For more information, please see full course syllabus of Trigonometry

For more information, please see full course syllabus of Trigonometry

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Related Articles:

- Navigating the High School Math Maze
- Unit Circle: How to Memorize & Use
- Unit Circle Chart & Blank Practice Chart (PDF)

### Sine and Cosine Values of Special Angles

**Main definitions and formulas**:

- A 45-45-90 triangle has side lengths in proportion to 1-1-√ 2.
- A 30-60-90 triangle has side lengths in proportion to 1-√ 3-2.
Degrees Radians Cosine Sine 0 0 1 0 30 π 6√ 3 21 245 π 4√ 2 2√ 2 260 π 31 2√ 3 290 π 20 1 - Use these values to find sines and cosines in other quadrants. The mnemonic ASTC (All Students Take Calculus) helps you remember which ones are positive in which quadrant. (All, Sine, Tangent, Cosine)

**Example 1**:

^{°}to radians, identify its quadrant, and find its cosine and sine.

**Example 2**:

^{R}to degrees, identify its quadrant, and find its cosine and sine.

**Example 3**:

**Example 4**:

^{°}to radians, identify its quadrant, and find its cosine and sine.

**Example 5**:

### Sine and Cosine Values of Special Angles

^{°}to radians, identify its quadrant, and find its cosine and sine.

- First convert 315
^{°}to radians by using the following formula: degree measure ×[(π)/(180^{°})] = radian measure - 315
^{°}×[(π)/(180^{°})] = [(315^{°}π)/(180^{°})] = [(7π)/4] - Locate [(7π)/4] on a unit circle (draw a picture). This will help you to determine the quadrant it is in and it will also help you to find the sine and cosine values.
- Looking at our picture, we see that [(7π)/4] is in quadrant IV. This will help us to solve for the sine and cosine values as well.
- Note that 315
^{°}is 45^{°}from the x - axis so we can use the 45^{°}− 45^{°}− 90^{°}triangle to find the sine and cosine. - Recall that cosine is the x coordinate and sine is the y coordinate of [(7π)/4] on the unit circle
- cos [(7π)/4] = [(√2 )/2] and sin [(7π)/4] = − [(√2 )/2]
- Using the Mnemonic ASTC (All Students Take Calculus) helps to determine which values are positive in which quadrant
- Since [(7π)/4] is in quadrant IV we know that cosine will be positive and sine will be negative

- First convert [(4π)/3] to degrees by using the following formula: radian measure ×[(180
^{°})/(π)] = degree measure - [(4π)/3] ×[(180
^{°})/(π)] = 4 ×60^{°}= 240^{°} - Locate 240
^{°}on a unit circle (draw a picture). This will help you to determine the quadrant it is in and it will also help you to find the sine and cosine values. - Looking at our picture, we see that 240
^{°}is in quadrant III. This will help us to solve for the sine and cosine values as well. - Note that [(4π)/3] is [(π)/3] past the x - axis so we can use the 30
^{°}− 60^{°}− 90^{°}triangle to find the sine and cosine. - Recall that cosine is the x coordinate and sine is the y coordinate of 240
^{°}on the unit circle - cos 240
^{°}= − [1/2] and sin 240^{°}= − [(√3 )/2] - Using the Mnemonic ASTC (All Students Take Calculus) helps to determine which values are positive in which quadrant
- Since 240
^{°}is in quadrant III we know that tangent will be positive so sine and cosine will be negative

^{°}is in quadrant III and cos 240

^{°}= − [1/2] and sin 240

^{°}= − [(√3 )/2]

^{°}to radians, identify its quadrant, and find its cosine and sine.

- First convert 210
^{°}to radians by using the following formula: degree measure ×[(π)/(180^{°})] = radian measure - 210
^{°}×[(π)/(180^{°})] = [(210^{°}π)/(180^{°})] = [(7π)/6] - Locate [(7π)/6] on a unit circle (draw a picture). This will help you to determine the quadrant it is in and it will also help you to find the sine and cosine values.
- Looking at our picture, we see that [(7π)/6] is in quadrant III. This will help us to solve for the sine and cosine values as well.
- Note that 210
^{°}is 30^{°}past the x - axis so we can use the 30^{°}− 60^{°}− 90^{°}triangle to find the sine and cosine. - Recall that cosine is the x coordinate and sine is the y coordinate of [(7π)/6] on the unit circle
- cos [(7π)/6] = − [(√3 )/2] and sin [(7π)/6] = − [1/2]
- Using the Mnemonic ASTC (All Students Take Calculus) helps to determine which values are positive in which quadrant
- Since [(7π)/6] is in quadrant III we know that tangent will be positive so cosine and sine will be negative

