BAM!!!
Mr. Tarrou.
Let's do some proofs involving sum and difference identities and... you know... see what they
look like.
Ok.
So our first proof in this example video is going to be the sine of x+3pi/2 and we are
going to verify that this is equal to the negative cosine of x.
This isn't really going to be much of a proof.
It almost more a practice of you knowing your unit circle.
So let's apply the sum and difference... or the sum identity for sine and see if we can
get the left hand side to match the negative cosine of x.
Well, the sine of two angles added together is the sine of the first times cosine of the
second plus... you keep the sign the same for the sum identity for sine... plus... the
sine of the first... cosine of the second...
plus cosine of the first... sine of the second.
So we apply that sum identity and start looking at the unit circle for what we have up here.
Now the cosine of 3pi/2, remember that is 1, 2, 3pi/2.
So the cosine down at 270 degrees is going to be 0.
So this is sin(x) times 0 which of course is going to completely cancel out.
And that is plus the cosine of x... and the sine of 3pi/2.
Remember we are down at 270 degrees.
The sine of 3pi/2 is -1.
And -1 times cosine is... well... -cos(x).
So there is our match and not much of a proof or a verification but we have proven that
identity.
Alright, let's do something with a bit more meat to it.
How about the sine of x+h minus the sine of x all over h.
That is supposed to equal the cosine of x times sine of h over h... plus sine of x times
cosine of h minus 1 over h.
Ok, that was a good bit to write out.
So whenever you do a proof remember you want to work your way down the tree.
You want to simplify.
You want to try come down that singular trunk if you can.
So both of these sides look kind of complicated.
But over here we have two term and over here we only have one.
So I am going to try to put these two terms together and see if we can get it to match
the left hand side.
I can't do that yet...
well I guess I kind of could... but we are going to take the cosine of x which is over
1 and multiply it to the first fraction.
We it kind of already looks like it is together and it really won't change hardly at all.
But now we have the sin(x) times cos(h)-1 over h.
You see this sine is really not part of the fraction.
It is out front.
So it is going to be placed over 1.
Also what you will have to remember is fraction bars act as a grouping symbol.
As well as meaning division, they act as a grouping symbol.
So it is going to be the sine of x, and I am going to have to distribute it through
those parenthesis when I put these together before I actually add the fractions.
You know, you multiply before you add.
Simple order of operations.
So I will put the cosine of x over 1.
Go ahead and multiply these together.
This is really basically done.
I am just going to rewrite it.
So we have the cos(x)sin(h) over h... plus... now I am going to take that sin(x) and distribute
it through the parenthesis I have added now because of the fraction bar.
We have sin(x)cos(h) minus sin(x).
That is all over h.
Now my two terms have exactly the same denominator.
So because I have limited space here, I have exactly matching denominators, so I am going
to go ahead and erase that and put this entire right hand side over a single denominator
of h.
Now something that I have not done that could be an issue is this plus sign.
Now there are two terms over here so if...
You know you should get in the habit of putting that second fraction or that second numerator
inside parenthesis to make sure that if you need to you distribute that negative sign
through.
Or of course, I could have just said I was taking positive sine of x and multiply it
through the parenthesis and getting positive times positive is positive sinxcosh and positive
sinx times -1 is negative sinx.
Ok, so this proof if fairly close to being done because... and I only have the one board
here to work on... so you need to make sure that you have your sum and difference identities
written on the side for notes...
The sin(h) times the cos(x) plus sin(x)cos(h), this is a little bit turned around from maybe
you have your identity written, but multiplication is commutative, so if I move these around
and reverse the order and say sin(h)cos(h) that would still be the same answer...
5 times 3 is 15 and 5 times 3 is 15.
So, this is the sum identity for sine.
So actually just by adding these two fractions together and making the common denominator,
I am already done.
This part of the identity is equal to the sin(x+h), and then minus the sin(x) over h.
And BAM!
I am done.
So three line proof.
Nice and short.
A lot of writing, but not a lot of steps.
