# Trigonometric Functions of Any Angle

BAM!!! Mr. Tarrou. Let’s start talking about

setting up trig functions for angles even if we don’t know them. Unit circle, we have

got the sixteen angles…we know all those angles. What about setting up trig functions

for angles that we don’t know such as this question? Given a point

on the terminal side

of negative two, negative five, set up all

six trigonometric functions. Ok, so let’s do that. We are talking about angles in standard

position so we are going to be using our sine of theta is y over r, cosine of theta is x

over r, and so on. And we are going to draw that angle in standard position. There we

go. I know that negative two, negative five is in quadrant three, the sketch is just going

to allow me to have a place to put the numbers. I am not going to bother trying to get these

drawn to scale. So negative x and negative y, that is going to be in quadrant three.

So I am going to draw an angle that is quadrant three. Now I am going to set up the reference

triangle and uh…that will allow us to work out our six trigonometric functions. We have

a point that is on the terminal side of negative two negative five. I am also going to put

those values along the x and the y movement just to make sure that I do not forget those

signs when I go to set up my six trig functions. Now I have a right triangle and it is a reference

triangle because it is agains the x axis. It is also marking off the number of degrees

between the terminal side and the x axis. So a little pythagorean theorem work here

and we got two squared plus five squared. I am not worrying about the negatives because

there really is no such thing as a negative distance and the negatives are going to get

squared away anyway. This all equals r squared. That is going to be four plus twenty-five

equals r squared. So r is going to end up being the square root of twenty-nine. We will

not be happy with a square root in the denominator so we will need to rationalize some of our

final answers. Well the sine of theta is y over r, so it is negative five over the square

root of twenty-nine. We will need to rationalize that by multiplying the top and bottom by

the square root of twenty-nine. And we get negative five times the square root of twenty-nine

over twenty-nine. The cosine of theta? The cosine of theta is x over y, so it is negative

two over the square root of twenty-nine. When you rationalize that you get negative two

times the square root of twenty-nine over twenty-nine. Remember that sine is y over

r and cosine is x over r. In quadrant three those x any y values are negative so make

sure these answers are both negative as well. Tangent of theta is y over x, so it will be

negative five over negative two, or five halves. And then the reciprocal of sine is cosecant.

The cosecant of theta is going to be this value flipped so it is going to be negative

square root of twenty-nine over five. We will not have to rationalize that. The reciprocal

of cosine is secant. That flipped is going to be the square root of twenty-nine over

two and it will be negative of course because we are still in quadrant three. The cotangent,

well that is going to be two-fifths. So we have evaluated all six trig functions in exact

form, there are no rounded decimals anywhere. We have evaluated and found the exact value

of six different trig functions without ever actually knowing what this rotation of theta

is. We just had to know a point on the terminal side of the standard position angle. Pretty

cool uh? Alright… BAM! Moving on:) Let’s do another example. Hope you are enjoying

these videos. If you are please spread the word about my now budding math channel. I

am really enjoying making these lessons. What if I just told you that the cosine of some

particular angle is less than zero and the cosecant of that same particular angle is

greater than zero. Could you tell me from just this little bit of information, now we

are not going to set up the six trig functions because I know nothing about the sides of

the reference triangles, all I know is that cosine is negative and cosecant is positive….

From just this information could you tell me what quadrant the angle actually is in?

We are trying to evaluate and work with trig function without the aid of a calculator telling

use what the actual measure of the angle is. Well, where is cosine going to be negative?

You put angle measures into a trig function and out of that trig function one more time

you get the sides of a right triangle. So the cosine of theta is x over r. Cosecant,

that trig function will have an angle put into it, and it will also give us the sides

of a triangle. Cosecant is the reciprocal of sine so it will give us r over y. Now when

we talk about angles in standard position it is impossible for the radius of a circle

to be negative. So we are only concerned about the sign changes of x and y. Well… So where

are the x coordinates negative? The x coordinates are negative in quadrants two and three. For

cosecant to be positive, the y values must be positive. In what quadrants are the y values

positive? Quadrants one and two. So if you have an angles whose cosine is negative and

the cosecant of that same angle is positive, then where is the overlap? The overlap is

going to answer the question, “What quadrant is theta in?” It looks like theta is in quadrant

two. So we are not evaluating a trig function necessarily, but are using their values to

identity this unknown angle… or at least the quadrant that this unknown angle terminates

in. So that is how you use trig functions and their signs. Go to the x and y ratios

and use them to evaluate what quadrant your angle is in. ALRIGHTY THEN! Now I kind of

did this in another video. I want go from cosine of theta, given that values and then

just find all the other five trig functions. I am going to say that the cosine of theta

is equal to negative three-fifths and that angle theta is in quadrant two. Well, when