- First convert [(3π)/4] to degrees by using the following formula: radian measure ×[(180
^{°})/(π)] = degree measure - [(3π)/4] ×[(180
^{°})/(π)] = 3 ×45^{°}= 135^{°} - Locate 135
^{°}on a unit circle (draw a picture). This will help you to determine the quadrant it is in and it will also help you to find the sine and cosine values. - Looking at our picture, we see that 135
^{°}is in quadrant II. This will help us to solve for the sine and cosine values as well. - Note that [(3π)/4] is [(π)/4] from the x - axis so we can use the 45
^{°}− 45^{°}− 90° triangle to find the sine and cosine. - Recall that cosine is the x coordinate and sine is the y coordinate of 135
^{°}on the unit circle - cos 135
^{°}= − [(√2 )/2] and sin 135^{°}= [(√2 )/2] - Since 135
^{°}is in quadrant II we know that sine will be positive and cosine will be negative

^{°}is in quadrant II and cos 135

^{°}= − [(√2 )/2] and sin 135

^{°}= [(√2 )/2]

^{°}to radians, identify its quadrant, and find its cosine and sine.

- First convert 150
^{°}to radians by using the following formula: degree measure ×[(π)/(180^{°})] = radian measure - 150
^{°}×[(π)/(180^{°})] = [(150^{°}π)/(180^{°})] = [(5π)/6] - Locate [(5π)/6] on a unit circle (draw a picture). This will help you to determine the quadrant it is in and it will also help you to find the sine and cosine values.
- Looking at our picture, we see that [(5π)/6] is in quadrant II. This will help us to solve for the sine and cosine values as well.
- Note that 150
^{°}is 30^{°}from the x - axis so we can use the 30^{°}- 60^{°}- 90^{°}triangle to find the sine and cosine. - Recall that cosine is the x coordinate and sine is the y coordinate of [(5π)/6] on the unit circle
- cos [(5π)/6] = − [(√3 )/2] and sin [(5π)/6] = [1/2]
- Since [(5π)/6] is in quadrant II we know that sine will be positive and cosine will be negative

- Locate [(3π)/4] on a unit circle (draw a picture). This will help you to determine the quadrant it is in and it will also help you to find the sine value.
- Looking at our picture, we see that [(3π)/4] is in quadrant II. This will help us to solve for the sine value.
- Note that [(3π)/4] is [(π)/4] from the x - axis so we can use the 45
^{°}− 45^{°}− 90^{°}triangle to find the sine value. - Recall that sine is the y coordinate of [(3π)/4] on the unit circle
- sin [(3π)/4] = [(√2 )/2]
- Since [(3π)/4] is in quadrant II we know that sine will be positive.

^{°}

- Locate 330
^{°}on a unit circle (draw a picture). This will help you to determine the quadrant it is in and it will also help you to find the cosine value. - Looking at our picture, we see that 330
^{°}is in quadrant IV. This will help us to solve for the cosine value. - Note that 330
^{°}is 30^{°}from the x - axis so we can use the 30^{°}- 60^{°}- 90^{°}triangle to find the cosine value. - Recall that cosine is the x coordinate of 330
^{°}on the unit circle - cos 330
^{°}= [(√3 )/2] - Since 330
^{°}is in quadrant IV we know that cosine will be positive.

^{°}= [(√3 )/2]

- Locate [(5π)/4] on a unit circle (draw a picture). This will help you to determine the quadrant it is in and it will also help you to find the sine value.
- Looking at our picture, we see that [(5π)/4] is in quadrant III. This will help us to solve for the sine value.
- Note that [(5π)/4] is [(π)/4] past the x - axis so we can use the 45
^{°}− 45^{°}− 90^{°}triangle to find the sine value. - Recall that sine is the y coordinate of [(5π)/4] on the unit circle
- sin [(5π)/4] = − [(√2 )/2]
- Since [(5π)/4] is in quadrant III we know that sine will be negative.

