BAM!!! Mr. Tarrou. Alright, we just got done setting up the unit circle, evaluating some
basic trig expressions using the six trig identities sine, cosine, tangent, cosecant,
secant, and cotangent. Now I want to use these in relation to the unit circle and hope to
explain where some of the small identities come from that you will be looking at in your
PreCalc or Trig book. The first three are the reciprocal identities. Ok, there is actually
six of them, but I am only going to have the ones set up for cosecant, secant, and cotangent
since those are.... they kind of work in both ways anyways. So the cosecant of theta is
one over the sine of theta. The secant of theta is one over the cosine of theta. The
cotangent of theta is one over tangent of theta. Again there is going to be six of these.
I could write the sine of theta is one over cosecant, cosine is one over secant, or that
tangent is equal to one over cotangent. Now, if somebody could explain to me why they decided
to change the first letter of these trig identities that are reciprocals, please let me know because
it is a little annoying trying to get those memorized:) Now the book just says, actually
not a lot of book but some books will just say that these are the three or six identities
and just say it is because I say so:) I want to prove just one of these four you. So let's
erase these bottom two and uh remind... I just used the sine and cosecant functions.
Why is it the cosecant of theta is equal to one over the sine of theta? Well, the one
thing I just preach in my class over and over and over and over again to try and remind
my students is, "What do you put in your trig function and what do you get out?" See really
almost all of trig is just studying a right triangle and it's rotation within a unit circle.
And of course I already said the unit circle is set up by your 60-30 triangles and your
45-45 triangles. So you put in an angle measure, either degrees or radians, and you get out
the sides of a right triangle. Well cosecant is r over y. It is the reciprocal of sine.
The cosecant of theta is going to be r over y. Now we have one over sine. It is supposedly
that cosecant is equal to one over sine. Well, sine is y over r. You might remember your
three basic trig functions sine, cosine, tangent from geometry as doing SOHCAHTOA! Sine is
opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over
adjacent. And I explained why we prefer these labels in trig. If you want to watch that
video just search Introduction to Trig Functions Setting up the Unit Circle Part 1 and 2. Now
when you have a number divided by a fraction, that is one divided by y over r. Well, we
can't really divide fractions we can only multiply them. So, if you remember fractions
at all when you try to divide fractions you actually multiply by the reciprocal of the
second fraction. Well that is always going to be in your denominator. So, when you have
a fraction over a fraction the denominator is always going to flip up and you get, well
let's see. r over y equals when this flips up it is going to be r over y. There is your
first ever, from me anyway, trig proof in terms of the ratios of the sides of a right
triangle that we are dealing with in trigonometry. Cool! So there. Cosecant is equal to one over
sine, as long as the angle measures of course are the same. Alright. Let's take a look at
the Pythagorean Identities. The Pythagorean Identities. I am not going to prove all three,
but I am going to prove a couple of them for you. The first one and most commonly used
is sine squared plus cosine squared equals one. Ok, well I want to do a proof of this.
I want to show that the sine squared of theta plus the cosine squared theta, or any matching
angles for that matter, are always going to equal one. When I am done, this proof is only
going to be good for sine and cosine. You cannot interchange them with any of the other
trig functions. Well again, what do you put into a trig function? An angle measure of
degrees or radians. What do you get out? The sides of a triangle and in the case of sine
that is y over r and it is squared. That is what the exponent of two means on the trig
function. Go get the ratio off of the unit circle and once you get it, square it. Plus
the cosine of theta. Well, cosine is x over r. We used to know that as cosine is equal
to adjacent over hypotenuse. So the cosine of theta is out, x over r in its place squared.
And that equals one. So we are going to square each one of these fractions and get y squared
over r squared plus x squared over r squared, and that is equal to one supposedly. We are
going to try and prove that in a second. Well that is what we are doing right now anyway.
We have common denominators. We have y squared plus x squared over r squared equals one.
And here in the proof, and it is a pretty straight forward small proof to be honest
for trigonometry, but in this stage of the proof I like to remind my students what is
it exactly that we are studying. In trigonometry, almost all of the studies are about right
triangles and their rotation about a circle. In this case I have drawn a unit circle. Well
what famous theorem do you know about right triangles? It starts with a P. Pythagorean
Theorem, a squared plus b squared equals c squared. Well, our a values are always going
to be the x values. The other leg is going to be the y value and the hypotenuse is always
going to be the radius of the circle. So where we might have remembered a squared plus b
squared is equal to c squared, now it is going to be x squared plus y squared equals r squared.
