Right Angle Trigonometry (Trigonometric Functions of Acute Angles)

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

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

00:00
hi everyone welcome back in this episode we’re going to introduce
the six trigonometric functions for an acute angle let’s do some math [Music]
hi everyone welcome back we’re going to begin uh by talking about
uh what was done previously on this uh series trigger met trigonometry is fun
so we had the two episodes so far where we talked about trigonometric angles
and then we also talked about radians and degrees and we talked about
converting between them and we also talked about the unit circle
and all the special angles that go on the unit circle and we talked about
you know that’s something that you should really work hard to memorize
how to draw a unit circle with all the special angles in less than five minutes
so those are the two previous episodes they’re part of this this episode is
part of this series so i recommend that you go check that out

00:01
here we go uh here’s the series uh link is below in the description trigonometry
is fun step-by-step tutorials for beginners
all right so for today uh this episode we’re going to start off with what are
the six trigonometric functions here so i’m going to begin by drawing a
triangle right here and this is a right triangle
so let’s put this right triangle right here and we’re going to say this is angle
theta right here and so let’s make this a positive angle theta
and i’m going to label this side right here hypotenuse
and i’m just going to abbreviate that and i’m going to label this side right
here called the side adjacent to theta so we have the angle theta right here
and this right here is the side adjacent to theta right so we’re looking at a
triangle three three sides and i’m going to label this right here

00:02
as the side opposite of theta side opposite of theta and using these three
uh sides here we’ll be able to define the six trigonometric functions of theta
so the first one is going to be i’ll put it over here so sine theta
we’re going to define as the ratio of and for the side opposite of theta i’m
going to abbreviate that with opp for opposite
and the side adjacent i’m going to abbreviate that to a adj
so sine theta will be equal to opposite over hypotenuse
and so that this is the uh sine now i’m abbreviating the sine also
sine is s i n is sine function the full name is spelled
out s-i-n-e but i’m just going to abbreviate it over here when we’re using

00:03
it in an equation like this and so also we’re going to have the
cosine of the angle theta and that’s going to be the adjacent side
now when i say the adjacent side what i really mean is the length of the
adjacent side so that’s an actual number so we’re going to say adjacent over
and again when i say hypotenuse here what i mean is the length of the
hypotenuse hypotenuse is always the longest side
and then we’re going to define uh one more so we’re going to define the tangent
of theta and the way we’re going to define the
tangent of theta is going to be sine over cosine
so it’s going to be opposite over hypotenuse
over adjacent over hypotenuse but when you do that you see that you actually
get a cancellation here and so long story short
because these right here cancel out because this is just a complex fraction

00:04
so we’re really going to define tangent as opposite over adjacent
so what we’re doing is we’re taking in a an angle a trigonometric angle and it’s
going to be measured in radians so we’re taking in a real number
and tangent is going to look at the side the length of the sides here and
it’s going to take the length of the opposite side here of theta and i’m
going to abbreviate that length by opp and the length of the side adjacent to
theta and i abbreviate that ag a dj and when i divide that when i make that
ratio i get an output right there all right and so
now i’m going to define six uh sorry three more so there’s six all total so
let’s define three more now the other three are going to be the reciprocals of
these so the cosecant function will be the reciprocal here hypotenuse
over opposite so the hypotenuse right here over the hypotenuse

00:05
and the secant will be the secant of theta will be the
reciprocal of the cosine so hypotenuse over adjacent
and then the cotangent will be the reciprocal of
the tangent so this will be adjacent over opposite
and again these are all abbreviations so this is the sine function
the cosine function the tangent function the cosecant function
and as you can see here some of them have co in front of it
and the secant function and the cotangent function
so this is all six of them right here this is how we spell out the names um
but then when we use them in mathematical um you know equations
we’ll just abbreviate them with the three letters here cot

00:06
but it’s just abbreviation for the cotangent that’s the full full name in
any case this works when theta is an acute angle
which is the subject of this video right here now
let’s look at an example let’s say let’s throw down another triangle right here
and let’s say this right here is 3 and so that’s the length of that side
and i’m not going to worry about units right so the units actually cancel out
notice that right here like if this is measured in inches and that’s measured
in inches then the inches would cancel out so you would input a real number and
you would do a computation right here sine of theta and you would do that
computation that ratio and then your output would be a real number here so
i’m going to measure this as 3 and i’m not going to worry about the units
and 4 here again not worrying about the units and so right here we can use the
pythagorean theorem and we’ll get this length of the side hypotenuse is five