- Recall that cosine is the x coordinate.
- Draw a unit circle and locate positive [1/2] along the x - axis. Draw a vertical line through that point. This will help you to identify the two points on the circle.
- Connect the points on the circle to the origin. Notice that a 30
^{°}- 60^{°}- 90^{°}triangle was formed. - The first angle is 60
^{°}past 0^{°}⇒ 0^{°}+ 60^{°}= 60^{°} - The second angle is 30
^{°}past of 270^{°}⇒ 270^{°}+ 30^{°}= 300^{°}

Angle 1 | Angle 2 | |

Degree | 60^{°} | 300^{°} |

Radian | [(π)/3] | [(5π)/3] |

Quadrant | I | IV |

- Recall that sine is the y coordinate.
- Draw a unit circle and locate − [(√3 )/2] along the y - axis. Draw a horizontal line through that point. This will help you to identify the two points on the circle.
- Connect the points on the circle to the origin. Notice that a 30
^{°}- 60^{°}- 90^{°}triangle was formed. - The first angle is 60
^{°}past 180^{°}⇒ 180^{°}+ 60^{°}= 240^{°} - The second angle is 60
^{°}short of 360^{°}⇒ 360^{°}− 60^{°}= 300^{°}

Angle 1 | Angle 2 | |

Degree | 240^{°} | 300^{°} |

Radian | [(4π)/3] | [(5π)/3] |

Quadrant | I | IV |

*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 Values of Special Angles

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 0:00
- 45-45-90 Triangle and 30-60-90 Triangle 0:08
- 45-45-90 Triangle
- 30-60-90 Triangle
- Mnemonic: All Students Take Calculus (ASTC) 5:21
- Using the Unit Circle
- New Angles
- Other Quadrants
- Mnemonic: All Students Take Calculus
- Example 1: Convert, Quadrant, Sine/Cosine 13:11
- Example 2: Convert, Quadrant, Sine/Cosine 16:48
- Example 3: All Angles and Quadrants 20:21
- Extra Example 1: Convert, Quadrant, Sine/Cosine
- Extra Example 2: All Angles and Quadrants

### Trigonometry Online Course

I. Trigonometric Functions | ||
---|---|---|

Angles | 39:05 | |

Sine and Cosine Functions | 43:16 | |

Sine and Cosine Values of Special Angles | 33:05 | |

Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D | 52:03 | |

Tangent and Cotangent Functions | 36:04 | |

Secant and Cosecant Functions | 27:18 | |

Inverse Trigonometric Functions | 32:58 | |

Computations of Inverse Trigonometric Functions | 31:08 | |

II. Trigonometric Identities | ||

Pythagorean Identity | 19:11 | |

Identity Tan(squared)x+1=Sec(squared)x | 23:16 | |

Addition and Subtraction Formulas | 52:52 | |

Double Angle Formulas | 29:05 | |

Half-Angle Formulas | 43:55 | |

III. Applications of Trigonometry | ||

Trigonometry in Right Angles | 25:43 | |

Law of Sines | 56:40 | |

Law of Cosines | 49:05 | |

Finding the Area of a Triangle | 27:37 | |

Word Problems and Applications of Trigonometry | 34:25 | |

Vectors | 46:42 | |

IV. Complex Numbers and Polar Coordinates | ||

Polar Coordinates | 1:07:35 | |

Complex Numbers | 35:59 | |

Polar Form of Complex Numbers | 40:43 | |

DeMoivre's Theorem | 57:37 |

### Transcription: Sine and Cosine Values of Special Angles

*Hi we are working out some examples of common values of sin and cos of common angles.*0000

*Remember what we learned, everything comes back to knowing those two key triangles.*0007

*There is the 45, 45, 90 triangle whose values are (root2)/2, (root 2)/2 and 1.*0014

*And then there is the 30, 60, 90 triangle whose values are 1/2, (root 3)/2 and 1.*0025

*If you remember those set of numbers, you can work out sin and cos of any common value anywhere on the unit circle.*0038

*That is what we are doing here, we have given the values 225 degrees, convert it to radians, let us start with that.*0046

*225 x pi/180 is equal to, well 225/180 simplifies down to 5/4, so that is 5pi/4.*0053

*Let us draw that on the unit circle and see where it lands.*0069

*My unit circle is 0, pi/2, which is the same as 90 degrees, pi radians is equal to 180 degrees, and 3pi/2 radians is 270 degrees, 2pi radians is equal to 360 degrees.*0085

*We have got 225 degrees or 5pi/4, 225 is between 180 and 270, in fact it is exactly half way between there because it is 45 degrees from either side.*0105

*It is right there, if you like that in terms of radians, 5pi/4 is just pi + pi/4, that is the angle we are looking at.*0118

*That is in the third quadrant, we found its quadrant, we converted it to radians, we want to find its cos and sin, that is the x and y coordinates.*0133