I am given cosine I am given an x and an r value. So, I am given two sides of the reference

triangle and it is in quadrant two. So theta looks like. There is a rough sketch of an

angle in quadrant two. Our reference triangle again drawing against the x axis is going

to be…. Always draw your reference angles or reference triangles, the same thing, against

the x axis. Cosine is x over r, so negative three and five. Again just putting that sign

in there as a place holder so I don’t forget it later. Pythagorean Theorem three squared

plus something squared equals five squared. Well three squared plus, I am again ignoring

the negative because there is no such thing as a negative distance…that is a directional

sign, y squared equals five squared. When you get done solving that you are going to

find out the y coordinate is four. Now that we have the three sides of our reference triangle,

I can set any trig function that I like and for time purposes I am only going to do one

of these. The sine of theta is equal to y, which is four, over r which is five. And in

quadrant two our y values are positive, so thus our sine function had better be positive

when I am done setting it up. The tangent function in quadrant two, the x’s are negative

and the y’s are positive so the tangent would be negative. Let me just write that down.

The tangent of theta is negative three-fourths…NO IT IS NOT!!! The tangent of theta is negative

four-thirds. Ok In quadrant two your x’s and y’s have different signs so your tangent which

is y over x should have a negative sign. So that is how you set up these little right

triangle drawings and show the rotation for all these trig function problems. That is

what we are studying, right triangles and their rotation about the origin, or a circle

if I were drawing a unit circle. ALL RIGHTY THEN!!! Are you still awake? Are you with

me? Let’s go to the next problem. The tangent of theta, forgot to write that in my notes,

is equal to five-twelves. I am not going to blatantly tell you what quadrant theta is

in, but I will say that the sine of theta is negative. My tangent ratio which is y over

x is positive. The only way you can divide two values that can have different signs,

the only way you can divide y and x and get a positive answer, is if either both are positive

which happens in quadrant one of if they are both negative. Negative five divided by negative

twelve would give you a positive five-twelves. This angle might be in quadrant three. Now

that I am told the sine of theta is negative, and the sine of theta being y over r, where

are my y values negative? Y values are negative in quadrants three and four…they are positive

in two. Well…HEY! There is my overlap so that is how I am going to draw my angle and

reference triangle, in quadrant three. So just a rough sketch of an angle in standard

form in quadrant three. A quick rough sketch of a reference triangle again against the

x axis. Reference angles is an acute angle and it is the number of degrees or radians

between the terminal side and the x axis, so here is your reference triangle. Tangent

is y over x, so y over x. Now I am in quadrant three so both the x movement and the y movement

are negative. Now the Pythagorean Theorem, five squared plus twelve squared equals 169.

The hypotenuse will be thirteen after you take the square root and finish working through

the Pythagorean Theorem. A squared plus b squared equals c squared. Now I have the three

sides and I can set up any trig function I like. Let’s just do one of these. I want to

get in one more example before the end of my time period. Let’s say I want the sine

of theta, which is y over r. The cosine of theta is x over r. And let’s just again hit

this idea of a reference angle. What if I did only rotate ten degrees? What would be

the reference angle? It would be ten degrees. What if I rotated 170 degrees? How many degrees

away from the x axis am I? I am ten degrees away from the x axis so the reference angle

is going to be ten degrees. What if I rotated negative twenty degrees? What would the reference

angle be? Well it is acute, it is the number of degrees between the terminal side and the

x axis, so the reference angle would be twenty degrees. Almost a matching number but a sign

change. Make sure that you remember that a reference angle is just the number of degrees

between the terminal side and the x axis. Sometimes you compare it to zero degrees,

sometimes you compare it to 180 degrees, sometimes you might calculate it from 360 degrees. If

you rotate 340 degrees, your close to the x axis… I am running out of time so I am

getting a little flustered. If you rotate 340 degrees your reference angle will be 360-340=

20 degrees. Those ideas of reference angles and their importance is all over the unit

circle. I am Mr. Tarrou. BAM!!! Thank you for watching. Go Do Your Homework!

i just love this video. it made things so much more understandable. can't get anything out of my teacher, so thanks for this!

