Trigonometric Functions of Any Angle (Including Reference Angles)

Video Series: Trigonometry is Fun (Step-by-step Tutorials for Beginners)

(D4M) — Here is the video transcript for this video.

00:00
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

00:01
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

00:02
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

00:03
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

00:04
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

00:05
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

00:06
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

00:07
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

00:08
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

00:09
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

00:10
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

00:11
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

00:12
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

00:13
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

00:14
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

00:15
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

00:16
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

00:17
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

00:18
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

00:19
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

00:20
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

00:21
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

00:22
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

00:23
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

00:24
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

00:25
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

00:26
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

00:27
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

00:28
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

00:29
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

00:30
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

00:31
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

00:32
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

00:33
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

00:36
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

00:38
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

00:40
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

00:41
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
enjoyed this video please like and subscribe to my channel
and click the bell icon to get new video updates

About The Author
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David A. Smith (Dave)

Mathematics Educator

David A. Smith is the CEO and founder of Dave4Math. His background is in mathematics (B.S. & M.S. in Mathematics), computer science, and undergraduate teaching (15+ years). With extensive experience in higher education and a passion for learning, his professional and academic careers revolve around advancing knowledge for himself and others. His work helps others learn about subjects that can help them in their personal and professional lives.

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