*Let me draw those in there, we want to figure out what those x and y coordinates are.*0146

*Look that is a common triangle and I remember what the values of those common triangles are.*0150

*That distance is (root 2 )/2, that distance is (root 2)/2.*0157

*I’m getting that because I remember this common 45, 45, 90 right triangle.*0162

*That means the sin and cos of both (root 2)/2 and we just need to figure out whether they are positive or negative.*0171

*All students take calculus, down there on the third quadrant only the tan is positive, both the sin and cos are negative.*0188

*Another way to remember that is just to remember that in the third quadrant both x and y values are negative.*0201

*Sin and cos of this angle are both negative (root 2)/2.*0207

*Again, what we are using over and over in these examples is these two key triangles.*0220

*The 45, 45, 90 triangle and the 30, 60, 90 triangle, you want to remember the values for those two triangles, (root 2)/2, (root 3)/2, and ½.*0228

*Remember those values and then when you have an angle that is in the other quadrants, it is just a matter of translating one of those triangles over there.*0244

*And then figuring out whether the sin and cos are positive and negative.*0252

*Let us try one more example here, we want to identify all the angles between 0 and 2 pi whose cos is –root 3/2.*0000

*I start by drawing my unit circle that is not quite straight, let me straighten that up a little bit.*0010

*Cos is – root 3/2, now the cos remember is the (x) value so I’m going to go on the (x) axis and I’m going to go to –root 3/2.*0037

*There it is.*0053

*That is –root 3/2 on the (x) axis and then I’m going to draw and see what angles I will get from that.*0058

*It looks like, remember that root 3/2 is one of my common values, that means that the y values are going to be ½.*0073

*I need to figure out which angles those are but that is one of my common values ½ root 3/2 that means that is a 30 degree angle, that is 60 and that is 30.*0083

*I just need to figure out what those angles are, if you remember we started 0, 90, 180, 270, and 360.*0097

*That first angle there is 30 degrees short of 180, the first angle is 150 degrees.*0111

*The second angle is 30 degrees past 180, so that is 210 degrees.*0120

*I have got my angles in degrees I will convert them into radians x pi/180 is equal to 5pi/6 to 10 x pi/180 is 7pi/6 radians.*0127

*I got those two angles in radians now, that is the first one 5pi/6, that is the second one 7pi/6.*0151

*And identify which quadrant each one is in, one of them is in the second quadrant, one of them is in the third quadrant, quadrant 2 and quadrant 3.*0159

*It all comes back to recognizing those common values, ½, square root of 3/2, square root of 2/2.*0184

*Once you recognize those common values, you can put these triangles in any position anywhere on the unit circle.*0191

*You just figure out where is your root 3/2, where is your ½, where is your root 2/2 and then you figure out which one is positive and which one is negative.*0198

*The whole point of this is you can figure out the sin and cos of any angle anywhere on the unit circle as long as it is a multiple of 30 or 45, or in terms of radians if it is a multiple of pi/6, pi/6, pi/4, pi/3.*0209

*You can figure out sin and cos of all these angles just by going back to those 3 common values and by figuring out whether their sin and cos are positive or negative.*0226

*Now you know how to find sin and cos of special angles, this is www.educator.com, thanks for watching.*0238

*Hi, these are the trigonometry lectures on educator.com.*0000

*Today we're going to learn about sine and cosine values of special angles.*0003

*When I say these special angles, there are certain angles that you really want to know by heart.*0009

*Those are the 45-45-90 triangle, and the 30-60-90 triangle.*0014

*Let me talk about the 45-45-90 triangle first.*0022

*I'll draw this in blue.*0028

*Here's a 45-45-90 triangle and I'm going to say that each side has length 1.*0041

*If each of the short sides has length 1, by the Pythagorean theorem, we can figure out that the long side, the hypotenuse, would have length square root of 2.*0052

*I'm going to scale this triangle down a little bit now.*0066

*I wanted to scale it down so the hypotenuse has length 1.*0071

*That means I have to divide all three sides by square root of 2.*0075

*If I scale this down, so the hypotenuse has length 1 that means the shorter sides has length 1 over the square root of 2, because I divided each side by square root of 2.*0079

*Then if you rationalize that, the way you learned in your algebra class, multiply top and bottom by square root of 2.*0095

*You get square root of 2 over 2, and square root of 2 over 2.*0102

*Those are very important values to remember because those are going to come up as sines and cosines of our 45-degree angles on the next slide.*0113

