Trigonometry Addition and Subtraction Formulas

Ok we are here to try more examples of the addition and subtraction formulas.0000

This time we are going to use the formula for cos(A-B) and the co function identities to derive the other three addition and subtraction formulas.0007

If you remember back in the previous set of examples, we proved the formula for cos(A-B).0016

We did it without using the other addition and subtraction formulas.0024

We are not getting trapped in any circular loops of logic.0028

We really did prove the cos(A-B) from scratch.0031

And now that we got that available to us, we are going to start with that formula and we are going to try to derive all the others.0035

Hopefully, it would be easier than the original proof of the cos(A-B) formula.0044

Let us remember what that formula is because we are allowed to use it now.0050

The cos(A+B) is equal to cos A cos B + sin A sin B, we are allowed use that.0055

I want to derive the other three formulas.0070

I'm going to start with cos(A+B) and I'm going to write that as cos((A-(-B)).0073

I'm gong to write in addition, in terms of a subtraction.0079

The point of that is now I can use my subtraction formula.0091

So, this is cos A. I'm just going to invoke this formula above except whenever I see a B, I will change it to (-B).0095

I have cos A cos(-B) + sin A sin(-B).0107

Remember, cosine is not even a function.0125

That means cos(-x) is the same as cos(x).0128

Sine is an odd function, sin(-x) is equal to -sin(x), I got cos A and cos(-B), but cos(-B) is the same as cos B.0137

Now sin A and sin -B, sine is odd so sin(-B) is -sin B.0156

But look, now I got cos(A+B) is equal to cos A cos B - sin A sin B, that is the formula for cos(A+B).0169

I was able to do that much more quickly than we were able to prove the original formula for cos(A-B).0180

Let us see how that works for sin(A+B).0188

Now, I'm going to have to bring in the co function identities, let me remind you what those are.0194

Those say that cos(pi/2)-x is the same as sin(X), sin(pi/2)-x is equal to cos(x).0198

Somehow we are going to use those to derive the sin formulas from the sin formulas.0215

The way we do that is I have sin(A+B), I'm going to use the first co function identity and write that as cos((pi/2-(A+B)).0221

That is by the first co function identity.0238

Now, that is cos(pi/2-A). I am going to group those two terms together and then -B, because it was minus the quantity of A+B.0241

I'm going to use my cos subtraction formula, this one that we started with.0255

cos, I am going to substitute n instead of A-B, I have (pi-2)-(A-B).0263

So, this is cos of the first term, cos(pi/2-A), cos of the second term is B + sin of the first term x sin of the second term.0273

But now, cos(pi/2-A) again using the co function identity is just sin A, sin A cos B.0293

Now, sin(pi/2-A) using the co function identity at the second co function identity is cos A x B.0305

Now we got the addition formula for sin, because we started with sin(A+B) and we reduced it down to sin A cos B + cos A sin B.0318

That is where the addition formula for sin comes in.0330

Finally, sin(A-B) we are going to do the same trick that we did for cos(A+B).0334

We will write this as sin, instead of writing it as a subtraction, we will think of it as adding a negative.0344

This is sin A - (-), I'm sorry A + (-B).0355

The point of that is that we can then invoke the sin formula that we just proved, we got sin(something +something).0365

According to the sin formula that we just proved, it is the sin of the first one x the cos of the second one which is (-B) + cos of the first one x sin of second one which is (-B).0372

Now again, we are going to use the odd and even properties.0390

This is sin A, cos (-B), cos does not even function so that is cos B + cos A.0394

Actually I should have said plus because look we have sin (-B) and sin (-x) is -sin(x). This is -cos A sin B.0406

But now, we started with sin(A-B) and we derived sin A cos B - cos A sin B.0418

That is exactly the subtraction formula for sin.0429

In each one of those identities, we did not use anything external.0434

We just started with the identity for cos(A-B) and then we made some clever substitutions to figure out cos(A+B), sin(A+B), sin(A-B).0438

Just making little substitutions into the one formula that we started with to get the formulas for the other three expressions.0454

Remember, it was a lot of work to prove that original formula for cos(A-B).0464

But once we have that one we can sort of milk it over and over again to get the other three formulas.0469

Extra example 2, which is to convert 75 degrees and -15 degrees to radians and we will use the addition and subtraction formulas to find the cos and sin.0000

So, 75 degrees, we will start out with that one.0014

Remember, the conversion formula is pi/180, that simplifies down to 5pi/12, that is not a common value.0020

I do not know the cos and sin of 5pi/12.0034

I‘m going to write that as a combination of two angles that I do know, that is (pi/4 + pi/6).0040