00:07
and the way you do that is just say three squared plus four squared equals
this side squared and so then you would take the square root of that
right so it’s a 9 plus 16 is 25 and this is the square root of the 25 right here
so check out the previous video in fact i have a whole series that can
get you started in that the series is functions in the graph step-by-step
tutorials for beginners where i go over the pythagorean theorem in in great
detail in that series but in any case here we go three four five and that’s a
very popular triangle right here so anyways sine of theta and so we’re
looking at this theta of this angle right here theta
and so sine of this angle right here i only need 1s sine of theta is
going to be opposite over hypotenuse so opposite so 4 over 5
and cosine will be 3 over 5. cosine of theta is 3 over 5. now you

00:08
have to specify the theta and here’s why is because there’s two
angles here there’s a theta and by the way sometimes we denote angles with
greek letters sometimes we’ll use capital um letters like a b or c or something
but if i’m using greek letter here i’ll use a greek letter up here so this is
angle theta so another popular one is alpha i’ll use a beta up here
so there’s two angles here and so if someone asks you to find cosine and they
give you this triangle here well then you have to ask them well which angle
right and so i’m saying cosine of theta right here
is 3 over 5. in other words you cannot leave off the theta right here that’s
not optional all right and then tangent of theta will be
four over five over three over five the fives will cancel it’s really just four
thirds and then we got the reciprocals so the cosecant of theta

00:09
is the reciprocal five over four and the secant of theta here is five over three
and the cotangent is the reciprocal of tangent so three over four and there’s
the six trig functions evaluated at this
angle right here and we didn’t know what the angle is although we could actually
go and find it but that’s not important for this episode right here this episode
right here is what are the trig functions right here so here’s what
there are in general and here’s what they are with an example right here
obviously the size of the triangle can change and so the outputs will change
all right so um now let’s look at some more uh examples
now let me give you some of the angles right here because in our previous
episode we talked about special angles so let’s make the connection with what
we did before uh let me erase this real quick for us and so we’re going to

00:10
get this set up here now now the question is evaluating the trig
functions of an angle and the first one we’re going to do here is
45 degrees right so what is sine 45 degrees cosine 45 degrees and
tangent of 45 degrees so this time i’m giving us the actual
angles and we’re looking for an output now to do this right here we’re going to
draw a triangle and so we’re going to have this right here as a 1 and a 1
and a 45 degrees right here and we can use the pythagorean theorem
to get this side over here would be 1 squared plus 1 squared square so to be
square root of 2 here all right good so um
what is sine of 45 degrees so sine of 45 degrees
is going to be the opposite over the hypotenuse and the cosine of 45 degrees

00:11
will be pretty much the same thing you know just 1 over square root of 2.
now some people like to rationalize this so if you haven’t seen
that before let me just refresh your memory really brief
so i’m going to multiply the numerator and denominator by square root
of 2 and i get square root of 2 over 2. so you could use this right here or this
right here depending upon you know what’s being asked of you
tangent of 45 degrees and so that’s gonna be one over one
that’s just one or if you wanted to write it out
it’s one over square root of two over one over square root of two
which is you know unnecessary to write that let’s just write a one
all right and so now we have the reciprocal functions it doesn’t actually
actually ask us to find them i’m not asking that over here but let’s just do
that over here anyways cosecant uh 45 degrees

00:12
will be the reciprocal which is just square root of two
and the secant of 45 degrees will be also square root of two and then
cotangent 45 degrees will be just one over one what is still one all right
now what about the 30 degrees here so let’s find sine 30 degrees cosine and
tangent of 30 degrees so for that we need a different triangle
so i’m going to put a triangle down right here
and let’s make this a little straighter and what we’re going to have here is 60
degrees and we’re going to come down here with a perpendicular right here and
we’re going to have 60 degrees we’re going to have 30 degrees
and we’re going to have a two and a two and a one and a one
and this missing side right here is gonna be square root of three right here
so that’s the square root of three and so now we can find the trig
functions uh for the for these angles right here and then also why we’re at it

