BAM!!! Mr. Tarrou. Let's do some more proofs. Ok, we have the tangent squared x over secant
of x is equal to secant minus cosine. How are we going to deal with this power of two
on the left hand side? You are not going to prove this identity if you can't get the degrees
to be the same. We can't divide it way, we can't square root it away because those two
functions require that you have to work with both sides of the identity. So what to do?
I don't know. You can, at this point... Remember these are proofs, not here is five steps and
follow them in the right order to get the right answer. I am going to try something,
if you don't like it then try something else. Maybe you will do the proof quicker than me,
maybe I will do the proof quicker than you. You know what, if the two sides match then
who cares. Just as long as the algebra is correct all the way through it. So, I see
this one...or this one tangent squared x and I don't see any tangent functions on the right
hand side. So, I am going to take out tangent squared x and one of our identities is 1 plus
tangent squared x equals secant squared x. I am going to use that Pythagorean identity
to rewrite the numerator and I am going to bring this one over to the other side and
I am going to write tangent squared x is equal to secant squared x minus 1. Now you first
step may have been to write this as sine over cosine and this as one over cosine to start
with and that might be perfectly just fantastic. Maybe it is going to be a good proof. But
that is what I have done first. Now I am looking at my problem and I still have a degree of
two where over all I have is degrees of one. I don't see any way of canceling anything
out. Remember you CANNOT cancel terms. My students like to something like, "Can't the
secants cancel out?" Canceling this secant out with this secant would be like saying
that this is ok. (4 + 9) over 4 My kids might be recognizing this if they are watching it
tonight...right!? So could I cancel these fours? I have a common thing on the top and
bottom of this fraction, but it is a common term. If I cancel out the fours, four divided
by four is equal to one. The last time I checked one plus nine which is ten does not equal
what is really four plus nine over four. Four plus none is thirteen, the denominator is
four, and the last time I checked thirteen over four is not equal to ten. You cannot
cancel out terms!!! YOU HEAR ME? ARE YOU PAYING ATTENTION!!! Ok:) Good. Now if it were multiplication
you could cancel out. You can factors, but you cannot cancel terms. That is not going
to cancel away. So what else is going on? Well this is one term, a fraction bar is a
grouping symbol. We have one term on the left and two terms on the right. How can we take
the fraction and break it up into two parts? Well, it has two terms in the numerator so
what you have is a subtraction of two fractions with a common denominator. You don't believe
me? How about 4/3 minus 1/3? You have a common denominator so don't you just subtract the
numerators and write the common denominator once? Yeah you do! So what I am going to do
is, I am going to take secant square minus one and split it those terms up and give each
one their own fraction with a common denominator of secant. We have got secant squared x over
secant minus one over secant. I am getting excited because NOW I can cancel the secants
out because this is secant times secant over secant. Secant times secant, with two things
are being multiplied together then they are factors and not terms. So this one secant
in the bottom will cancel out with one of the two secants on top, and you can think..when
you divide like bases you subtract the exponents as well. We get secant minus, one over secant...one
over secant is a reciprocal function...yeah you are right, that is cosine x and that is
match!!! WHOO! Alright, going on to the next one. Okie dokie. So the next proof we are
going to look at or the next identity we are going to try and verify is the cosine of negative
theta over one plus the sine of negative equals the secant of theta plus the tangent of theta.
Well in this particular proof, or identity, that we are going to verify my angle measures
are not the same. Now they are very close, but they are not equal. Instead of theta we
have negative theta so we have opposite angles. Now we need to talk about a chapter 1 concept
in my book. You might remember it from the previous video if you are watching these in
sequence, we need to discuss the idea of even and odd functions. Cosine, the cosine graph
starts on the y axis at 1 and falls in both directions. The cosine graph is symmetric
to the y axis which means that if you put in opposite x's you get out the same y value.
So, cosine is an even function. If you are not sure what even functions are, or odd functions,
you will need to go back and review that concept. You have already covered in your class if
you are doing PreCalculus. The cosine of an angle is equal to the cosine of that opposite
angle. It is an even function. So the cosine of negative theta is equal to the cosine of
theta. Cosine again is symmetric to the y axis so it is even. Where as sine is not symmetric
to the y axis. Sine has origin symmetry. If you were to take a look in book for a parent
function of sine and turn you book 180 degrees, you know what the difference is...all of the
numbers are upside down and besides that nothing! Sine is an odd function and that means that
when you plug in an opposite angle, so I want to write this as the sine of theta not negative
theta... So I am going to plug in an opposite angle. When you plug in opposite values, you
get out opposite answers. The sine of negative theta is going to be the opposite of the sine
of positive theta. So if I plug in an opposite value, I am going to get out an opposite answer.
Alright? So now we have got our two sides matching. And I am going to stop working with
the left hand side because sine and cosine are the building block trig functions. There
is nothing else I can do, well I guess I could, but I am going to try and simplify instead
of making these more complicated. I am going to write secant and tangent in terms of sine
and cosine and hopefully something nice happens. Let's see what happens...SHALL WE!!! So secant
is one over cosine of theta. Oh, I have got to just laugh at my own joke, that is terrible.