*First, I'd like to look also at the 30-60-90 triangle.*0122

*I have to do a little geometry to work this out for you.*0129

*I'm going to start with an equilateral triangle, a triangle where all three sides have 60 degrees.*0133

*I'm going to have assume that each side has length 2.*0142

*The reason I'm going to do that is because I'm going to divide that triangle in half.*0145

*If we divide that triangle in half, then we get a right angle here and each one of these pieces will have length 1.*0150

*Now, if I just look at the right hand triangle.*0160

*Remember that each one of the corners of the original triangle was 60 degrees.*0169

*That means that the small corner is 30 degrees, and I have a right angle here.*0175

*Now, the short side has length 1, the long side has length 2.*0182

*I'm going to figure out what the other side is using the Pythagorean theorem.*0186

*Let me call that x for now.*0191

*I know that x ^{2} + 1^{2} = 2^{2}, which is 4.*0194

*So, x ^{2} = 4 - 1, which is 3.*0202

*So, x is the square root of 3.*0209

*That's where I got this relationship, 1, square root of 3, 2.*0213

*There's the 1, there's the square root of 3, and there's the 2.*0219

*Now, I'm going to turn this triangle on its side, and I want to scale it down.*0223

*Originally, it was 1, square root of 3, 2.*0232

*But again, I want to scale the triangle down so that the hypotenuse has length 1.*0238

*To do that, I have to divide everything by 2.*0244

*So the short side now has length 1/2.*0247

*The longer of the two short sides has length square root of 3 over 2.*0251

*Remember that's the side adjacent to the 30-degree angle.*0258

*That's the side adjacent to the 60-degree angle,*0262

*That's the right hand side.*0264

*These two triangles are very key to remember, in remembering all the sines and cosines.*0267

*In fact, if you can remember these lengths of these two triangles, you can work out everything else just from these two triangles.*0273

*Let me emphasize again, the 45-45-90 triangle, its sides have length 1, square root of 2 over 2, and square root of 2 over 2.*0282

*The 30-60-90 triangle, has sides of length 1, 1/2, and root 3 over 2.*0297

*Those are the values that you need to remember.*0310

*If you can remember those, you can work out all the sines and cosines you need to know for every trigonometry class ever.*0312

*Let's explore those a little bit.*0321

*We already figured out, let me draw a unit circle.*0324

*We know that sines and cosines occur as the x and y coordinates of different angles.*0328

*Well, if you're at 0...*0338

*Let me just draw in some key angles here, 0, here's 90, here's 45, here's 30, and here's 60.*0341

*If you're at 0 degrees, which is the same as 0 radians, then the cosine and sine, the x and y coordinates are just 1 and 0.*0359

*We already figured those out before.*0368

*The other easy one is the 90-degree angle up here.*0370

*We figured out that that's π/2 radians, and the cosine and sine are 0, and 1 there.*0374

*Now, the new ones, let me start with 45, because I think that one's a little bit easier.*0380

*The 45-degree angle, there it is right there.*0386

*We want to figure out what the x and y coordinates are because those give us the sine and cosine.*0391

*Well, we just figured out that a triangle that has 1 as its hypotenuse has square root of 2 over 2, as both its x and y sides.*0397

*That's where we get the square root of 2 over 2 as the cosine and sine of the 45-degree angle, also known as π/4 radians.*0407

*For the 30-degree angle, I'll do this one in blue.*0418

*The 30-degree angle, we have again, hypotenuse has length 1.*0422

*Remember, the length of the long side is root 3 over 2.*0431

*And the length of the short side is 1/2.*0436

*That's how you know the sine and cosine of the 30-degree angle or π/6.*0439

*The cosine, the x-coordinate, root 3 over 2, sine is 1/2.*0447

*The 60-degree angle, that's just the same triangle but it's flipped the other way so that the long side is on the vertical part and the short side is on the horizontal axis.*0452

*The short side is 1/2.*0468

*The long side is now the y-axis, that's root 3 over 2.*0473

*That's how we get 1/2 being the cosine of 60 degrees, root 3 over 2 being the sine of 60 degrees.*0478

*These values are really worth memorizing but you remember that you figure all out from those two triangles.*0488

*All you need to know is that one triangle has length 1, has hypotenuse 1, and sides root 2 over 2, that's the 45-45-90 triangle.*0494

*The other triangle has hypotenuse 1, and then the long side is root 3 over 2, short side is 1/2, that's the 30-60-90 triangle.*0509

*Just take that triangle and you flip it whichever way you need to, to get the angle that you're looking for.*0526