00:13
we’re going to do the 60 degrees so if you remember the previous episode these
were the special angles 30 45 degrees and 60 in the first quadrant right there
so what will be the sine of 30 degrees so now i’m looking at 30 degrees right
here and so i need to go with opposite over hypotenuse so that’ll be one half
and then cosine of 30 degrees which will be adjacent which is square
root of 3 over the hypotenuse and then tangent of 30 degrees here will be
this one over this one the twos will cancel so we just get one over square
root of three and then if you want to rationalize that
i don’t think that you should rationalize necessarily every time but
at least once if we rationalize this we’re gonna get square root of three
over three and so that’s one way that you could uh
look at that you should recognize and know that one over square root of three

00:14
is the same thing as square root of three over three but i’ll leave it like this
all right so this will be cosine sorry not cosine cosecant of 30 degrees if we
want to find that that’ll be just 2 and then the secant of 30 degrees
will be 2 square roots of 3 and then the cotangent of 30 degrees
will just be square root of 3 and going to get smaller
so i’ll move up there so cotangent of 30 degrees is square root of 3.
and then the last but not least we can find these 60s because we got a 60 in
here also right that’s 90 and the whole thing is 180 so we got a 60 right here
in any case so let’s go over here and find these now let’s go here sine
of 60 degrees now i’ll be looking at this angle right here so it’s going to
be opposite over adjacent so it’s going to be square to 3 over 2
and then cosine 60 degrees will be the one half and then the tangent will be

00:15
the ratio of these two so i’ll get square root of three over one or square
root of three and then we can take the reciprocals cosecant of 60 degrees
and secant of 60 degrees and cotangent of 60 degrees
and so just taking the reciprocals of these 2 or square root 3 2 and
1 over square root of 3. and so there we go we did them all
excellent um so now let’s look at um can can we do something more elaborate or
you know can we if what if we’re not just given numbers right sine of 45
degrees right we’re given the input what if we were given some kind of
geometric information maybe we’ll have to figure that out right so let’s look
at something like that let’s look at another example here
and in this example i’m going to draw us
a triangle for us and we’re going to you know continue to use our

00:16
trig functions here so let’s go here with this
in fact let me see if i can move over here
and i’m going to put the triangle uh up here so let’s see here we got a triangle
right here and inside of here we have a another triangle
and this is our angle theta and this is going to be a b and c and
this is going to come down here this is going to be a 4 a 5
and a 3. so that’s a 3 4 5 triangle right there this is a right angle
i can try to make that a little bit more right anglish and
all right that’ll be good enough right there almost looks like a christmas tree
falling down all right anyways um i’ll try one more time
all right that one’s good enough and this angle is called phi

00:17
and this one will be square root of 41 out here and this will be a 2.
and this is a 4. all right so there’s all the geometric information that we have
and our goal is to find sine of theta first
so what would be sine of theta so here’s theta right here so i’m looking at sine
let me let me put it down here sine theta so sine will be opposite over
hypotenuse so this would be oh yeah this is b um let me put that
point down here b right here all right that’s good so let’s see here we’re
going to get bc over db and i mean the length of bc right here opposite over
hypotenuse and so this is d this is a d right here sorry um so db and bc

00:18
anyways this is just four fifths and then cosine of theta um
well let’s work with let’s find cosecant of phi actually so
you know this is different angle so i can’t just take the reciprocal sure we
could take the reciprocal here cosecant of theta is 5 over 4 but i’m not asking
you that i’m asking you cosecant here right and so what will the sign of
this right here be so the sign here will be three over five
so this will be five over three there we go what about cosine of angle a
so angle a is right in here and so this angle right here you know we need to
uh know the triangle right here so we’re looking at this angle right here this
angle a right here or we could call it angle um you know bac