Ok, the sum and difference identities as well...
they tend to be somewhat shorter in length, at least with the PreCalculus books that I
am using.
So our next example is going to be the sine of alpha plus beta.
That is over cosine alpha and cosine beta.
That is supposed to equal the tangent of alpha plus tangent of beta.
Ok.
So let's see if we can get these two sides to match.
Well, Lordy Lordy...
Let's see one term... two terms.
Right now as it stands there really is no algebra that we can work on.
This proof is really not going to be any longer either way.
We can start on the left hand side and apply the sum identity for sine, or because the
left hand side is all in terms of sine and cosine we can take tangent and write those
in terms of sine and cosine and just all the algebra process of finding common denominators
and what not... just allow the proof to fall out.
I am going to work on this side because I have all this room to fill out.
But with this proof I would say that working with the left or right hand side first, either
one is just pretty much the same.
So the tangent of alpha is going to be sinA/cosA plus the tangent of beta is sinB/cosB.
And here we have two terms and over here we only have one.
Remember as you are doing these identities you need to keep making sure that you look
at the goal.
You know, look at the other side of the identity and make sure that you know where you are
going.
Well, if I have two terms and I only need one I need common denominators.
These denominators, even though they both say cosine, have nothing in common because
of the different angle size.
So we are going to have to go over and multiply top and bottom by cosA.
Over here I am going to multiply top and bottom by cosB.
So again, these are just the value of 1 right.
I am just making common denominators.
So we have sinAcosB plus cosAsinB.
Trying to write this like you might see in your identity formula sheet.
Now over cosBcosA.
And sine of the first cosine of the second plus cosine of the first sine of the second.
Does that ring a bell?
This is the identity, the summation identity for sine.
So actually just writing tangent in terms of sine and cosine and finding those common
denominators, this proof is already done.
sinAcosB+cosAsinB with the alpha and betas is equal to sine(A+B) over cosBcosA.
And that matches exactly the left hand side so I have gotten this identity verified.
So BAM!
There you go.
Let's try one more and see if we have enough time.
Well let's see here.
We have the cot(A+B) equals the cotAcotB minus 1 over cotB+cotA.
Now you might be saying to yourself, I did not learn any sum or difference identities
for cotangent.
Remember that cotangent is just cosine over sine, so this side is equal to... or can be
rewritten as cos(A+B) over sin(A+B).
Let's see if we can get the right hand side to match that.
Well these are factors of sine and cosine, these are not.
So this is going to be cosine of A over the sine of A... cosB/sinB minus 1 all over again
cosB/sinB plus cosA/sinA.
Ok.
We are going to have to again...
Any time you have a bunch of... you start getting fractions over fractions... you are
going to make one fraction on top, one fraction on the bottom, and then flip it up.
So we are going to take sinAsinB and multiply it to the numerator over here and denominator.
Do the same thing with the denominator.
SinB... sinA... nothing in common.
So sinB comes over here top and bottom.
Same with the sinA, top and bottom.
And let's see what happens.
Ok cleaning all this up we have the cosAcosB-sinAsinB all over the common denominator of sinAsinB.
I really need to make sure I stop and pay attention to what I am writing.
I say so many words and variables in these.
Let me get these lights out of the way.
This over sinAcosB+cosAsinB all over our common denominator of sinAsinB.
AHHH look what is going to happen.
Our top fraction and our bottom fraction have exactly the same denominator.
So when I flip the bottom up, I don't actually need to show that step...
I mean that is what is going to happen.
We are going to flip the bottom fraction up and cancel, but because our top and bottom
denominators are EXACTLY the same those are just going to cancel out.
Nothing else will because you cannot cancel out terms.
So cosine of two angles minus the sine of those same two angles multiplied together.
This is the identity for the sum of two angles for the cosine function.
The cos(A+B).
And sinAcosB+cosAsinB, that is the pattern for the summation in sine.
So this is going to be the sin(A+B).
The last time I checked, that is what I had on the left hand side.
BAM!!!
Mr. Tarrou.
Go Do Your Homework!