I'm a bit confused about the reference angle. I know for example sin 150 = (180-30) = sin 30 = 1/2 But for my question, how would you solve for cos 190? is it cos 10?

Thanks for the upload. I appreciate the help as my prof. is quiet and german.

Loved your video ! Come teach in the Bahamas

Thanks, gonna pass trig b/c of your videos

What do you get if you mix Bruce Springstein, George Micheal, and a super nerd? Prof Rob Bob!

I love yer videos!

Excellent

Hahaha

thanks for the help 🙂

How I am spending my Saturday afternoon…. BAM!!! Yes I am typing Closed Captions.

i got an exam tommorow and this vid is helping me a lot… you sir deserve a sub.

This video could be one of the best pre-calc i hve ever watched and it really solved my problem with the trig Angles…

Keep it up ProfRobBob

You are great! THANK YOU!!!!!

Thank you for all of your help!

My finals for precalc are coming up and I've been watching your videos to study better. I'd just like to say that they are very helpful and you are such a wonderful teacher, thank you for uploading these videos! On that note, I was watching this video before going to sleep and right when my eyes were closing, I heard your voice go "Are you still awake?! Are you with me?!" That was the most perfect timing ever, it did wake me up lol once again, thank you so much!

Thank you so much for these videos. I have a math test tomorrow; your videos are a great way to study. I saw a bearing video of yours. Could you make more bearing videos with different types of questions? They give me a hard time.

Thanks

when you use the term csc do you mean cos^-1. I am learning mathematics in UK I don't think we use the term csc?

Another great video!

I'm very confused about negative angles, negative ratios, and the different quadrants. I have watched this video several times, but I still don't understand… One thing that should help is if someone could tell me the ratios (in terms of x, y, and r)? Thank you in advance!

how do you know where the angle goes? Like, for the first example you put it at the opposite of -5. Why dont you put it at the bottom to where its the opposite of the -2?

This helped me SO much!! Thank you!!!

This is so helpful and I love the shirt

Super..Man

Sir .. Thanks very much for your time and effort .. You are really helping millions around the world .. I hope you continue doing this great work 🌹

I have a question ??

What is the origin of the trig functions how did we calculate them .. For examble If I am a computer programmer and i wanna make a programme to calculate these functions .. What is the algorithm I should use ???

I appreciate your time and effort and thank you 😄

Can u tell how to find trigonometric functions of angle 225 degrees plzzz

Wow!!! Thank you for this video Prof. it help me catch up to class…cause lately i'm absent in class for one week… and i study at home and watching video like these.thank you it helps me very much..

Still makes absolutely no sense. Trig is soo stupid

I enjoy this video but still learning 🙂

went back to school at 31 to be a software engineer and trig is kicking my butt, your videos are a huge help though and hopefully i can pull it off this semester and move on to calc. thanks for the help!

how can I calculate that the value of sin30 is 1/2?

thank you

BAM!! i found new idea here in this this. yeah!!

Choose the point on the terminal side of -45°.

I need help with this problem Asap. I have no idea how to solve it. PLEASE HELP.

Well done, thanks.

+ProfRobBob you have no idea how useful these videos were to me. I have a final tomorrow and I've searched a while to find a video that makes he material as comprehensive and interesting as yours. Again, thank you. Keep up the good work!

his energy and bams kept me awake ty

Love how enthusiastic you are. It makes my online class much easier and entertaining when I use your videos. Thanks man 🙂

This video probably saved my life. Thank you so much.

Also, awesome handwriting! Finally, a fellow cursive-ist. 🙂

Thank you so much! I have spent the entire unit so confused and my test is in Two days and now I am finally starting to understand it because of you.

I'm still confused on how to use it, but then again I skipped precal for IB math SL…

This is so amazing, thank you.

What about angles greater than 360 degrees?

bam! You're blocking my view😄

Robert you bad.

