*1177

*This one, even though I haven't labeled it on the x-axis, is actually (5π/2,0), because it's π/2 beyond 2π,(-π/2,0).*1182

*Those are the zeros of the cosine graph.*1197

*The maxima, the high points, remember, cosine and sine never get bigger` than 1 or less than -1.*1202

*Any time it actually hits 1, it's a maximum.*1207

*They're the two maxima, at 0 and 2π.*1213

*The minimum value is -1, so there is (π,-1) and the next one would be at (3π,-1), there is one at (-π,-1).*1218

*Now, let me do the zeros of this sine graph.*1240

*There is one (0,0), (-π,0), (π,0), and (2π,0).*1245

*The maximum value would be 1 and that occurs at π/2, and again at 5π/2.*1258

*The minimum value would be at 3π/2, where the sine is -1,*1273

*Remember sine and cosine never go outside that range, -1 to 1,*1278

*and at -π/2.*1282

*All these values you should pretty much have memorized there the sort of simplest values, the easiest ones to figure out of sine and cosine.*1289

*Let's try an example where we're using this trigonometric functions in a triangle.*1301

*What we're told is that a right triangle has short sides of length 3 and 4.*1307

*We're asked to find the sine, cosine, and tangent of all angles in the triangle.*1317

*Remember, I haven't really explained what tangent is yet, but we did learn that formula SOH CAH TOA.*1323

*That's what we're going to be using here.*1327

*The first thing we need to figure out here is what the hypotenuse of this triangle is.*1330

*We have the Pythagorean theorem that says, h ^{2} = 3^{2} + 4^{2}, which is 9 plus 16, which is 25.*1334

*That tells us that the hypotenuse must be 5.*1349

*Now we're going to find the sine, cosine and tangent of each one of these angles.*1354

*Let's figure out this angle first, so I'll call it θ, sin(θ), remember SOH CAH TOA,*1364

*let me write that down for reference here, SOH CAH TOA,*1372

*sin(θ) is equal to opposite over adjacent*1381

*That's 4, that's the opposite side from θ over...*1386

*Sorry, I said sin(θ) is opposite over adjacent, of course, sin(θ) is opposite over hypotenuse, and the hypotenuse is 5.*1394

*So, sin(θ) there is 4/5.*1401

*Cos(θ) is equal to adjacent over hypotenuse.*1404

*Well, the side adjacent to θ is 3, hypotenuse is still 5.*1409

*Tan(θ) is equal to opposite over adjacent.*1416

*Again, we haven't really learned what a tangent means yet, but we can still use SOH CAH TOA.*1423

*The opposite over adjacent is 4/3.*1427

*Let me call the other angle here φ.*1435

*Sin(φ) is equal to the opposite over the hypotenuse, so 3/5.*1442

*Cos(φ) is equal to adjacent over hypotenuse, the adjacent angle beside φ is 4.*1451

*Tan(θ) is equal to the opposite over adjacent, so that's 3/4.*1463

*Finally, we have the right angle here, I'll call that α.*1472

*We can't really use SOH CAH TOA on that, but I know that sin(α),*1476

*α is a 90-degree angle, or in terms of radians, it's π/2.*1480

*The sin(π/2), we learned before, is 1.*1489

*Cos(α) is cos(π/2), and we learned that the cos(π/2) before was 0.*1494

*Finally, tan(α) is tan(π/2), and we haven't really learned about tangent yet.*1505

*In particular, we haven't learned what to do with the tan(π/2).*1514

*But we'll get to that in a later lecture, and we'll learn that that's actually not defined.*1518

*So we can't give a value to the tangent of π/2.*1523

*All of these angles were things we worked out just using this one master formula, SOH CAH TOA.*1532

*That tells you the sine, cosine and tangent of the small angles in the triangle.*1541

*The SOH CAH TOA does not really apply to the right angle.*1548

*But we already know the sine and cosine of a right angle, of a 90-degree angle, because we figured them out before.*1550

*We'll try some more examples here.*1558

*I want to try graphing the function sin(x + π/2) and cos(x - π/2).*1562

*Then, I want to determine whether these functions are odd or even, or neither one.*1570

*Well, something that's really good to remember here from your algebra class, or from the algebra lectures here on educator.com,*1576

*is that you have a function, and you try to graph f(x) minus a constant, *1584

*what that does is it moves the graph of the function over by the amount of the constant.*1593

