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in this episode we’ll extend the definitions of the six trigonometric

functions we’ll also work some examples explaining how to find the reference

angle and how to use it to evaluate the six trigonometric functions let’s do

some math all right everyone welcome back um in this we’re going to begin by

looking back at what we did in the previous episodes uh on trigonometric is

trigonometry is fun so we looked at the radians and degrees

trigonometric angles we talked about the unit circle

we talked about all the special angles on the unit circle

and we and we also talked about reference angles in this episode right here

and in the previous one and the very last one we talked about the

trigonometric angles of an acute angle so i recommend checking those three

episodes out before getting to this one right here

all right and so the question we’re going to start with on this episode is

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how do we extend the trig functions to a real variable in other words how can we

input um you know into our trig functions more

than just an acute angle like 30 degrees how do we find for example sine of you

know 702 degrees or how would you find sine of 280 pi over three

all right and so that’s the question that we’re going to start off with is

and now to do this here we’re going to make a unit circle right here let’s draw

unit circle and let’s say this right here is the radius is one

and or in fact let’s just put it over here

let’s extend this out and let’s put it right here let’s just put a tick mark

right there one and i’m going to extend this out here

also and this up and this down and i’m putting this on the

in the cartesian plane there unit circle and let me just chop it off about right

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there and so yeah we did a whole episode uh on the unit circle and

so let me just point that uh series out here trigonometry is fun

step-by-step tutorials for beginners a link is below in the description

um in this episode here what we want to do now is to let me get a red

to take an angle let’s say this angle right here

and we’re going to get this terminal side of the angle

we get this point right here so let’s call this here angle t

and this point right here we’re going to call it p of x y

and the reason why is because it’s on the x y axis right so it’s a point it’s

got an x y coordinate to it and we want to find and figure out what

are the six trigonometric functions for this angle here t

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um and so what we’re going to do is we’re going to

you know t is a real number let me just write that down here t is a real number

so we’re thinking about t as in a radian an angle measured in radians

and this right here p of x y is a point on the terminal side so point on

the t on the unit circle there’s two things to pay attention to

there uh on the unit circle determined by t

so that’s how they’re related to each other

and now we have these six trigonometric functions

so i’ll start putting down here so sine of t is is the y right there

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and then cosine of t is the x right there and tangent of t

as we saw in our previous episode is the ratio y over x

of course we have to assume here in order for this to be defined is that x

is not zero i’ll put that in parentheses and then we have the reciprocals

cosecant of t is defined as one over y assuming y is not zero

and the secant of t is reciprocal so one over x assuming x is not zero

and then the cotangent of t is defined as the reciprocal here of

tangent which will be x over y and this is assuming that y is not zero

there we go the this is the way that we’re going to

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define the six trigonometric functions of an angle t

so t is our angle and now i drew it in quadrant two but we can do this for any

quadrant right here and so if we have 30 degrees then we can use this right here

or if we have we could go around multiple times 360 360 360 and then we

could go more in any case the six trigonometric functions are defined

by looking at this point here and corresponding this point here we

have a right triangle right here i’ll just dash that in and draw the

right triangle right here so we got this little x here and this little y here

that gives us the point here x y and that’s why we can make this

definition here and that’s why this definition extends the definition that

we had when t was an acute angle when t is an acute angle then we get a triangle

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over here so i’ll just draw t over here now in quadrant one

and this is the point here p of x y and we have this right triangle right here

and this is the x and this is the y and in the previous episode we called

this right here uh angle theta and we said this was going to be opposite over

hypotenuse for the sine and so it turns out to be the same exact definition

when the t is in quadrant one but this definition right here allows us to use

any quadrant we want all right so um we’re going to talk about reference

angles now the reason why is because we’re going to be able to um

work with very large angles and we’ll be able to

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kick back to our reference angles and have as we as i mentioned before

we’ll be able to know a very small number of reference angles like 30 45

and 90 degrees but we’ll be able to extend our knowledge base using our

reference angles let me show you how to do that now so reference angles

now i talked about reference angles in a previous episode but i want to kind of

uh cover this again so reference angles are always positive

now i’m going to abbreviate reference angles with r a so our a reference

angles are always positive so positive and they’re between 90 and 0.