*These angles, these sines and cosines are the key ones to remember, root 2 over 2, root 3 over 2, and 1/2.*0534

*From that, what you're going to do is figure out the sines and cosines of all the other angles all over the unit circle.*0546

*Here's my unit circle.*0573

*We figured out all the sines and cosines of all the angles in the first quadrant.*0575

*All we have to do now is figure out all the sines and cosines of the angles in the other quadrants.*0583

*Let me draw them out.*0589

*But it's the same numbers every time.*0590

*All you have to do is figure out whether those numbers are positive or negative, and that just depends on which quadrant you're in.*0593

*All you have to do is remember those key numbers, root 2 over 2, root 3 over 2 and 1/2.*0601

*Then you're going to figure out which ones are positive in which quadrants.*0607

*Let me show you how you'll remember that, 1, 2, 3, 4.*0613

*Remember that the sine is the y value, and the cosine is the x value.*0622

*In the first quadrant, both the x's and y's are positive, and so sines and cosines are going to positive.*0630

*I'm going to write that as an a, which stands for all the values of everything is positive.*0636

*In the second quadrant, over here, the x values are negative, the y values are positive.*0643

*Now, the x values correspond to the cosines, the cosines are negative and the sines are positive.*0653

*I'm going to write as s here, for the sines being positive.*0658

*In the third quadrant, both x's and y's are negative.*0667

*X and y are both negative, that means cosines and sines are both negative.*0669

*Tangent, we haven't learned the details of tangent yet but we're going to learn later that tangent is sine over cosine.*0675

*Both sine and cosine are negative, that means sine over cosine is positive.*0683

*It turns out that tangent is positive down that third quadrant.*0687

*We'll learn the details of tangent later.*0691

*In the meantime, we'll just remember that tangent is positive in the third quadrant.*0693

*In the fourth quadrant here, that was the third quadrant that we just talked about, now we're moving on to the fourth quadrant.*0698

*The x's are positive now, the y's are negative.*0705

*That means the cosine is positive, but the sine is negative.*0708

*I'll list the cosine here because I'm listing the positive ones.*0713

*First quadrant, they're all positive.*0718

*Second quadrant, sines are positive.*0720

*Third quadrant, tangents are positive.*0722

*Fourth quadrant, cosines are positive.*0724

*The way you remember that is with this little acronym, All Students Take Calculus.*0727

*That shows you as you go around the four quadrants, All Students Take Calculus.*0733

*It shows you which ones are positive in each quadrant.*0741

*First quadrant are all positive.*0744

*Second quadrant sines are positive.*0746

*Third quadrant, tangents are positive.*0748

*Fourth quadrant, cosines are positive.*0750

*That's how you'll remember what the signs are in each quadrant.*0753

*That means the positive signs and the negative signs in each quadrant.*0756

*The numbers are just all these values that you've just memorized, root 3 over 2, root 2 over 2, and 1/2.*0758

*We'll do some practice finding sines and cosines of values in other quadrants, based on this table that we remember.*0768

*And these common values in common triangles that we remember.*0775

*Then we'll take those values, introduce some positive and negative signs, and we'll come up with the sines and cosines of angles in other quadrants.*0780

*Let's try some examples.*0787

*First example here is a 120 degrees, you want to convert that to radians, identify it's quadrant, and find it's cosine and sine.*0791

*First things first, let's convert it to radians, 120 times π/180 is 2π/3, because 120/180 is 2/3 so that's 2π/3 radians.*0800

*Identify it's quadrant.*0806

*Well, let me graph out my unit circle here.*0814

*There's 0, there's π/2, π, 3π/2, and then 2π.*0820

*Now, 2π/3 is between π/2 and π.*0840

*In fact, it's closer to π/2.*0845

*It's 2/3 of the way around from 0 to π.*0848

*If you like that in terms of degrees, π is 180 degrees, and so 2π/3 is 2/3 of the way over to 180.*0852

*Now, we're going to find cosine and sine.*0862

*Let me show you how to do this.*0867

*You draw a triangle here.*0868

*Remember, we're looking for the x and y coordinates.*0870

*Draw a triangle here.*0873

*That's a 30-60 triangle.*0875

*This is 60, that's a 60-degree angle.*0877

*That's a 30-degree angle.*0887

*We know what the lengths of these different sides are.*0891

*We know that the long side there is root 3 over 2 and the short side there is 1/2.*0897

*We know, remember, the cosines and sine are the x and y coordinates.*0907

*The cosine of 120 or 2π/3 is 1/2, except that we're going to have to check whether that's positive or negative.*0910