So, x squared plus y squared comes out, or vise verse when you are adding it does not
make any difference of the order. So we have r squared over r squared, and any time you
do divide a number by itself you do indeed get the value of one. So... BOOM... BAM...
SHMAM!!! There is your proof. Sine squared plus cosine squared of matching angle measures
theta or whatever you want to call it will always equal one no matter what the angle
measure. I used a generic angle measure, i used the generic ratios of y over r and x
over r, and I got that yes indeed we have one equals one. So it does not matter what
the angle measure is, sine square plus cosine squared will always equal one. AWESOME! Another
Pythagorean Identity involves tangent, it is one plus tangent squared theta equals secant
squared theta. Let's just try that proof one more time. Tangent is, you put in an angle
measure and you get out the sides of a triangle of y over x squared, equals secant which is
the reciprocal of cosine so it is r over x. And let's see if that happens again. Square
all of this out. We have one plus y squared over x squared equals r squared over x squared.
So far it looks like we are going to have a common denominator that matches which is
good. We are going to multiply this on the top and bottom by x squared. Write these two
terms as a single term, so x squared plus y squared over x squared equals r squared
over x squared. Do those indeed equal each other again? Yes, x squared, which is one
leg, plus y squared, the other leg, will be equal to r squared, the hypotenuse. So, r
squared over x squared equals r squared over x squared. BAM!!! That always works. One plus
tangent squared always equals secant squared, as long as the angle measures are the same.
I am not going to prove this last Pythagorean Identity, but let me write it up here for
you. It is one plus cotangent squared theta is equal to cosecant theta. I am glancing
over to the side board because if you watch these videos enough you know that sometimes
my brain and my hand don't always agree. I think one thing and the hand writes another.
So this is your third and last Pythagorean Identity. So, let's check this with an angle
measure. Let's not prove it, but let's just try to work out an answer. Let's say that
one plus cotangent squared of pi over three equals cosecant squared of pi over three.
Well these two sided indeed equal each other. Well, let's see. The cotangent of pi over
three, the cotangent is x over y, so at pi over three radians or 60 degrees my x over
y value is one-half over the square root of three over two. X over y, good. This equals the cosecant squared
of pi over three. Well, cosecant is the reciprocal of sine. If you rotate pi over three or 60
degrees, the cosecant of pi over three, the reciprocal of sine of pi over three, is going
to be two over the square root of three squared. You see I took the y coordinate, the sine
value and just flipped it because cosecant and sine are reciprocals. Now, let's see if
we can figure this out and get the same thing on both sides. Well, when you have a common
denominator on the top and bottom of your fractions, if you have a fraction over a fraction
and they have exactly the same denominators, those will cancel out. If you show the step
where you flip the bottom fraction up, again you will see that. So we have...oops I forget
the power of two. So we have one plus one over the square root of three squared. Let
me go ahead and square this other side out. Two squared is four and the square root of
three squared is three. Ok, so normally we would rationalize our denominator but because
I am going to square the fraction, I don't need to do that step. It is just going to
make more work for me. So we have one plus on squared is one, the square root of three
squared is three, and because I am running out of space I am going to run this sideways.
I need a common denominator adding one plus one-third. So that is going to be three over
three. Indeed three plus one is four and my my common denominator is three. Look at there.
I have matching answers. So yes, and this is not a proof now, this is only an example.
But for the angle of pi over three, one plus cotangent squared of pi over three does indeed
equal cosecant squared of pi over three. A little bit of work to show you how to use
the unit circle to evaluate algebraic expressions or equations, or trig expressions or equations.
Ok, so we are almost done with our identities that might be in your Precalc book or trig
book at this point. Another set of identities are the quotient identities. This is a good
example to show you how sine and cosine are the building blocks of all the other trig
identities. The tangent of theta is sine theta over cosine theta. At least that is what the
book says. And again, if you go to these if you go to these building blocks, these little
tiny formulas over here reminding yourself that we do put in angle measure, degrees or
radians, and get out the sides of a triangle. Let's use these sides of these triangles to
see if indeed this one particular of two quotient identities actually works. Well the tangent
of theta is y over x. The sine of theta is y over r and the cosine of theta is x over
r. Well, if I take this bottom fraction and flip it up making it a multiplication of r
over x, we can see that the r's are going to cancel with the multiplication and we do
get y over x equals y over x. So indeed, in another little tiny trig proof tangent of
theta is equal to sine theta over cosine theta...of course if all the angles are the same. The
last quotient identity, and I am nearly out of time, is the cotangent of theta is equals
sine. Oops! I am not doing tangent. Cotangent is cosine theta over sine theta. I am Mr.
Tarrou and I am OUT!!! BAM! Go Do Your Homework:D