00:19
or just angle a for short but in any case i’m looking at this
right triangle right here the big one and you know the opposite is 4 and the
hypotenuse is 41 right here so we’re going to call this right here the
cosine oh we’re looking for cosine so i need the adjacent so that’ll be 2 plus 3
so 2 plus 3 over the square root of 41 here so 5 over square root of 41.
and so now what about the last one what about is the tangent of angle b
so what is angle b actually right so let’s just call it angle abc just to
be clear so what is the angle abc here um so to to find out the tangent of that
we’re not looking for the angle we’re looking for the tangent of that

00:20
so that’s the big triangle again the big right triangle
and this angle right here is what we’re looking for and remember tangent is
going to be the um opposite over the adjacent right here so
we’re looking at five over four so two plus three
which is the whole side opposite over four so 5 over 4 here are very good
so let’s do another one let’s do another example
again we’re going to be given some information here and our goal will be to
find out you know some of the missing stuff here so let’s go this triangle
right here i’m going to try to put the triangle right here so
on this one right here we’re looking at a right triangle we have a and we have b
we have c c is the right angle and then this is a side b right here

00:21
and this is a six and this is a 10. now when i’m writing a b here i actually
mean the length of the side opposite of the angle b
all right so when i use a b here i’m really talking about our number here
okay so a is 6 and c is 10. find the six trigonometric functions right can we
find you know sine of a cosine of a and tangent of a and
everything else right so what will sine of a be so this will be 6 over 10
and that reduces to 3 3 over 5 and then cosine of angle a
well we’ll need the missing side here won’t we we need the adjacent over 10 so
this will be b over 10 whatever b is can we actually go find the b if we know
two sides of a right triangle can we just find the third the answer is yes of

00:22
course let’s do that over here so we have 6 squared plus b squared equals
10 squared and so this is b squared equals you know
100 minus 36. i’ll take the square root of that it’ll be 64.
and so b here is eight so we’re looking at a right triangle right here and
um yeah so b is eight right here so we got a six eight and a ten
all right and so the cosine of a here now we can go back to this this will be
b over 10 so 8 over 10 and 8 over 10 reduces to 4 over 5. and then we can do
tangent of a just the ratio the 5’s will cancel so we get three fourths
and now that we found the first three here now we can go find the other three
so the cosecant of angle a will be the reciprocal five thirds
and the secant uh will be the reciprocal of cosine which will be five fourths

00:23
and then last the cotangent of a will be four over three
all right so there we go so we got some geometric information and
we can find the six trigonometric functions there
all right so um now i want to while i’m erasing this right here i wanted to
mention to you that something really amazing is going on
and it’s the reason why we have these um cofunctions why some of them are
called co like tangent and cotangent right and so they have something to do
with complementary angles and so let’s look at that right now so what we have
going on here is um let’s see here we have um yeah so
what are cofunctions right that’s what i
want to get to so um sine and cosine are cofunctions and let me just write them

00:24
down here sine and cosine and tangent and cotangent and secant and cosecant
right so these are all cofunctions these two are cofunctions these two are
co-functions and these two are co-functions right here um and so
what we can do is let’s let’s draw a triangle to to see to see what’s going
on here so let’s call this triangle abc again
and yeah so this is c this is a little b this is a here
and so let’s call this let’s write these down here now so sine of angle a
is going to be a over c over c and cosine of angle a is b over c
and tangent of angle a is a over b and we have the um

00:25
the reciprocals of those so the reciprocal of sine is cosecant angle a
which is the reciprocal c over a and then we have the secant which is the
reciprocal of cosine c over b and we have the cotangent of angle a
is the reciprocal of tangent all right so that’s all all that we know already
that’s everything in terms of angle a what if we think of the six
trigonometric functions in terms of angle b in terms of angle b
what is the a over c so now in terms of angle b here that’s the adjacent over
the hypotenuse so this right here is cosine of angle b
in other words cosine of angle a is equal to cosine of angle b
and that happens when a and b are complementary angles in other words
they add up to 90 degrees and the reason why they add up to 90 degrees here is
because this is already 90 and we get 180 in a circle uh sorry triangle

00:26
and so cosine of angle a here would be of angle a would be adjacent but if i’m
looking with respect to angle b here this will
be opposite so this would be sine of of angle a uh sorry angle b
and yeah so what about tangent um so tangent of angle a um
and that’s looking but with respect to angle a it’s opposite over adjacent
now if i look with respect to b it’s opposite over adjacent and that
one’s right here so this is tangent of angle b
and this will be cotangent of angle b and
if we look at the reciprocal right here of sine this will be c over b
which is right here and so this will be cosecant of angle b and then
to get the c over a here it’s the reciprocal of this right here so that’s