*Remember All Students Take Calculus.*0930

*In the second quadrant, only the sine is positive, so the cosine must be negative.*0935

*The sine of 2π/3, the y value is root 3 over 2.*0942

*In the second quadrant, sines are positive, so that's positive.*0950

*Our cosine and sine are -1/2 and root 3 over 2.*0954

*If you didn't remember the All Students Take Calculus thing, you can also just work it out once you know what quadrant it's in.*0958

*It's in quadrant 2 and we know there that the x coordinates are negative, and the y coordinates are positive.*0964

*The cosine must be negative and the sine must be positive.*0974

*The whole point of this is that you only really need to memorize the values of the triangles, root 2 over 2, root 3 over 2 and 1/2.*0981

*Once you know those basic triangles, you can work out what the sines and cosines are in any different quadrant just by drawing in those triangles and then figuring out which ones have to be positive, and which ones are to be negative.*0991

*Let's try another one.*1005

*This one is converting 5π/3 radians to degrees, identifying it's quadrant, and finding its cosine and sine.*1009

*5π/3 times 180/π, the π's cancel, and the we have 5/3 of 180, 180 over 3 is 80, so this is 5 times 60 is 300 degrees.*1018

*Let's try and find that in the unit circle.*1040

*We have 0, π/2 which is 90, π which is 180, 3π/2 which is 270, and 2π which is the same as 360 degrees.*1053

*Now, 5π/3, that's bigger than π and that's smaller than 2π.*1074

*In fact, that's π + 2π/3.*1080

*That's π, which is right here, plus 2π/3.*1090

*So, 2/3 the way around from π to 2π.*1099

*There it is right there.*1103

*That's in the fourth quadrant.*1104

*So, we figured out what quadrant it's in.*1108

*If you like degrees better, 300 degrees is a little bigger than 270, in fact, it's 30 degrees past 270, and 60 degrees short of 360.*1110

*That's how you know that that angle is in that quadrant.*1121

*Now we have to find its cosine and sine.*1126

*That's the x and y coordinates.*1128

*We set up our triangle there.*1134

*We already know that that's a 60-degree angle, because it's 60 degrees short of 360.*1137

*That's a 30-degree angle and we just remember our common values.*1142

*The horizontal value, that's the short one, that's 1/2, that's the long one, root 3 over 2.*1147

*We know our values are going to be 1/2 and root 3 over 2, we'll just have to figure out which one's positive and which one's negative.*1154

*I know my cosine of 5π/3 is going to be either positive or negative 1/2.*1162

*The sine of 5π/3 is positive or negative root 3/2.*1171

*Remember All Students Take Calculus.*1179

*Down there in the fourth quadrant, the cosine is positive and the sine is negative.*1184

*If you don't remember All Students Take Calculus, you just look that you're in the fourth quadrant, x coordinates are positive, y coordinates are negative.*1189

*You know which one's positive or negative.*1199

*All you have to remember are those key values, 1/2, root 3 over 2, root 2 over 2.*1202

*Remember those key values for the key triangles.*1209

*Then it's just a matter of drawing the right triangle in the right place and figuring out which one is positive, and which one is negative.*1212

*Let's try another one.*1219

*This one is kind of tricky.*1223

*This one's going to be challenging us to go backwards from the sine.*1224

*We have to find all angles between 0 and 2π whose sine is -1/2.*1229

*This is kind of foreshadowing the arc sine function that we'll be studying later on and one of the later lectures.*1242

*In the meantime, the sine is -1/2.*1249

*Remember now, the sine is the y-coordinate.*1252

*We want things whose y coordinates are -1/2.*1256

*I'm going to draw -1/2 on the y-axis, -1/2.*1259

*I'm going to look for all angles whose y-coordinate is -1/2.*1266

*Look, there's one right there.*1272

*And there's one right there.*1276

*I'm going to draw those in.*1277

*I'm going to draw those triangles in.*1280

*I know now that if we have a vertical component of 1/2, the horizontal component has to be root 3 over 2.*1289

*That's because we remember those common triangles, 1/2, root 3 over 2, root 2 over 2.*1298