so between 0 and 90 degrees and the reference angle is always found with

reference angles uh with respect to i like to abbreviate with respect to

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also with respect to the x-axis so when we say reference angle we’re

we’re referring to the x-axis um and yeah so let’s look at some

examples uh let’s just put this down here reference angles and then

always with respect to the x-axis and let’s look at our first example

right here 218 degrees so what’s the reference angle for 218 degrees

all right so let’s just sketch a graph up here real quick what’s 218 degrees

look like um so it’s more than 180 but it’s not quite 270.

so let’s just put it right there so let’s say this is 218 degrees right here

and so what is the reference angle going to be so let’s put the reference angle

in red so this is the reference angle right here

and we’re going to draw this triangle right here

there’s the reference angle sitting right there inside of it

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and so what is this referencing over here so it’s going to be so this angle

right here 218 is 180 plus a little bit more and that little bit more

is not quite reaching the 270 so it’s going to be 270

minus 218 and that will give us the reference angle so 270 minus actually

the reference angle is going to be 218 sorry 218 and then take away the first

180 degrees and that gives us what’s left over

so 218 and then minus this part right here and so that gives us what 38 degrees

so that’s the reference angle so this so the reference angle right here is

38 degrees so 218 but the reference angle is 38 degrees

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and we’ll see how to use that in a couple of minutes

let’s go do one more example real quick what about the reference angle of 1387

so now we got to go around multiple times so i like to sketch a graph though

so i’m going to go around once that’s 360 go around it twice that’s 720. can

we go around more if we went around three times that would be 1080

and we still got more to go how much more to go do we have

so we’re looking at this right here we’re looking at 13 87 degrees

minus three times and i went around three head three times

so i went around 336 times and so what’s going to be left here right here

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so this is going to be 1387 minus 1080 and that’s 307. and so

you know we got to go 307 more right so that’s past the 270 but not quite

around to a 360. and so there’s the terminal side right there

and now we’re going to go reference the x-axis right here now

so when we reference the x-axis right here we’re looking at going right here

and what is that angle right there so to find this angle right here we’re

going to look at 360 degrees and then minus 307

so we’re going to to find this angle right here in red we’re going to do all

of it 360 and then we’re going to take away that last piece that last part we

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went around which was 307 and that gives us the remainder right there

so this is 53 degrees right here 53 degrees so the reference angle right

here is 53 degrees all right and so now let’s see what we

can do with these reference angles right here so now the question is going to

change for us and it’s going to be why do we need

reference angles so let’s look at that now all right so

let’s let’s look at a general situation right now let’s look at the different

quadrants so let’s say i’m going to draw a quad

let’s say quadratic one’s over here and so let me

put that over here so here’s an angle right here in quadrant one

here’s the angle right here let’s call the angle theta

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where’s the reference angle so the reference angle is the same angle

right so that’s quadrant one any angle in quadrant one here

is its own reference angle right here and so theta is theta prime

now what about quadrant two how do we find a quadrant two

um right so it’s not just for a quadrant one angle but any angle even if it went

around 360 degrees and then a little bit more

but it ended up in this quadrant then you could find the reference angle um

part of it right there but if it’s quadrant one angle

then you know that works right there um so what about quadrant two

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so quadrant two now are angles over here and now the reference angle is right

here so how would we find a quadrant two ref uh

angle how would we find the reference angle for quadrant two so now we would say

theta prime is 180 degrees minus the theta so here’s the theta right here

and here’s the theta prime right here and the way to find the reference angle

right here would be just to go all the way to 180

and then take away this part right here and that will give us the reference

angle right there so let’s put quadrant two in quadrant two [Music]

in quadrant one diagram in quadrant one and then let’s do a quadrant three

diagram let’s do that right here so now we have a quadrant three angle

let’s call it theta quadrant three and where’s the reference angle so the

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reference angle is right here so how would we find the reference angle here

