For more information, please see full course syllabus of Trigonometry

For more information, please see full course syllabus of Trigonometry

### Related Articles:

### Tangent and Cotangent Functions

**Main definitions and formulas**:

- The
*tangent function*is defined by

for values of θ where cosθ≠ 0. (For values of θ where cosθ = 0, the tangent is undefined.)tanθ = sinθ cosθ - The
*cotangent function*is defined by

for values of θ where sinθ≠ 0. (For values of θ where sinθ = 0, the cotangent is undefined.)cotθ = cosθ sinθ - Master formula for right triangles: SOHCAHTOA!
sinθ = opposite side hypotenusecosθ = adjacent side hypotenusetanθ = opposite side adjacent side Degrees Radians Cosine Sine Tangent Cotangent 0 0 1 0 0 undefined 30 π 6√ 3 21 2√ 3 3√ 3 45 π 4√ 2 2√ 2 21 1 60 π 31 2√ 3 2√ 3 √ 3 390 π 20 1 undefined 0 - The slope of a line is the tangent of the angle that the line makes with the
*x*-axis. - 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**:

**Example 2**:

**Example 3**:

**Example 4**:

**Example 5**:

**Example 6**:

### Tangent and Cotangent Functions

- Recall: SOHCAHTOA, tanθ = [Opposite/Adjacent], cotθ = [Adjacent/Opposite]
- First calculate the missing leg of the triangle by using the Pythagorean Theorem: a
^{2}+ b^{2}= c^{2} - 5
^{2}+ 6^{2}= x^{2}⇒ 25 + 36 = x^{2}⇒ 61 = x^{2}⇒√{61} = x - Draw a triangle in which the sides and angles are labeled so tangent and cotangent can be determined.

- Recall: SOHCAHTOA, tanθ = [Opposite/Adjacent], cotθ = [Adjacent/Opposite]
- First calculate the missing leg of the triangle by using the Pythagorean Theorem: a
^{2}+ b^{2}= c^{2} - 9
^{2}+ 11^{2}= x^{2}⇒ 81 + 121 = x^{2}⇒ 202 = x^{2}⇒√{202} = x - Draw a triangle in which the sides and angles are labeled so tangent and cotangent can be determined.

- Recall: SOHCAHTOA, tanθ = [Opposite/Adjacent], cotθ = [Adjacent/Opposite]
- First calculate the missing leg of the triangle by using the Pythagorean Theorem: a
^{2}+ b^{2}= c^{2} - x
^{2}+ 7^{2}= 12^{2}⇒ x^{2}+ 49 = 144 ⇒ 95 = x^{2}⇒√{95} = x - Draw a triangle in which the sides and angles are labeled so tangent and cotangent can be determined.

- Recall: SOHCAHTOA, tanθ = [Opposite/Adjacent], cotθ = [Adjacent/Opposite]
- First calculate the missing leg of the triangle by using the Pythagorean Theorem: a
^{2}+ b^{2}= c^{2} - x
^{2}+ 5^{2}= 8^{2}⇒ x^{2}+ 25 = 64 ⇒ 39 = x^{2}⇒√{39} = x

- Recall: SOHCAHTOA, tanθ = [Opposite/Adjacent], cotθ = [Adjacent/Opposite]
- First calculate the missing leg of the triangle by using the Pythagorean Theorem: a
^{2}+ b^{2}= c^{2} - x
^{2}+ 4^{2}= 6^{2}⇒ x^{2}+ 16 = 36 ⇒ 20 = x^{2}⇒ 2√5 = x

- First plot all three angles on the unit circle. Notice that [(3π)/4] uses the 45
^{°}− 45^{°}− 90^{°}triangle and [(4π)/3],[(11π)/6] both use the 30^{°}− 60^{°}− 90^{°}triangle. - Use the mnemonic ASTC to determine which ones are positive and which are negative

Angle | sin | cos | tan | cot |

[(3π)/4] | [(√2 )/2] | [(√2 )/2] | − 1 | − 1 |

[(4π)/3] | − [(√3 )/2] | − [1/2] | √3 | [(√3 )/3] |

[(11π)/6] | − [1/2] | [(√3 )/2] | − [(√3 )/3] | − √3 |

- First find two consecutive asymptotes by solving the following equations: 2x = − [(π)/2] and 2x = [(π)/2]
- So, x = − [(π)/4] and x = [(π)/4] are two asymptotes
- Use the asymptotes and sketch the tangent graph
- Zeroes are − [(π)/2], 0, [(π)/2], π, [(3π)/2], 2π, Asymptotes are [( − π)/4], [(π)/4], [(3π)/4], [(5π)/4], [(7π)/4], [(9π)/4], ...