00:27
secant of angle b and so you can see the relationships
right here between all six of them if the a and b add up to 90 degrees in fact
let’s write that down here um if a plus b equals 90 degrees then
sine of angle a equals cosine of b and secant of angle a
equals cosecant of angle b and tangent of angle a is equal to
cotangent of angle b and so yeah that’s just complementary
angles there and it just has to do with the fact that
we have the six all six are defined uh with the three sides all
possibilities are defined and so then if we change our perspective from angle a
to angle b we still end up with the same six though

00:28
still end up with the same six all right so um
now let’s put it all together and look at what this has to do
with the unit circle that we talked about before so i want to bring that
into perspective here real quick so let me get rid of this real quick
um so in the last episode we talked about the unit circle and we did all the
uh constructing and putting all the special angles in here and now i’m just
going to kind of bring it all together because we didn’t actually label the
sine and cosine so here i want to put degrees
and i want to say theta right here in radians
and we are going to have the sine column the cosine column
and the tangent column and we could make three more columns but
i think that’s going to be enough to see the pattern we’re going to start off
with 30 degrees uh sorry 0 3 30 and then 45 and then 60 and then 90

00:29
and those are our special angles there and we know what they are in radians pi
over six pi over four pi over three pi over two
now if you know how to make a unit circle in five minutes or less then you
should be comfortable with everything that i’m doing here
if you’re not then you should go back and definitely practice that right there
so now we’re now it’s time for these columns here right we we talked about
these columns right here uh in this episode and so but i wanted to show you the
pattern the pattern that we talked about in the last episode here let me move
this down here a little bit so we know the sine of angle theta is zero degrees
here and we know its cosine is one here and we know tangent is the reciprocal
of zero over one and which is zero and so we have this pattern here we found

00:30
this to be one half and we found this to be square root of three over two
and we found this to be square root of three over three and
you know you see the pattern the pattern is um
yeah and so 45 degrees we found this to be 1 over square root of 2
and we found this one right here to be 1 over square root of 2
and we found this to be a 1 here and then we found this one to be square
root of 3 over 2. and so this is really just summarizing
what we’ve already found here reciprocal here um
one zero you know when we look at the reciprocal here
it’s not defined right here so we’ll just put a put a dash in there like that so
another way to think about these values right here is the pattern
and this is what i really love here is we can replace this one-half and you can

00:31
think about it as a square root of one over 2 and
we can think about this right here as a 30 degrees right here square root of
yeah square root of one over two and this one right here we can think
about it as a square root of two over two and this one square root of three over
two and we can think about this one as a square root of four over two
and so and you can think about zero as square root of zero over two
so you can kind of see a pattern here emerging is zero and then that’s one half
and then it’s one over square root of 2 and then square root of 3 over 2
and that’s just 1. but if you see the pat if you write it
like this then you can see the pattern and then over here the pattern reverses
that would be square root of 4 over 2 square root of 3 over 2

00:32
square root of 2 over 2 square root of 1 over 2 square root of 0 over 2.
and you know this is just the ratio right here so so really here’s the
pattern right here um sometimes i like to write it like this instead if we put
the angles up here 0 and then let’s go with radians pi over 3
pi over 6 pi over 4 pi over 3 pi over 2 and then we have sine of theta
and cosine of theta and this is theta so i fill this in right here
and this is going to be square root of 0 over 2 square root of 1 over two square
root of two over two square root of three over two square
root of four over two and these go backwards square root of four over two
square root of three over two square root of two over two square root of one
over two square root of zero over two and then you just simplify that right

00:33
that’s really a zero that’s a one-half and then you can leave it like this or
you could say 1 over square root of 2 and then that square root of 3 over 2
already simplified and that’s a 1. and then you could just go backwards
right there like that i just i just love this
right there it just looks so pretty to me
um okay so that’s it for this episode i want to say thank you for watching and
i’ll see you next time 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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