theta prime so to for this right here theta prime will be equal to theta

right here all the way to theta and then take away the 180 degrees right so take

away that part right there and then we get what’s left over which is the theta

prime that’s the reference angle right there

so for quadrant three angle you’re going to take the angle

and then you’re going to subtract 180 degrees from it and you get the

reference angle right there and then for quadrant four here

so now let’s draw an angle in quadrant four so let’s call this here

uh quadrant four theta and where’s the reference angle

it’s right here straight straight to the x-axis

and then now how would we find the reference angle so theta prime here would be

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you know go the whole way go 360 and then take away the angle theta right here

and so that’s how we would find that reference angle right here theta prime

right there all right very good now let’s ask the question here this is

an important question here when is the sine function positive and

then we’re going to ask also the cosine function because remember

when we’re when we’re looking at the unit circle right here we’re looking at

the p of x y point so here here’s the p of x y point here’s

the p of x y point here it is over here it’s along the unit circle

and here it is down here it’s along the unit circle and so we’re going to define

remember we’re going to define the sine of t

or i’m using theta in these diagrams so let’s say theta

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sine of theta is just simply the y right so when i’m looking at the y here um

and so let’s let’s call this one here um let’s put it down here because we’re

asking for the sine function right here let’s put it right here actually so in

quadrant one the y value is positive in quadrant two the y value is still

positive in quadrant three the y value is negative

quadrant four the y value is negative because you come down to find the y value

so these this is where the sign so where’s the sign positive it’s going to

be in quadrant one and two and in quadrant three and four the sine

will be negative what about cosine of theta and that was given by remember by

definition the x so now i’m looking at the x so the x is

positive right here in quadrant one so quadrant one positive

quadrant two now the x is negative in quadrant three

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the x is still negative because i went to the left right and then quadrant four

x is positive right here so here’s the uh point x p of x y and so the x will be

positive right there i’ll have to go positive to get to that point right

there so here the quadrant four is positive so

here’s where sine is positive and here’s

where sine is negative in other words if your angle lies in this quadrant right

here the terminal side of your angle and then the sine has to be positive or

if the terminal side is down here in quadrant three

then we know that the sine is negative and the cosine is negative so

this is really important here this really helps guide us

and it helps us understand how to use reference angles to our advantage right

here and so we’re going to say we need reference angles to find

values for the six trigonometric functions and we can do so we can do

that by also looking at where sine and cosine are positive and where they’re

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negative so um let’s see some examples of doing this right here

i think it’ll become more clear when we see some examples so here we go

we’re going to look at this example here 210 degrees

so let me sketch that right here and let’s go right here

210 degrees so where is 210 degrees well it’s going to be past 180 but not all

the way to 270. so let’s say 210 is about right here so there’s 210 degrees

what quadrant is it in it’s in quadrant three

we know that sine and cosine are both negative in this quadrant

so what we’re going to do is we’re going to form this triangle right here and

the reference angle right here i’ll put in red

how much is that reference angle right so it’s going to be 210

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take away the 180 and what we get left over is 30 degrees this is 30 degrees

angle right there and so this is a right angle this is 30

degrees so that means this is 60 degrees right here so this is 60 degrees

and so now we can find the six trigonometric functions so

this is the x right here and this is the y and this is the p of x y

and you can just imagine a unit circle going on through here

but we don’t need to draw that so anyways this distance right here we

call it r right there so what do we have going on right here

for these for these values right here so we know that x is

minus square root of three that’s the x value right here it’s

sitting opposite of the 60 degrees and we know the y right here is minus

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one we have to go down to minus one right there and the radius right here is two

so we can find that by looking at the pythagorean theorem right there

all right so what we can do now is we can find the six trigonometric functions

so we have sine of 210 degrees and we have the cosine of 210 degrees

and we have the tangent of 210 degrees so once we find these two we can find

the tangent immediately just by looking at the ratio of these two and once we

find these three then we can find the other three just by taking the

reciprocal so really comes down to knowing these these two right here further

we can find these values right here by looking at the reference angle so sine

of 210 degrees will be equal to sine of the reference angle the reference angle

is 30 degrees and this will be cosine of 30 degrees

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now the caveat is we don’t know if it’s a plus or minus yet until we actually