- First find two consecutive asymptotes by solving the following equations: [x/2] = − [(π)/2] and [x/2] = [(π)/2]
- So, x = - π and x = π are two asymptotes
- Use the asymptotes and sketch the tangent graph
- Zeroes are 0, 2π, Asymptotes are - π, π, 3π

- First find two consecutive asymptotes by solving the following equations: [x/3] = 0 and [x/3] = π
- So, x = 0 and x = 3π are two asymptotes
- Use the asymptotes and sketch the tangent graph
- Zeroes are − [(3π)/2], [(3π)/2], [(9π)/2], Asymptotes are 0, π, 3π

- First find two consecutive asymptotes by solving the following equations: 4x = 0 and 4x = π
- So, x = 0 and x = [(π)/4] are two asymptotes
- Use the asymptotes and sketch the tangent graph
- Zeroes are − [(π)/8], [(π)/8], [(π)/2], Asymptotes are − [(π)/4], 0, [(π)/4], [(3π)/4]

*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

### Tangent and Cotangent 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
- Tangent and Cotangent Definitions
- Master Formula: SOHCAHTOA
- Tangent and Cotangent Values
- Slope and Menmonic: ASTC
- Example 1: Graph Tangent and Cotangent Functions
- Example 2: Tangent and Cotangent of Angles
- Example 3: Odd, Even, or Neither
- Extra Example 1: Tangent and Cotangent of Angles
- Extra Example 2: Tangent and Cotangent of Angles

- Intro 0:00
- Tangent and Cotangent Definitions 0:21
- Tangent Definition
- Cotangent Definition
- Master Formula: SOHCAHTOA 1:01
- Mnemonic
- Tangent and Cotangent Values 2:29
- Remember Common Values of Sine and Cosine
- 90 Degrees Undefined
- Slope and Menmonic: ASTC 5:47
- Uses of Tangent
- Example: Tangent of Angle is Slope
- Sign of Tangent in Quadrants
- Example 1: Graph Tangent and Cotangent Functions 10:42
- Example 2: Tangent and Cotangent of Angles 16:09
- Example 3: Odd, Even, or Neither 18:56
- Extra Example 1: Tangent and Cotangent of Angles
- Extra Example 2: Tangent and Cotangent of Angles

### 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: Tangent and Cotangent Functions

*We are back with some extra examples of tan and cot functions, we are given here a right triangle with short sides of length 5 and 12.*0000

*Let me try to draw that and what we have to do is find the tan and cot of all the angles in the triangle.*0009

*Our first step there is to figure out what the hypotenuse is, hypotenuse ^{2} is 5^{2} + 12^{2}, which is 25 + 144 which is 169.*0022

*That is exactly 13 ^{2}, this triangle is rigged up to have a nice hypotenuse of 13 and so now let me label these angles (theta) and (phi), and we will figure out what the tan and cot of each one is.*0037

*The key point here, it all comes back to SOHCAHTOA.*0052

*We will be using SOHCAHTOA over and over again in your trigonometry so it is really worth memorizing that one.*0060

*If you have a hard time remembering the word, then remember the acronym Some Old Horse Caught Another Horse Taking Oats Away.*0067

*In particular, the tan is opposite of adjacent, tan(theta), the opposite side is 5, the adjacent side has length 12.*0073

*Cot(theta) is just the opposite of that, so it is 12/5.*0086

*Tan(phi), the opposite side for angle(phi) is 12 and the adjacent side is 5.*0097

*Cot(phi) is just the other way around it is 5/12.*0106

*Finally, the 90 degree angle a bit of special case there, let me give it in radians as pi/2.*0112

*Remember, tan(pi/2) is undefined and cot(pi/2) cos/sin which is 0/1 is just 0.*0119