*That's very useful in the trigonometric setting.*1599

*Let me start out by graphing f(x)=sin(x).*1602

*And we did that in the previous example, and I remember what the sine graph looks like.*1608

*It starts at 0, it peaks at π/2, it goes back to 0 at π, it bottoms out at 3π/2 at -1, and then it goes back to 0 at 2π.*1615

*What I graphed there was just sin(x), I have not introduced this change yet.*1635

*What I'm going to do, in blue now, is the sin(x + π/2).*1643

*What that's going to do is going to move the graph over π/2 units.*1651

*But remember there was a negative sign in there that I don't have here.*1656

*This is really like, sin[x - (-π/2)].*1661

*It moves the graph over -π/2 units, which means it moves it to the left π/2 units.*1672

*I'm going to take this graph and I'm going to move it over to the left π/2 units.*1680

*Now it's going to start at -π/2, come back down at π/2, bottom out at π, and come back to 0 at 3π/2.*1691

*So there is -π/2, 0, π/2, and it comes back at 3π/2.*1709

*That's what our graph of sin(x + π/2) looks like.*1719

*Then the question is, is that odd or even, or neither?*1725

*Well, remember there is a graphic way to look at the graph of a function in determining whether it's odd or even, or neither.*1730

*An odd function, remember, has rotational symmetry, and even function has mirror symmetry across the y-axis.*1738

*Well, if you look at this at this graph, it certainly does not have rotational symmetry.*1750

*If you tried to rotate it around the origin, it would end up down here, and that would be a different graph.*1757

*However, it does have mirror symmetry around the y-axis.*1762

*So because it's mirror symmetric around the y-axis, sin(x + π/2).*1772

*F(x) is an even function because it has mirror symmetry, mirror symmetry across the y-axis.*1781

*OK, let us move on to the next one, cos(x - π/2).*1812

*Again, I'm going to start with the basic cosine function that we learned how to graph in a previous example.*1816

*So that my mark's here, there is π, there is 2π, π/2, 3π/2, 0, -π/2.*1829

*Now, cosine had zero at 1, then it comes down to 0 at π/2, bottoms out at -1 at π, comes back to 0 at 3π/2, and by 2π, it's back up at 1.*1841

*What I have just graphed there in black is cos(x).*1862

*I have not tried to introduce the shift yet.*1865

*But what we want to do is to graph cos(x - π/2).*1871

*That's like saying, you see up here is π/2, that's going to shift the graph π/2 units to the right,*1878

*because it's -π/2, it shifts it to the right.*1888

*We take this graph and we move it over to the right π/2 units.*1890

*I drew that a little too high, let me flatten that out a little bit.*1904

*What we have there is the graph of cos(x - π/2), and of course that keeps going in the other direction there.*1910

*We see, actually if you look carefully, cos(x - π/2) actually turns out be the same graph as sin(x).*1921

*That's a familiar function if you remember those graphs.*1930

*Again, we're being asked whether the function is odd or even, or neither.*1935

*For odd, we're checking rotational symmetry around the origin.*1939

*Look at that.*1944

*If you rotate the graph 180 degrees around the origin, what you'll end up with is exactly the same picture.*1946

*g(x) is an odd function because it has rotational symmetry around the origin.*1953

*Is it an even function?*1982

*Does it have mirror symmetry around the y-axis?*1984

*No, it does not because it has this kind of bump on the right hand side, and it does not have the same bump on the left hand side, so it's not an odd function.*1987

*Sorry, it's not an even function.*1996

*It's just an odd function.*1998

*What we did there was we started with the sine and cosine graphs that we remembered.*2000

*it's worth memorizing the basic sine and cosine graphs.*2004

*Then we examined the shift that each one introduced.*2008

*Each one got shifted π/2 units to the right or left.*2013

*Then we drew the new graphs.*2016

*Then we looked back at them and we checked what kind of symmetry do they have.*2019

*Do they have rotational symmetry or mirror symmetry?*2022

*And that tells us whether they are odd or even.*2027

1 answer

Last reply by: Dr. William Murray

Thu Feb 19, 2015 3:36 PM

Post by patrick guerin on February 17 at 09:14:02 AM

On practice question 2 it said the sin of theta was sqroot of 63 over 12 when i thought it was 9 over 12. Could you check it out please. Thanks.