look at the actual quadrant sine is negative in quadrant three

so this will be a negative right here cosine is also negative right so 210 is

in quadrant three and cosine is negative in quadrant three

so these will both be negatives right here so let’s finish this out

now we should know these uh quickly because we’ve memorized the the special

angles we know what sine of 30 degrees is so this will be a one-half so this

will be a minus one-half cosine 30 degrees so this will be a minus

and then square root of three over two and then now we look at the ratio

and when we look at the ratio the minus 2’s cancel is we just get 1 over square

root of 3. and now once we found these 3 now we can find the other three

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immediately so cosecant of 210 degrees secant of 210 degrees

and cotangent of 210 degrees let’s see if i left enough room here

all right so we found these right here immediately just by looking at the

reciprocal of these this is minus two and this is minus two over square root of

three and then this is just square root of three

and so there they are there’s the six trigonometric functions of 210 degrees

and the way that we did that is we looked at what the reference angle is here

now um we didn’t actually need to know these values right here

because we already knew these values right here we already knew what sine of

30 degrees was we already know what cosine of 30 degrees was we know those

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immediately all right let’s look at more examples

so our first one is going to be um 240 minus 240 degrees

so let’s look at where is minus 240 degrees uh and we’re just going to find the

cosine of it we don’t need to go find all six every time right so let’s just

find the the cosine of this right but where is minus 240 degrees um before i

try to figure out the reference angle i usually like to draw it

so minus 240 degrees so first off it’s minus right so it’s not going to be

quite 270 minus 270. so if i went around to right here that’d be minus 270

and we’re not quite there and we’re also past minus 180 though so we’ve passed

mine minus 180 we’re not going to 270 so i’d put it about right there

so there’s -240 degrees right there and now what is the reference angle so

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this is not the reference angle it’s always reference angle with respect to

the x-axis so here’s the reference angle right there

so how do we get this reference angle right here

so this is going to be minus 180 and then we got a little bit more um so

you know this is going to be minus 180 and then we need to go minus

60 more right so the reference angle is minus 60 degrees

minus 60 degrees that’s the reference angle right there oh no sorry

we need to go minus 60 degrees more however the reference angle is 60 degrees

right it’s always positive so let’s just put that down over here reference angle

of minus 240 degrees is 60 degrees which is positive

so again i go minus 180 and then i have to go minus 60 more to get to the minus

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240 and so then we ask what is the acute angle right here that’s that’s with the

x-axis and that’s 60 degrees so now we can find the cosine of minus 240 degrees

and that will be cosine i usually save a little space for my sine

that’ll be equal to cosine of the reference angle and then i need to

figure out if i get a positive or negative here now this is in a quadrant

two angle right here minus 240 is a quadrant two angle right there so that’s

quadrant two and cosine is negative in quadrant two so i’m going to put a

negative here now we know what cosine is 60 degrees is

that’s one of the things we have memorized that’s a half so this will be

a minus one half right here so this answer right here is minus one

half right there cosine minus 240 degrees is minus one half

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all right let’s look at one more uh what is the tangent of 675 degrees

so let’s uh let’s draw that over here um

maybe i’ll make that a little straighter all right so where is

675 degrees right so we’re going to go around 360 degrees but we still got more

to go so how much more do we have we got um so let’s look at you know 675

and then we took away minus the 360. and so that’s what 315 right

so i’m going to go around 315 more and there’s my terminal side of the angle

and so now i ask what is the reference angle

and so we have to go around some more to get to the 360. we have to go around 45

degrees more that’s the reference angle right there

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so i usually like to put that down reference angle of 675 degrees is 45 degrees