*Finding cot and tan in right triangles is just a matter of remembering that mnemonics SOHCAHTOA, tan=opposite/adjacent.*0136

*Ok we are given here a bunch of angles and we want to find the tan and cot of each one.*0000

*A good first step is to graph this angles on a unit circle.*0006

*I got a unit circle here, I will label the common values pi/2, pi, 3pi/2, and 2pi and I want to figure out where these angles are.*0019

*2pi/3 is a little bit past pi/2, so we are going to find the tan and cot there.*0031

*7pi/6 is a little bit past pi, and 7pi/4 is in between 3pi/2 and 2pi.*0038

*Those are the three angles and to find their tan and cot, I will draw my triangles in and write down the sin and cos of each one.*0048

*For each angle I’m going to find the sin, cos, and I will use those to find the tan and cot.*0061

*Let us start with 2pi/3, that is over here, I will draw a triangle there, that is a 30, 60, 90 triangle.*0072

*I know that the sin is long one there so that must be (root 3)/2, cos of the short one is ½ except that is negative because the x coordinate is negative.*0082

*For tan, it is (sin/cos) that is just –root 3, cot=(cos/sin) that is -1/root 3, but that rationalizes to -root 3/3.*0094

*Next one is 7pi/6 so that is this angle right there, and that is a 30, 60, 90 triangle.*0110

*7pi/6 the (sin) is ½ but that is negative because the y coordinate is negative.*0121

*(Cos) the big one is root 3/2 but again that is negative because the x coordinate is negative.*0128

*(tan) if you divide those together, sin/cos=1/(root 3)/3 and it is positive because both of those are negative, so that 2 negative cancel.*0134

*(cot) is( cos/sin) that is positive root 3 because they are both negative and so the two negative is cancelled.*0146

*7pi/4 that is this angle down here and that is a 45, 45, 90 triangle, (sin) and (cos) are both root 2/2 for 45, 45, 90 triangle.*0157

*We got to figure out which one is negative and its (sin) because the (y) coordinate is negative here, we are down below the (x) axis.*0177

*(cos) is positive because the (x) coordinate is positive.*0185

*(tan) you divide those together and you get -1, and (cot) you divide (cos/sin) but you still just get -1.*0188

*It is probably not worth memorizing the (tan) and (cot) of those angles but it is worth knowing the common values for the 30,60,90 triangles.*0202

*For 30, 60, 90 it is ½, root 3/2, for the 45, 45, 90 triangles the (sin) and (cos) would be root 2/2 and root 2/2.*0213

*If you remember those common values then you can draw a triangle anywhere else in the unit circle and just figure out which of these values is the (sin) and which one is the (cos).*0226

*You can figure out which one is positive and which one is negative, to get the (tan) you just divide the (sin) and (cos) together, same for the (cot).*0238

*The only tricky part there is figuring the (tan) and (cot) are positive or negative, but you just look back at whether the sin and cos are positive or negative.*0249

*That tells you right away whether the (tan) and (cot) are positive and negative.*0259

*You can also remember the mnemonic all students take calculus which stands for all the values positive in the second quadrant, only the sin is positive, that is why the 2pi/3 (cot) where negative.*0264

*In the third quadrant, (tan) is positive which means the (cot) will be as well.*0283

*In the fourth quadrant, only (cos) is positive which means the (tan) will be negative.*0288

*A lot of different ways to remember that, choose which one works for you.*0294

*We will come back later with more lectures on trigonometry at www.educator.com.*0298

*Hi this is Will Murray for educator.com and we're talking about trigonometry.*0000

*We're finally ready to learn about the tangent and cotangent functions.*0004

*We've been eluding to those in some of the earlier lectures but this is the lecture where we're going to formally define what the tangent and cotangent means and really get some practice with them.*0010

*The definition of tangent is very simple, it's just tangent, by definition, the tangent of an angle is sin/cos.*0020

*You can only talk about that when the cosine of the angle is not 0.*0032

*If cosine of a particular angle is 0, we just say the tangent is undefined.*0039

*The cotangent is just the same thing except you flip them the other way up.*0043

*The cotangent of an angle is cosθ/sinθ.*0047

*If the sinθ happens to be 0, then we say the cotangent is undefined.*0053

*Now, the master formula that I've already mentioned before in a previous lecture for right triangles is SOH CAH TOA.*0061