That is because pi/4 is 3pi/12 and pi/6 is 2pi/12 and you put them together and you got 5pi/12.0044

The key point of that is the pi/4 and pi/6 are common values.0060

I know the sin and cos(pi/4 and pi/6), I have memorized them and hopefully, you have memorized them as well.0066

Once I use my addition and subtraction formulas I can figure out the sin and cos of 5pi/12.0074

Let me remind you the addition and subtraction formulas we will be using.0081

Here, we are going to find cos(A+B) which is(cos A cos B) – (sin A sin B).0085

I’m going to go ahead and write the formula for sin(A+B).0103

It is equal to (sin A cos B) +(cos A sin B).0110

What is invoked those here we are trying to find the cos(5pi/12) which is the same as the cos(pi/4 + pi/6).0122

I’m going to use the cos addition formula cos(pi/4) cos(pi/6) – sin(pi/4) sin(pi/6).0138

All of those are common values, I have got those all computed to memory.0159

This will be very quick to finish from here.0163

This is cos(pi/4) I remember is square root of 2/2, cos(pi/6), I remember is square root of 3/2 – sin(pi/4) is root 2/2, sin(pi/6) is just ½.0166

If I put those together the common denominator there is 4, (root 2 x root 3 is 6) – (root 2 x 1).0182

That is my cos(5pi/12) which is the same as the cos of 75 degrees.0195

Let us find the sin now, sin(5pi/12) is equal to sin(pi/4) + pi/6, which by the addition formula for sin is sin of the first one pi/4, (cos of the second one pi/6) + (cos of the first one x sin of the second one).0202

And now again those are common values, I remember them all.0234

Sin(pi/4) is root 2/2, cos(pi/6) is root 3/2, cos(pi/4) is also root 2/2, sin(pi/6) is just ½.0237

I put these together over common denominator 4 and I get (root 6 + root 2/4).0257

What to remember those two values because we are actually going to use them in the next part.0271

The next part is to figure out -15 degrees, we want to start out by converting that to radians.0277

-15 degrees we multiply that by our conversion factor pi/180, that gives us 15/180, simplifies down to 112, so we get –pi/12 radians.0283

Now, there are two ways we could proceed from here. We can write –pi/12 as (pi/6 – pi/4) and that is because pi/6 is 2pi/12, pi/4 is 3pi/12. You subtract them and you will get –pi/12.0307

We could do at that way or we can write –pi/12 as (5pi/12 – pi/2- 6pi/12).0334

I want to do it that way because I want to practice that plus I think the sin and cos of pi/2 are a little bit easier to remember, I want to practice that method.0341

Let me write the formulas for sin and cos because we are going to be using those.0351

Cos(A-B) is(cos A cos B) + (sin A sin B) and sin(A-B) is equal to (sin A cos B) – (cos A sin B).0356

I’m going to be using those subtraction formulas the cos(-pi/12).0390

If we use the second version that is cos(5pi/12-pi/2).0399

And now by the subtraction formula that is (cos(5pi/12) x cos(pi/2)) +(sin(5pi/12) x sin(pi/2)).0408

Now look at this, the cos(pi/2), remember that is cos 90 degrees, the x coordinate of 90 degrees angle that is 0.0431

That whole term drops out, sin(pi/2) is 1.0441

This whole thing simplifies down to sin(5pi/12) and we worked that out on the previous page.0447

The sin(pi/12) we did this work before, that was the (square root of 2) + (square root of 6)/4.0460

We are invoking previous work there, this would be something that I would not have remembered but because I just work that out in the previous problem I remember the answer now.0470

We are going to try to figure out the sin(-pi/12) the same way.0485

So, sin(-pi/12) is the same as sin(5pi/12) – pi/2, because it is (5pi/12 – 6pi/12).0489

Using the subtraction formula for sin, that is sin of the first one, which is sin(5pi/12) x cos of the second one (pi/2) – cos of the first one (5pi/12) x sin of the second one (pi/2).0506

The point of that is that the pi/2 values are very easy.0526

I know that the cos, just like before is 0 and the sin is 1.0529

This whole thing simplifies down to –cos(5pi/12).0539

Again, I worked out the cos(5pi/12) on the previous page, the cos(5pi/12) in the previous page was (root 6 – root 2)/4.0550

Or we want the negative of that this time, I will just switch those around and I will get root 2 – root 6 divided by 4.0565

The key to doing that problem, well first of all, converting those angle to radians, that is a simple conversion factor of pi/180, that part was fairly easy.0583