*We're going to figure out what those angles are.*1307

*I know that's a 30-degree angle.*1312

*I know that that is 180.*1318

*The whole thing is 210 degrees.*1322

*I know that that is 30.*1330

*I know that that must be 60.*1334

*Remember, this is 270 degrees down here.*1337

*We have 270 degrees plus 60 degrees.*1342

*This is getting a little messy, so I'm going to redraw it over here.*1345

*That's the angle we're trying to chase down here.*1357

*We know that's 60.*1360

*That much is 270.*1362

*So, 270 plus 60 is 330 degrees.*1366

*Those are the two angles that we're after, 210 degrees, 330 degrees.*1372

*Let me convert those into radians.*1378

*If you multiply that by π/180 then that's equal to...*1382

*Let's see, that's 7π/6 radians.*1390

*This one times π/180 is equal to 11π/6 radians.*1397

*We've got our two angles in degrees and radians.*1409

*The quadrants, the first one was in the third quadrant, quadrant 3.*1413

*The second one was in the fourth quadrant, quadrant 4.*1420

*Those are the two angles in both degrees and radians that had a sign of -1/2.*1426

*Their y value was -1/2.*1436

*What this comes down to is knowing those common values, 1/2 root 3 over 2, root 2 over 2.*1442

*Once you know those common values, it's a matter of looking at the different quadrants and figuring out whether the x and y values are positive or negative.*1454

*In this case, we had the sine, sine remember is the y-coordinate.*1461

*Since it was negative, we knew that we had to be at -1/2 on the y-axis.*1468

*We found -1/2 on the y-axis, drew in the triangles, recognized the 30-60 triangles that we've been practicing and then we were able to work out the angles.*1473

*We'll try some more examples of that later.*1484

1 answer

Last reply by: Dr. William Murray

Fri Jan 8, 2016 1:23 PM

Post by sania sarwar on January 7, 2016

Hi Sir

thanks for the lectures they are really helpful but can you please suggest me on which lecture to watch for this problem because I cant get my head around it.

If sin(theta)=0.3,cos(x)=0.7 and tan (alpha)=0.4 then find sin(3pi/2+theta)

1 answer

Last reply by: Dr. William Murray

Thu Feb 19, 2015 5:27 PM

Post by patrick guerin on February 19, 2015

Why is it that when you type in the cosine or sine of a number on a calculator, you get something like .0679 or something?

3 answers

Last reply by: Dr. William Murray

Mon Nov 17, 2014 8:12 PM

Post by Tami Cummins on September 20, 2013

I know this is so simple but I really do not understand how 5pi/3 equals pi + 2pi/3.

1 answer

Last reply by: Dr. William Murray

Fri Jun 21, 2013 6:22 PM

Post by A C on June 17, 2013

Please help: why, if I use a calculator to find the angle value of sin/cos/tan, does sin and tan give me the angle (neg. or pos. depending on the quadrant obviously) but cos gives me are larger angle (>90degrees), what I assume is the standard position angle?

1 answer

Last reply by: Dr. William Murray

Wed May 22, 2013 3:25 PM

Post by Monis Mirza on May 17, 2013

hi,

In the extra example I, how did you know that the special triangle is

45-45-90 triangle?

Thanks

1 answer

Last reply by: Dr. William Murray

Thu Nov 15, 2012 6:21 PM

Post by peter chrysanthopoulos on November 14, 2012

nevermind I got it

1 answer

Last reply by: Dr. William Murray

Thu Nov 15, 2012 6:20 PM

Post by peter chrysanthopoulos on November 14, 2012

how did you know that it was a 30/60/90 triangle in example 1?

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 11:34 AM

Post by Dr. William Murray on October 17, 2012

Hi Chin,

It depends on which direction you're going. If it's radians to degrees, multiply by 180/pi. If it's degrees to radians, multiply by pi/180.

This makes sense if you follow the rules that you learn in physics and chemistry about units: 180 degrees = pi radians, so (180 degrees)/(pi radians) = 1. Then when you want to convert in either direction, you multiply by 1:

(3 pi/4 radians) x (180 degrees)/(pi radians) = 135 degrees.

90 degrees x (pi radians)/(180 degrees) = pi/2 radians.

Hope this helps. Thanks for studying trigonometry!

Will Murray

1 answer

Last reply by: Dr. William Murray

Thu Apr 18, 2013 11:35 AM

Post by chin chang on October 15, 2012

On the example problems you converted the degrees or radians into one or the other by multiplying pi/180 or 180/pi. How do you know what to multiply for each situation? Does it make sense? For instance on the second example problem, how did you to multiply 5pi/3 radians by 180/pi?

3 answers

Last reply by: Dr. William Murray

Fri Sep 28, 2012 4:54 PM

Post by Ivon Nieto Ivon Nieto on September 25, 2012

Is there an easier way to memorize the unit circle sine/cosine?