all right so now we can find the tangent of 675 degrees tangent of 675 degrees

is equal and i’m going to save a little space for my sign right here

positive or negative but then it’s going to be tangent of the reference angle

which is 45 degrees and then 45 degrees here we know what tangent of 45 degrees

is it’s just a one all right but this is in quadrant four right

and in quadrant four is tangent positive or negative right so

tangent’s negative in quadrant four the way that you can remember that is sine

is negative here but cosine is positive here and tangent is sine over cosine so

what’s a negative over a positive it’s a negative so that’s why i’m putting

negative here for tan for for 675 degrees it’ll be minus tangent to

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reference angle now tangent of 45 degrees is

if you want to think about it like this sine of 45 degrees which is might be

something you have memorized over cosine of 45 degrees

which you also may have memorized so now you know these are the same values

so this is just a minus one right so however you want to think about

that as an intermediate step but you know hopefully by now you know

tangent 45 degrees is just one because sine and cosine have the same

value at 45 degrees all right so let’s do uh one more

let’s look at this one right here all of

this right here so we got to piece these together and then do some and then do

some algebra on it so i want to find cosine 120 i want to find sine of 60 i

want to find tangent of 30 and then i’m going to plug them all into this number

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right here and finish the deal so let’s do that real quick

um let me play this one right here for us i appreciate if you like

and subscribe to the channel that really helps out me

making uh other videos all right so let’s look at so we already know tangent

30 degrees and sine is 60 degrees we already know those what we don’t know is

120 degrees let’s chop it down and bring

it to an acute angle and see what we can do there so let’s do that up here let’s

look at 120 degrees it’s somewhere up here in quadrant two

and so what’s left over to get to the 180 and that’s 60. so here’s 120 degrees

and we’re in quadrant two right we’re past 90

and then we still have we would still have 60 more to go to get to the 180 so

remember reference angles always with respect to the um x axis

so once we draw this once we see the reference angle of 60 now i’m ready to

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come over here and say i’ll come over here and say cosine of 120 degrees

that’s we’re trying to find right here so cosine of 120 degrees is cosine of 60

degrees the reference angle you can always go down to the reference angle

that’s why reference angles are so important however you have to pay

attention to the sign right here this is quadrant two

and when we have a quadrant two angle we end up over here we’re going to need a

negative x and cosine are the x values so we’re going to need a minus sign

right here cosine is negative in quadrant two

so now we remember what is cosine is 60 degrees so cosine is 60 degrees is one

half so this is minus one half actually i guess i’ll put it over

here all right so now we have to figure out what is sine

of 60 degrees but we already know that sine is 60 degrees the reference angle

is 60 degrees it’s in quadrant one sine is positive in quadrant one

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but in fact we all know that the square root of 3 over 2.

and then now we need to know tangent of 30 degrees

now this one isn’t on the unit circle so

you could think about it as sine over 30 degrees divided by cosine of 30 degrees

and you know these values right here you know what sine of 30 degrees is and you

know what cosine of 30 degrees is when you simplify that fraction right there

we’re going to get square root of 3 over 2 over

all right and so then we can just simplify we can write it like that

or if you wanted to you could write it as 1 over square root of 3 either way

it’s fine you know the 2’s cancel when you simplify that

all right so now we know this one this one and this one now we can find out

what all this is equal to so i’ll put an equal sign right here cosine 120 that’s

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minus one halves so this part right here is minus one half plus two times