*You can just remember the words SOH CAH TOA or there's a little mnemonic to help you remember that if you want.*0071

*It's Some Old Horse Caught Another Horse Taking Oats Away.*0076

*If that's easier for you to remember than SOH CAH TOA, then by all means remember that.*0082

*The key point is that this tells you, the SOH CAH TOA formula, tells you how to interpret the sine and cosine and tangent of an angle in terms of the lengths of sides of a right triangle.*0086

*I've drawn a θ down here in one of the angles of a right triangle.*0105

*Then we talked about the side adjacent to θ and the side opposite to θ, then the hypotenuse of the right triangle.*0109

*The master formula that you want to remember is that the sinθ is given by the opposite side over the hypotenuse, the cosθ by the adjacent side over the hypotenuse.*0128

*The tanθ, which we're learning about today, is the opposite side over the adjacent side.*0140

*Let me run over the common values of the tangent and the cotangent function.*0149

*The way to remember these is to remember the common values of the sine and cosine because remember tangent is just sin/cos and cotangent is just cos/sin.*0156

*If you can remember the common values of sine and cosine, you can always work out the common values of tangent and cotangent.*0165

*I've listed the particular, the common values in the first quadrant here, in degrees, we have 0, 30, 45, 60, 90 and those correspond to radians of 0, π/6, π/4, π/3 and π/2.*0173

*You should really have the sines and cosines of these angles memorized to be able to reproduce them very quickly.*0190

*If you can, then you can immediately figure out the tangent and cotangent without really having to memorize anything extra.*0199

*For 0 degrees, the cosine is 1 and the sine is 0.*0205

*Tangent is sin/cos, that's 0/1, that gives you 0.*0212

*Cotangent is cos/sin, which would give you a division by 0, that's why we say it's undefined.*0216

*For 30 degrees, the same as π/6 radians, the cosine and sine are root 3 over 2, and 1/2.*0225

*Tangent, if you divide those together, you get 1 over 3.*0234

*If you rationalize that, that rationalizes into root 3 over 3, so that's the tangent of π/6.*0244

*Cotangent, if you divide them the other way, cos/sin, you just get root 3.*0248

*For 45, the cosine and the sine are both root 2 over 2, so the tangent and cotangent are both 1.*0254

*For 60 degrees or π/3, you just get the opposite you had for 30 degrees or π/6, the tangent turns out to be root 3 and the cotangent turns out to be root 3 over 3.*0263

*For 90 degrees or π/2, the cosine and sine are the opposite of what they were for 0 degrees, and so the tangent is undefined and the cotangent is 0.*0276

*The values that you get for tangent and cotangent, there's 0, 1, root 3 over 3, and root 3, and undefined.*0288

*Those are the common values you get for tangent and cotangent.*0305

*When you have one of these common angles, one of these multiples of π/6 or π/4, in degrees, that's 30 degrees or 45 degrees, it's really.*0311

*When you want to know the tangent or cotangent, you know it's going to be one of these common values and it's just a matter of figuring out which one.*0323

*It's good to remember that these common values come up over and over again because when you're working out values for any particular angle you expect it to be one of these common values.*0333

*There's a couple more facts that I want to talk about, uses for the tangent function.*0347

*One is that the slope of the line is the tangent of the angle that the line makes with the x-axis.*0354

*Let me try to draw that.*0360

*Suppose you have just a random line in a plane like this, what I'm going to do is look at the angle θ that the line makes with the x-axis.*0366

*Now, I'm going to move this line over.*0379

*I'll draw it in blue, the translated version over to the origin.*0382

*That's meant to be a line with the same slope that just moved over to the origin.*0392

*If I make a triangle like this, we still have θ in one corner of the triangle.*0400

*The adjacent side is equal to the amount that the line is running over, so that's the run.*0411

*The opposite side is equal to the rise that the line makes in that triangle.*0421

*If we remember SOH CAH TOA, tanθ is equal to the opposite side over the adjacent side.*0432

*That's the TOA part of SOH CAH TOA, which is equal to the rise over the run, which is equal to the slope of a line.*0443

*That tells you that for any line, the tangent of this angle θ that it makes with the x-axis is equal to the slope of that line.*0455