Once we figured out how to convert those angle to radians, it was a matter of writing them as either sums or differences using addition or subtraction of common values, pi/6, pi/4, pi/3.0593

Things that you already know the sin and cos of by heart.0608

75 degrees 5pi/12, the key there was to figure out that was pi/4 + pi/6 and then know that you remember the common values, the sin and cos(pi/4) and pi/6.0614

So you can work out the sin and cos of 5pi/12, the -15 degrees converted into –pi/12 and then we can write that as pi/6 – pi/4, that would be one way to do it.0630

Or since we already know the sin and cos of 5pi/12, it is a little bit easier to write it as 5pi/12 – pi/2.0646

Then we can use the addition and subtraction formulas which because of the pi/2, essentially reduced it down to knowing the sin and cos of 5pi/12, which we figured out on the previous page.0654

So, that is how you use the addition and subtraction formulas to find the values of sin and cos of other angles when you already know the sin and cos of the common values.0668

That is the end of the lecture on addition and subtraction formulas.0681

We will use these formulas later on to find the double and half angle formulas that is coming next in the trigonometry lectures on www.educator.com.0684

Hi this is Will Murray for educator.com and we're talking about the addition and subtraction formulas for the sine and cosine functions.0000

The basic formulas are all listed here.0009

We have a formula for cos(a-b), cos(a+b), sin(a-b), and sin(a+b).0012

Unfortunately, you really need to memorize these formulas but it is not quite as bad as it looks.0021

In fact, if you can just remember one each for the cosine and the sine, maybe if you can remember cos(a+b) and sin(a+b), we'll learn later on in the lecture that you can work out the other formulas just by making the right substitution into those starter formulas.0026

If you remember what cos(a+b) is then you can substitute in -b in the place of b, and you can work out what the cos(a-b) is.0047

The same for sin(a+b), if you can remember the formula for sin(a+b), you can substitute in -b for b and find out the formula for sin(a-b).0058

You do have to remember a couple of formulas to get started, but after that you can work out the other formulas by some basic substitutions.0070

It's not as bad as it might sound in terms of memorization here.0078

There's a couple of cofunction identities that we're going to be using as we prove and apply the addition and subtraction formulas.0082

It's good to remember that cos(π/2 - x) is the same as sin(x).0092

The similar identity sin(π/2 - x) is equal to cos(x).0098

Those aren't too hard to remember if you kind of keep a graphical picture in your head.0104

Let me show you how those work out.0111

Let me draw an angle x here.0113

Then the cosine and sine, remember the x and y coordinates of that angle.0117

That's the cosine, and that's the sine.0125

And π/2 - x, well π/2 - x, remember of course is a 90-degree angle, so π/2 - x, we just go back x from π/2.0128

There's x and then that right there is π/2 - x.0139

If we write down the cosine and sine of π/2 - x, this is the same angle except we just switch the x and y coordinates.0147

When you go from x to π/2 - x, you're just switching the sine and cosine.0157

That's kind of how I remember that cos(π/2 - x)=sin(x) and sin(π/2 - x)=cos(x).0164

We'll be using those cofunction identities, both to prove the addition and subtraction formulas later on, and also to figure out the sines and cosines of new angles as a quicker in using these addition and subtraction formulas.0172

Let's get some examples here.0189

The first example is to derive the formula for cos(a-b) without using the other addition and subtraction formulas.0191

There's a key phrase here, it says, without using the other formulas.0199

The point of that is that once you figure out that one of these formulas, you can figure out a lot of the other formulas from the first one.0204

If you can figure out one formula, you need one formula to get started because otherwise you kind of get in a circular logic clue.0213

You need one of these formulas to get started and we'll have to go a bit of work to prove that.0221

Figuring out the other formulas from the first one turns out not to be so difficult.0227

What we'll do is we'll work out the formula for cos(a-b).0233

Then in our later example, we'll show how you can work out all the others just from knowing the cos(a-b).0238

This is a bit of a trick, it's probably not something that you would easily think about.0246

It really takes a little bit of ingenuity to prove this.0250

We'll start with a unit circle.0254

There's my unit circle.0269

I'm going to draw an angle a and an angle b.0272

I'm going to draw an angle a over here, so there's a, this big arc here.0274

I'll draw a b a little bit smaller, so there's b.0283

Then (a-b) is the difference between them.0293

This arc between them is going to be (a-b).0296

That's (a-b) in there.0299

Now, I want to write down the coordinates of each of those points.0302

The coordinates there, I'll write them in blue, are cos(a), the x-coordinate, and cos(b), the y-coordinate.0307

That's the coordinates of endpoint of angle a.0318

In red, I'm going to write down the coordinates of the endpoint of angle b, cos(b), sin(b).0324