and then then we need sine of sixty which is square root three over two so

square root of three over two now this squared right here means that this is

squared and then minus and then now we have tangent of 30 degrees tangent of 30

degrees we found it to be this square root of 3 over 3

and then that also takes a squared and so now we can write this as minus

one half plus uh three over four and then minus and then three over nine

and so then when we put the fractions together which we’ll do right here

we’re just going to get 2 over 3. just put them together and simplify

and we get 2 over 3. and so there we have it all right so now let’s look at the

domain now that we’ve extended these six trig functions uh to real numbers uh

00:34

let’s take a look at the domains and be focused on that

uh this real important to keep in mind the domain

so let’s just refresh our memory real quick we have the unit circle

let me move that down just a hair of the unit circle and

we had a point here on the unit circle i uh let’s call it here p of x y

and here’s our angle t and so here’s where our definition is sine of t is just y

cosine of t is just x and then we have the other tangent of t is y over x

and cosecant is one over y and the secant of t is one over x

and the cotangent of t is x over y now this is assuming certain things here

00:35

x is not zero and this is talking about the domain here right we need to

understand what the domain is x is not zero y is not zero right

so what t can we input here well t is the angle right here and we

can go around as many times as we’d like and we can use negative values also so

here this t can be any real number right here

in fact the same is true for cosine this t can be any real number right here and

we’ll get out an x so let’s put the domains here so domains and we have three

that we’re going to think about the sine and cosine

so the domain is all real numbers all real numbers

and then the other two we’re going to talk about are the

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tangent and secant and the reason why is because they’re both divided by x here

and that’s going to pose a problem we cannot input those t

that make 0 as an output so for tangent and secant the domain is going to be

pi over 2 plus n pi and then where innocent integer where n is an integer

and so the reason why we write it like this is because this is representing an

infinite number of values so n can be an integer so n can be zero in which case

we just have pi over two and the reason why we do that is because at pi over two

when we input pi over two that would give us the cosine is zero

remember cosine of pi over two is zero and that x giving us a zero down here

00:37

will mean we can’t do tangent and the same thing for secant so

and then if you do another uh n for example a one or a minus one or a

two or a minus two and you plug in those values that you’re

getting out there for whatever n you choose zero one or two or whatever integer

the x you input that t in here and the cosine will make it a zero

and that’s okay for cosine but it’s not okay for tangent and secant

and then and then in a similar case we have to worry about where the y is

zero because we have two cases for those two so we have

cosecant and cotangent so i’ll just abbreviate those cosecant and cotangent here

and it’s going to be wherever the tan wherever the sine is zero where the

outputs are zero we cannot use the inputs where the sign outputs a zero

because we’ll be dividing by zero in that case so we have to look and see hey

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where is the sign outputting zero right and that was when t was 0. so this will

be 0 now we’ll just say n pi where n is a positive where n is an integer

n can be any integer um and so you know all real numbers

except these values right here and all real numbers except these values right

here and then sine and cosine is just all real numbers so i’ll put um the double

quotes here if i could squeeze them in here

in other words all real numbers and then all real numbers except these and all

real numbers except these so we want to be able to do any y over any x

except for when x is zero and the same thing for these three right here

okay so that helps us talk about the domains and now

00:39

when we’re talking about evaluating the trig functions right here so evaluating

the trig functions now though the way that we did this was

we noticed that and i wanted to kind of show you this right here

is that this is where all of the trig functions are positive

and this is where the sine function is positive because the y’s are positive

and this is where tangent is positive because it’s sine over cosine

and this is where cosine is positive and so some people will make up a

story about the s and the a and the t and the c so you can you know see all

t see whatever you want to you know make up to help memoric device

that you want to use so but anyways how are we evaluating the trig functions

right here so step number one is to find the reference angle

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find the reference angle and then step two find the sign

and that’s what we you know have some kind of device to help us with that find

the sign and then find the value put those together

find the value and so this is what we did those exam this is why we did those

examples to help us make sure that we got that right there but um we

did them in degrees so now let’s do them in radians right here

so let’s uh um just to make sure let’s find um

each one here and the first one is going to be cosine of 2 pi over 3 and now

let’s look at them in radians right so where is 2 pi over 3 at

right so let’s draw let’s draw 2 pi over 3 up here now

you know we’re thinking in terms of degrees right so we’re thinking like 90

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and then 180 into 270 and 360 and we think about where the angle is

right but now this is over three this is

pi over three so we’re thinking about in

terms of pi over 3’s so remember pi over 3 is 60 degrees about right there

and then this right here is 180 degrees which you can think about it as 3 pi

over 3 and so 2 pi over 3 is going to be a 60

and then another 60. so it’s about 120 degrees

but i’m not going to label it in degrees it’s 2 pi over 3.