*Another thing that you want to remember about sines and cosines is that once you know the common values of tangent and cotangent.*0470

*If you remember which quadrants the sine and cosine are positive and negative, you can figure out which quadrants the tangent and cotangent are positive and negative.*0480

*Let me label the quadrants here, 1, 2, 3, and 4.*0491

*Let me label the positive ones.*0500

*Remember sine corresponds to the y value, so sine is positive in the top two quadrants, negative in the bottom two.*0502

*Cosine corresponds to x values, so cosine is positive in the first quadrant and in the fourth quadrant, and negative in the other two quadrants.*0510

*If you want to figure out when if tangent is positive, it's positive either when sine or cosine are both positive or both negative.*0520

*Remember, the tangent is equal to the sin/cos.*0529

*Tangent is going to be positive in the first quadrant, negative in the second quadrant because sine is positive, cosine is negative.*0535

*Tangent is positive in the third quadrant, because both sine and cosine are negative, and tangent is negative in the fourth quadrant because cosine is positive but sine is negative.*0544

*If you put this together, tangent is positive in the first and third quadrants, negative in the second and fourth, which is kind of the origin of this mnemonics that you can use to remember, ASTC.*0557

*The way you remember that is All Students Take Calculus, that stands for all the sine, cosine, tangent functions are positive in the first quadrant.*0573

*In the second quadrant, only sine is, in the third quadrant, only tangent is, and in the fourth quadrant, only cosine is.*0585

*Now, you might wonder how does cotangent fit in to all of this.*0594

*Cotangent, remember, is just the flip, the reciprocal of tangent, tangent was sin/cos, cotangent is cos/sin.*0597

*Cotangent is going to be positive whenever tangent is positive.*0611

*It'll be positive in the first quadrant and in the third quadrant.*0614

*Now we know our common values of tangent and cotangent.*0622

*Remember that 0, 1, square root of 3, and root 3 over 3, and we know which quadrants is going to be positive or negative, we can work out the tangent and cotangent of any common angle around the unit circle.*0625

*Let's try some problems now.*0642

*Let's start out by drawing graphs of the tangent and cotangent functions, and we're going to label the zeros and asymptotes of each and try to figure out what the periods are.*0645

*I'm going to start with the tangent function, and I'll do that one in blue.*0662

*For the tangent function, I know that the tangent of 0 is 0.*0682

*I'm going to put a dot at 0.*0687

*Let me mark my axis a bit, π/2, π, 3π/2, 2π.*0690

*Tangent starts out at 0.*0703

*The tanπ/4 we said is 1, so let me put a dot there.*0710

*When tangent gets close to π/2, remember that's really getting close to dividing by zero so it goes up to positive infinity there.*0717

*Tangent has an asymptote at π/2 and it goes up to positive infinity there, so it looks like that.*0730

*In the second quadrant, tangent is negative, at near π/2 it has an asymptote going down to negative infinity.*0745

*The tangent of π is 0 because it's sin/cos.*0757

*As tangent approaches 3π/2, it has another asymptote.*0764

*That's what the graph of tan(x) looks like.*0790

*That should be in blue because I'm drawing my tangent graph in blue.*0799

*tan(x) has zeros at 0, π, 2π, and so on.*0805

*It has asymptotes, basically, at the places where you're trying to divide by zero, and those are all the places where the cosine is zero.*0823

*Remember tangent is sin/cos.*0835

*Those are at π/2, 3π/2 and so on.*0839

*That's what the tangent function looks like.*0849

*Let me try to graph the cotangent function.*0853

*Now, the cotangent function, remember that's cos/sin, that has asymptotes and zeros exactly the opposite of where tangent did.*0859

*It has zeros wherever cosine is 0, which is π/2, 3π/2 and so on, because cosine is 0 at those places, 3π/2, 5π/2.*0885

*It has asymptotes wherever sine has 0, because we're trying to divide by 0, and that's 0, π, 2π, and so on.*0907

*Cotangent looks kind of just the opposite of tangent, like that.*0922

*That's the cotangent function in red.*0965

*For the next example, we're going to look at right angle triangles, and try to work out some trigonometry there.*0972

*We're given that a right a triangle has short sides of length 3 and 4.*0977

*We're trying to find the tangents and cotangents of all the angles in the triangle.*0989