Tangent is sine over cosine. What is worse is that it was not really a joke. Now we have
common denominators. So, on this side we have got one plus sine of theta over cosine theta...
hmm. Well, now both sides have the building block trig core functions and wow... I don't
know. Do you all have an ideas? Yeah.. Ok. I like that idea. We have got one minus sine
on the right side of the proof...right side? I have got one minus sine on the bottom of
the fraction on the left side of my identity. Let's start working with the right side. I
have got more room and I think I am going to need a few more steps. Thank you for the
suggestion;) I want a one minus sine in this fraction. Now I just said that you cannot
move things over from one side of this equation, excuse me identity, to the other. So I am
not going to multiply both sides by one minus sine theta because that would be moving it
from one side to the other. But what I am going to do is multiply this fraction by one.
Just a funny looking one. I am going to multiply it by one minus sine theta over one minus
sine theta. That is one right? See I multiplied it by one, right? Ok, what happens? Well,
this is what happens. We are going to have one times one. One times negative sine and
positive sine times one. Then we have a positive sine times negative sine. This is a difference
of squares pattern. So when I distribute these I am going to get 1 times 1 is 1. Then I am
going to get positive sine and negative sine and those are going to cancel out. And positive
sine and negative sine is negative sine squared theta all over cosine theta times one minus
sine theta. I am not going to distribute this stuff together because I need to do some canceling.
See that power of two that is showing up. I am going to do some kind of canceling to
make it match this left hand side. So I am if I leave the denominator in factored form,
that is going to save me some work. My numerator is full of the sine function and this numerator
is only cosine. I am not going to get these to match if I don't get the top of the fractions
to have the same trig functions in them. So the first Pythagorean theorem that says sine
squared plus cosine squared equals one, if the angles are the same...I should be probably
not be leaving theta out, says if I take one minus sine squared theta and substitute using
that first pythagorean identity...one minus sine square theta is equal to cosine squared
theta. Now look. We have one term on top and it is cosine. I have one term on top and it
is cosine squared. I have a one minus sine on the bottom, and I have a one minus sine
on the bottom. We are sooooo close! Oh, I can cancel these factors out. One of those
cosines on the bottom, that factor...it is multiplied to something else and not added,
that cosine on the bottom is going to cancel with one of the two cosines on top. LOOK AT
THERE! Cosine theta over one minus sine theta. Is that a match? Is that a match? YEAH IT
IS!!! Alright, let's do another one. I love this stuff. You just have to try different
things, do some different substitutions, and then all of a sudden the algebra starts going...
hey...hey...hey... aren't you getting close to the answer. We have got one plus cosecant
theta over secant theta minus cotangent of theta equals cosine. Now I have a lot going
on in the left hand side. Work your way down the tree and you will land or come down that
one trunk. Don't make it more complicated. Don't try and make this one term into two
terms and somehow add in cotangents and secants from just the simple cosine function. So start
with the left and figure it out. I only have cosine. I don't have cosecants, secants, or
cotangent so I am going to rewrite the left hand side in terms of sine and cosine right
off the bat. We have one plus one over sine theta all over one over cosine theta minus
cotangent which is cosine over sine. Now, that looks worse. We are going to have to
start cleaning up this first fraction and see if anything nice happens with this second
fraction when we are done. We are going to find a common denominator between one and
sine and that is going to be sine. So we are going to multiply this by sine theta on the
top and bottom. We have a common denominator now so I am going to do sine theta plus one
over sine all over cosine theta minus cosine theta over sine theta. Now this bottom fraction
is going to get flipped up and we are going to get, and this is going to sit here, we
are going to get sine theta plus one over sine and the cosine got flipped up...oops.
I mis-copied it. That was one over cosine theta, sorry about that. That is going to
get flipped up and be sine theta plus one times cosine over sine. Again, this bottom
gets flipped up. The second fraction already has a sine in the denominator. Now look. We
have got sine theta plus one times cosine minus cosine over sine and our final goal
is to have nothing but cosine left. If I take out the cosine from these two terms because
these are going to come together as one big fraction. Let me just write it like that since
we have a common denominator of sine. These two terms both have a cosine, so if I take
that cosine and I factor it out... Sine plus one times cosine with the cosine out is going
to be sine theta plus one. And if I take the cosine out of this term, cosine again divided
by cosine is one but it is minus one all over sine. Well I am running out of room be we
are also almost done. One minus one is zero. Now we have cosine times sine over sine of
theta. I almost made a mistake, but we did finish in time. BAM! The sines cancel out
and we get the cosine of theta. Which now that I am in a hurry I am making a couple
of mistakes here, let me fill that back in. BAM!!! Cosine theta from the left equals cosine
theta from the right. I am Mr. Tarrou. Go do your homework:D