1 answer

Last reply by: Dr. William Murray

Fri Dec 19, 2014 9:31 AM

Post by katrina williams on December 17, 2014

In the second to last practice problem what amount does n represent? I was able to draw the original graph but got lost by how far to move it over.

1 answer

Last reply by: Dr. William Murray

Mon Aug 4, 2014 7:48 PM

Post by patrick guerin on July 16, 2014

You placed a theta in the triangle that you created when you were defining sine cosine. Could you give me a small explanation on what theta is?

1 answer

Last reply by: Dr. William Murray

Tue Jun 17, 2014 12:08 PM

Post by Austin Cunningham on June 9, 2014

How come at around 8:00, he says that x^2 is the same thing as f(x)?

3 answers

Last reply by: Dr. William Murray

Thu Jun 5, 2014 11:43 AM

Post by Govind Balaji Srinivasa Raghavan on May 29, 2014

I dont know why. But all videos pause after sometime. Then it restarts again instead of continue. Also I cant skip to other part of video. Suppose tonight I watch half of video and go to sleep. Tomorrow morning, I have to watch from first, if i click on the play-head from where I should see, It automatically restarts.

1 answer

Last reply by: Dr. William Murray

Tue Mar 4, 2014 5:00 PM

Post by Damien O Byrne on February 28, 2014

Does sin cos and tan formulas only apply to right angled triangles

1 answer

Last reply by: Dr. William Murray

Wed Jan 22, 2014 3:04 PM

Post by Carroll Fields on January 16, 2014

In extra example I, at 3:30, why is the 'sine, cosine, and tangent of the right angle (alpha) , 1,0 , and undefined?And why for sine is it sin pi/2 ?

1 answer

Last reply by: Dr. William Murray

Mon Oct 21, 2013 7:15 PM

Post by yannick Haberkorn on October 12, 2013

i have to congratulate you Dr.william murray because i actually really feel i am in a learning environment and it feels great . Much thanks

1 answer

Last reply by: Dr. William Murray

Tue Apr 16, 2013 8:35 PM

Post by Dr. William Murray on January 27, 2013

Hi Emily,

Good question. As Jacob says in his post above, it's because we know that the graph of f(x-c) is like the graph of f(x), but shifted c units to the right. But you have to have the negative sign in there for this to work, so when we have f(x+(something)), we write it as f(x-(-something)). Then it's clear that the shift is (-something) units to the right, that is, (something) units to the left.

It's also worth reading Jacob's answer above -- same basic idea, but sometimes having a different person's phrasing helps.

Thanks for taking trigonometry!

Will Murray

1 answer

Last reply by: Dr. William Murray

Tue Apr 16, 2013 8:34 PM

Post by Emily Engle on January 27, 2013

At 28:10 Why do you change Sin (x+ Pi/2) to Sin (x-(- Pi/2)) ?

1 answer

Last reply by: Dr. William Murray

Fri Aug 31, 2012 5:45 PM

Post by Andraa Cram on June 25, 2012

@ 21:18, why, when going in the negative direction while graphing for sine (in red), does he draw the graph as (-Pi/2,-1) instead of (-Pi/2,1)? I'm very confused by this.

1 answer

Last reply by: Dr. William Murray

Sun Jan 27, 2013 4:18 PM

Post by Lourdes Johnson on June 3, 2012

Why does the lecture restart around a quarter in?

1 answer

Last reply by: Dr. William Murray

Sun Jan 27, 2013 4:16 PM

Post by Callistus Elue on May 23, 2012

the lecture reverts to the beginning almost as soon as it starts

1 answer

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Sun Jan 27, 2013 4:12 PM

Post by Jacob Burley on April 25, 2011

At 26:33 Professor Murray gave the algebraic equation for a graph that has a constant which was f(x-c). Our equation sin(x+pi/2) has a positive where the negative is in the original equation. In order to get the correct sign there we must change the + sign into two - signs because two negatives make a positive.

I know I'm not the greatest at explaining things but hopefully this helps a little bit.

1 answer

Last reply by: Dr. William Murray

Sun Jan 27, 2013 4:10 PM

Post by Shannon Bryington on February 28, 2011

At 27:40 on the video: Why was x + pi over 2 changed to x - neg pi over 2?

1 answer

Last reply by: Dr. William Murray

Sun Jan 27, 2013 3:59 PM

Post by Santhini Dheenathayalan on January 19, 2011

Great!