*The first thing we need to do is figure out what the hypotenuse of this triangle is.*0995

*We know that the hypotenuse squared is 3 ^{2}+4^{2}, which is 9+16, which is 25.*1000

*The hypotenuse must be 5 units long.*1010

*Let me figure out first of all the tangents of all the angles in the triangle and we're going to use the SOH CAH TOA.*1019

*Remember, tangent equals opposite over adjacent.*1023

*The tanθ is the opposite over the adjacent which is 3/4.*1030

*The tanφ is the opposite side of φ which is 4/3.*1040

*The last angle in the triangle is 90 degrees, that's the tangent of π/2 radians, remember the tangent is undefined for π/2, so I'm just going to leave that as undefined.*1050

*That's because tangent is sin/cos, and cos of π/2 is 0.*1069

*Now, cotangent of each of these angles, cotangent is just the opposite of tangent in the sense that when tan is sin/cos, cot is cos/sin.*1076

*The cotθ, instead of being 3/4, will be 4/3.*1089

*The cotφ, instead of being 4/3, will be 3/4.*1097

*The cot(π/2) which is the cos/sin, that's 0/1, is just 0.*1106

*The secret to working out tangents and cotangents of angles in right triangles, is just to remember the SOH CAH TOA formula.*1117

*In particular, the TOA part, says tangent equals opposite over adjacent.*1125

*That quickly helps you figure out the tangents of angles and right triangles.*1134

*For our next example, we are asked to determine if the tangent and cotangent functions are odd, even, or neither.*1138

*Let's recall the definitions of odd and even functions.*1144

*Odd function is where f(-x)=-f(x).*1149

*Let me draw that a little more clearly, f(-x)=-f(x).*1160

*It's the kind of function that has rotational symmetry around the origin when you look at the graph.*1169

*An even function satisfies f(-x)=f(x) with no negative sign and that has mirror symmetry across the y-axis.*1185

*Those are the two definitions we want to check here.*1209

*Let's try that out for tangent and cotangent.*1213

*Tan(-x), let's see what happens.*1215

*Tangent by definition is sin/cos, so that's sin(-x)/cos(-x).*1222

*Sine remember is an odd function, sin(-x) is -sin(x), cosine is an even function so cos(-x)=cos(x).*1226

*You put this together, you get -sin/cos, you get -tan(x).*1241

*In particular, that tells you that tangent is odd.*1248

*We can check with the graph of tangent, it's probably worth memorizing what the graph of tangent looks like.*1261

*The graph of tangent look like this, it crosses the x-axis right at the origin, and then it has asymptotes at -π/2 and π/2.*1271

*In particular, the graph of tangent has rotational symmetry around the origin.*1283

*If you spun it around 180 degrees, it would look exactly the same.*1290

*It has that rotational symmetry around the origin, the graph of tangent definitely confirms that tangent is an odd function.*1295

*Let's check that out for cotangent.*1304

*Cot(-x), cotangent is just cos/sin, that's cos(-x)/sin(-x).*1305

*Just like before cos(-x)=cos(x), because cosine is an even function, sin(-x)=-sin(x), so what we get is -cos/sin, -cot(x).*1318

*Cotangent is odd also, it's also an odd function.*1336

*Again, we can check that looking at the graph.*1346

*Let me draw a quick graph of cot(x), probably worth remembering but not quite as essential as the tangent function.*1352

*If you're going to remember one of them, remember the graph of the tangent function.*1360

*The cotangent function looks like this, it crosses 0 at π/2, and it has asymptotes at 0 and π.*1366

*The cotangent function, if you draw a dot on the origin, and then rotate the thing, 180 degrees.*1384

*If you rotate the thing 180 degrees around the origin, the graph would look the same.*1396

*It has this mirror symmetry, sorry, not mirror symmetry but rotational symmetry around the origin, so the cotangent function is an odd function.*1402

*It does not have mirror symmetry because if you flipped it across a mirror on the y-axis, it would not look the same so it's not an even function.*1415

*But it does have a rotational symmetry around the origin.*1425

*We got another couple of examples coming up later.*1429

*Why don't you try them out and then we'll work them out together.*1432

*Okay, another example of cotangents and tangents, we're given some common values here 5π/6, 5π/4 and 5π/3.*1437