and so now we have to ask the question what is the reference angle and now

we’re going to talk about the reference angle but we’re going to leave it in

radians so the reference angle right here is one more pi over three

so the reason why i got that is it’s going to be 1 pi over 3

2 pi over three and then three pi over three in other words those are pi over

three missing right here to get to the x axis

00:42

or you could think about it like this pi minus two pi over three

so it would be all of pi and then take away the pi and then take

away the 2 pi over 3 and then you get the remaining part right in here

and so this is 3 pi over 3 [Music] minus 2 pi over 3

and so you see that’s just pi over three all right so now we can use our method

right here so cosine of two pi over three is cosine of the reference angle which

is pi over three so i found the reference angle now we need to find the sine

now cosine is negative over here because the x will be and then

we’ll need a negative x to get to to this point over here

on the terminal side right so this is cosine’s negative in this quadrant and

now we know cosine of pi over 3 we just happen to have that memorized so that’ll

be minus one half right there cosine pi over three cosine is 60

degrees that’s a special angle so minus one half right here

00:43

and so that’s it for part a for part b tangent of minus pi over three

so let me label this as part a this is part b here

and let’s go look at where is minus pi over 3 at so minus pi over 3 is down here

it’s minus 60 degrees minus pi over 3 but anyways where is the

reference angle so the reference angle is the acute angle right there pi over

three so reference angle is is pi over three

and so now we can go try to find tangent of minus pi over three

so let’s do that here tangent of minus pi over three

is tangent of the reference angle which is pi over three

but then we need to choose the sign here is it positive or negative

so this is in quadrant four and we know that sine is negative here

00:44

and the cosine is positive so this right here will be a negative

right there negative over a positive which turns out to be negative now

if you don’t know tangent of pi over three that’s okay

we’ll write this as sine of pi over three over cosine of pi over three

now you should know what these two are what is sine of pi over three and what

is cosine of pi over 3 and so we’re going to get square root of

3 over 2 and we’re going to get one half and so we’re going to just get minus

square root of 3 right here all right so we found tangent of minus pi over three

it’s just minus square root of three all right and now for the last one right

here let’s see if we can fit it in right here part c what about 19 pi over four

what does that look like all right so now when we go around the

00:45

circle we’re going to go around 360 degrees or that’s two pi

and maybe we can go around another two pi and another two pi eventually we’ll

go too far how many times can can four go into here right so

we want to take away let’s see if we can fit it in here 19 pi over four minus

4 pi wait why 4 pi well 4 pi is what um 8 pi over 2 um so you know

we want to think uh in terms of 2 pi but you know this is over 4 right so if

we put it over 4 that’s really 8 pi over 4. but we can still do another eight pi

over four another one would be sixteen pi over four

and that would give us what um three pi over four

00:46

left over so when i’m going around i’m going to go around 8 pi over 4 or in

other words 2 pi and another 8 pi over 4 in other words 4 pi

but then i still have 3 pi over 4 left where’s 3 pi over 4 at

well here’s 1 pi over 4 2 pi over 4 and 3 pi over 4. so it’s right about here

so this will be 19 pi over four we went around and round and round we

ended up right here in quadrant two quadrant two and we’re looking for the

reference angle right here and as you can imagine it’s going to be pi over four

wait why pi over four well we went when we decided how much further to go

we were uh looking at the pi over force one pi over four two pi over 4 3 pi over

00:47

4 is left so we need 4 pi over 4 to get here that’s a pi over 4 left there okay

all right so now now we get now we can do now we can do this so sine

of 19 pi over four will be equal to sine of the reference angle

which is pi over four and then we have to worry about is it

positive or negative this is quadrant two sine is positive here because we’ll go

up to get the sine which is the y coordinate and so this will be a positive here

and so now we just need start to know what sine of pi over four is and that is

the square root of two over two which you should know already all right

so that does it for this episode and um i want to say thank you for

watching and uh have a great day uh check out in the next episode if you

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