SUPERMAN IS IN THE HOUSE 🙂

I cant thank you enough for all your videos.

Just awesome videos!!!!

Thank you! Good teaching video!

pythagorean theorem hha

I think this is gonna save me from an impending test my teacher didn't teach anything on, thank you so much

Omg thank you sooooo much 🙂

thank u sooo much

This is awesome! I watched this video 15mins before a quiz and boom! Got an A for my quiz! Thanks!

Biggest meme at 00:4

Thank you so much .

Hello ProfRobBob i dont know if you read comments anymore but i literally cried on the way out of my precal class today and just this video taught me everything we learned in class that i didnt get. I truly appreciate u. Thank you

I was feeling really bummed out that I wasn't understanding the material in class. I was afraid I wasn't smart enough… but this made it clear to me! Thank you!

BAM subbed!! i love ur energy, wish i could have u for a professor . thanks 4 the vid!

That entire one second of the beginning of this video was spent getting ready to jump into the video.

BAM!!! i understand now.. thanks prof.. im studying online. it really helps me a lot.

Hi, i am 40 years trying to understand trig to help my daughter in maths who is 13. It proved to be helpful. Thanks.

What an incredible feeling it is to hear the explosion of a "BAM" coming from MY mouth as I'm studying for tomorrow's test. Thank you so much brother! So grateful for you. You've helped a confused student in south Florida really understand what he's doing. Keep it up!!! I'll look forward to you in my future classes:)

what? Can anyone explain at the first example he said cos is x/r on the second example he said y/x.

twenty-naaine

you r awesome

You're amazing🤗🤗🤗🤗You have no idea how your lessons are helping me right now!!!)))

just one word- Awesome 😎

This honestly makes so much sense now. I appreciate it

god bless. most of the vids out there are useless as they start of with the basic stuff which i know.

THIS EXACT QUESTION CAME IN MY ENTRANCE EXAM (but different numbers of course) THANK YOUUUUUUUUUU!!!!!

This video helped me a lot thank you sir!! 🙂 and yes keep up the energy you are gr8 I will subscribe you sir and share this among my college teachers and friends…it'll be very helpful..honestly speaking u r simply gr8!

Can you give your email sir?

i like how he randomly switched to a near british accent and then seamlessly back XD 3:03

these are so helpful i wish i would of found your videos a long time ago

This video was extremely helpful! I've learned more in these 15 minutes than I have in my half semester of my

boringpre-cal class. And the closed captions help SO MUCH!!! Thank you times a million– already subbed 🙂Have a quiz coming up for Pre-cal and I've had a hard time grasping this chapter, these videos are helping so much. Great job, very entertaining and brain stimulating. You have a new subscriber!

I love his handwriting!

God bless you

mentally enriching, thank you

First of all THIS HELPED ME SO MUCH, thank you!!!! Secondly, thank you for completely erasing the board, instead of leaving bits of writing there like all other teachers seem to do, and thank you for having wonderful, legible handwriting.

i like your outfit

You've helped me soo much, thank u <3 <3

this helped so much and you have really nice handwriting

Professor Rob this was extremely helpful. Thanks! Where is that accent from?

he's superman . only superman wears a tucked in shirt and a belt. lol

Good video

Hindi me btaiye sir ji

its amazin when u can teach these lessons under 15mins. U helped me alot. Thank u for makin these video lol.

Wouldnt the angle theta equal to 180 – arcsin(4/5)? The theta angle that is being calculated is the in the triangle between the -3 and 5, no?

ALL RIGHTY THEN

Still clutch 8 years later

I am finally entertained by math

Thank God your here, my math teacher wasn't at school today, and my EOC is tommorow!

Someone get this man a Nobel Peace Prize.

8 years later and still very helpful, makes math fun😎

thank u

AMAZING!

Nobody:

Mr. Tarrou:

(5 mins in) I'm running out of time here guys! LOL

Thank you for everything!

THANK YOU SO MUCH! I am taking College Trig online, and the modules consist of a powerpoint and video clips from Pearson. There's very little explanation, so when something doesn't click I get really stuck. This helped me make sense of things.

Saved my life. Thanks!! Great vid.

Looks like andrei in pbb