*Those are all angles in the unit circle.*1447

*We want to figure out what the tangents and the cotangents are.*1451

*A good way to start this is to draw the unit circle and to figure out where those angles are.*1458

*There's my unit circle, and let me label some easy common values, there's 0, π/2, π, 3π/2 and 2π.*1478

*The angles we're being asked to find, 5π/6 is just a little bit short of π, 5π/4 is in between π and 3π/2, and 5π/3 is 5/6 away around the unit circle, it's down there.*1490

*Those are the three angles we're being asked to find.*1514

*It's probably easiest if we just write down the sines and cosines of those angles, because presumably we know those pretty well.*1518

*Then we can figure out the tangent and the cotangent.*1522

*Let's make a little chart.*1527

*We'll find the angle, I'll find the sine and the cosine, then we'll find the tangent just by dividing those together, and the cotangent.*1530

*If we start with 5π/6, that's this angle right here.*1547

*Remember the common values here.*1554

*The sine is the small one, that's 1/2, it's positive.*1558

*Cosine is the bigger one, square root of 3 over 2, and that's negative because the x value is negative there.*1560

*The tangent is sin/cos, so that's 1 over the square root of 3, which rationalizes to square of 3 over 3, and it's negative because the cosine was negative.*1568

*The cotangent is just cos/sin, so that's root 3, that's also negative.*1580

*Now, 5π/4, that's a 45 degree angle, South of the x-axis*1588

*The sine and the cosine, because it's 45 degrees, they're both root 2 over 2, but we've got to figure out whether they're positive or negative.*1599

*Since it's in the third quadrant there, they're both negative.*1608

*That means when you divide them together to get the tangent, you get a positive one.*1613

*When you check the cotangent, you also get a positive one.*1616

*Now 5π/3, that's another 30-60-90 degree angle down here.*1620

*The y value is the big one this time, so it's root 3 over 2, cosine is 1/2.*1629

*The sine is negative because we're still south of the x-axis, but the cosine is positive because we're to the right of the y-axis, meaning the x value is positive.*1634

*Tangent is sin/cos, that's root 3 and that's negative.*1648

*The cotangent is cos/sin, that's 1 over root 3, that rationalizes to root 3 over 3, that's also negative.*1655

*Those are our answers right there.*1671

*Let me just emphasize that the key to figuring this problem out is to remember the sines and cosines using the common values and those common 30-60-90 triangles and the 45-45-90 triangle.*1675

*As long as you remember those common values, then you can work out the sines and cosines, figure which ones are positive and which ones are negative, then divide them together to get the tangents and cotangents.*1687

*Of course it helps that you remember that the common values for tangent and cotangent, their always going to be 1, square root of 3, and square root of 3 over 3.*1699

*Then, it's just a matter of figuring out whether they're positive or whether they're negative, and which is which, for any given angle.*1710

1 answer

Last reply by: Dr. William Murray

Tue Aug 5, 2014 3:47 PM

Post by Jason Wilson on July 23, 2014

In example 4, TAN = SIN/COS, is that 1/2 over negative sqrt3 / 2 which evaluates to "- sqrt3 over 4" ?? you said one over sqrt3 that evaluates to negative sqrt3/3 please help thx

1 answer

Last reply by: Dr. William Murray

Fri Jul 5, 2013 9:52 AM

Post by Matthew Chantry on June 30, 2013

The practice question for this is a secant/cosecant question, which hasn't been covered yet. This may be an error.

Thanks.

1 answer

Last reply by: Dr. William Murray

Fri Jun 21, 2013 6:16 PM

Post by HAFSA Ahmad on June 12, 2013

I still don't understand how to figure out the angle in radians like in this example 5pi/6 , 5pi/4 etc.

2 answers

Last reply by: Dr. William Murray

Tue Jan 1, 2013 11:34 AM

Post by Jorge Sardinas on December 31, 2012

how did you turn sin[-x]/cos[-x into -sin x/cos x ; ->

1 answer

Last reply by: Dr. William Murray

Tue Jan 1, 2013 11:25 AM

Post by Ahmed Shiran on April 18, 2011

This is interesting lecture ! thanks :-)