Now, I'm going to connect those two points up with a straight line.0338

I want to figure out what the distance of that line is, and I'm going to use the Pythagorean formula.0345

Remember, the distance formula that comes from the Pythagorean formula is you look at the differences in the x-coordinates.0350

So, (x₂-x₁)²+(y₂-y₁)², you add those together and you take the square root of the whole thing.0365

That's the distance formula.0379

Here, the x₂ and the x₁ are the cosines, so my distance is cos(b)-cos(a).0384

Actually, I think I'm going to write that the other way around, this cos(a)-cos(b).0402

It doesn't matter which way I write it because it's going to be squared anyway.0411

Plus [sin(a)-sin(b)]², then I'll have to take the square root of the whole thing.0416

To get rid of the square root, I'm going to square both sides.0429

I get d²=(cos(a)-cos(b))²+(sin(a)-sin(b))².0435

That's one way of calculating that distance.0452

Now, I'm going to do something a little different.0455

I'm going to take this line segment d and I'm going to move it over, move it around the circle so that it starts down here at the point (1,0).0458

There's that line segment again.0475

Remember, the line segment was cutting off an arc of the circle exactly equal to (a-b), exactly equal to an angle of the size (a-b), which means that that point right there has coordinates (cos(a-b),sin(a-b)).0481

That point has coordinates (cos(a-b),sin(a-b)).0513

Now, I'm going to apply the distance formula, again, to the new line segment in the new place.0519

That says, again, the change in the x coordinates plus the change in the y coordinates, square each one of those and add them up and take the square root.0525

So, d is equal to change in x coordinates, that's cos(a-b).0535

Now, the old x-coordinate is just 1 because I'm looking at the point (1,0).0543

That quantity squared plus the change in y coordinates, sin(a-b) minus, the old y-coordinate is 0, squared.0547

Then I take the square root of the whole thing.0562

I'm going to square both sides, d²=(cos(a)-1)²+(sin²(a-b)).0566

What I'm going to do is look at these two different expressions here for d².0590

Well, they're both describing the same d², they must be equal to each other.0597

That was kind of the geometric insight to figure out to get me an algebraic equation setting a bunch of things equal to each other.0608

From here on, it's just algebra, so we're going to set these two equations equal to each other.0616

The first one is (cos(a)-cos(b))²+(sin(a)-sin(b))² is equal to the second one, (cos(a-b)-1)²+(sin²(a-b)).0622

Now, I'm just going to manipulate this expression expanded out, cancel few things and it should give us the identity that we want.0652

Remember, the square formula (a-b)²=a²-2ab+b².0659

We're going to be using that a lot because we have a lot of squares of differences.0671

On the first term we have cos²(a)-2cos(a)cos(b)+cos²(b)+sin²(a)-2sin(a)sin(b)+sin²(b)=cos²(a-b)-2×1×cos(a-b)+sin²(a-b).0679

Now, there's a lot of nice ways to invoke the Pythagorean identity here.0727

If you look at this term, and this term, cos²(a) and sin²(a), that gives me 1-2cos(a)cos(b).0735

Now I have a cos²(b) and a sin²(b), so that's another 1-2sin(a)sin(b), is equal to, now look, cos²(a-b) and sin²(a-b), that's another 1.0749

It looks like I forgot one term on the line above when I was squaring out cos(a-b)-1², I got cos²(a-b)-2cos(a-b), then there should be +1², there's another 1 in there.0772

There's another 1 in there, -2cos(a-b).0793

That's it because we already took care of the sin²(a-b) that got absorbed with the cos(a-b).0800

There's a lot of terms that will cancel now.0807

The 1s will cancel, 1, 1, 1 and 1, those will cancel.0809

We're left with -2, I'll factor that out, cos(a)×cos(b)+sin(a), because I factored out the -2, sin(b), is equal to -2cos(a-b).0814

Now, if we cancel the -2s, look what we have.0838

We have exactly cos(a)×cos(b)+sin(a)×sin(b)=cos(a-b).0840

That's the formula for cos(a-b).0860

That was really quite tricky.0864

The key element to that is that we did not use the other addition and subtraction formulas.0867

We really derived this from scratch, which means that we can use this as our starting point.0873

Later on, we'll derive the other addition and subtraction formulas but we'll be able to use this one to get started.0878

The others will be a lot easier.0884

This one was trickier because we really had to later on geometric ideas from scratch.0887

What we did was we graphed this angle a and angle b.0892

We connected them up with this line segment d, and we found the length of that line segment using the Pythagorean distance formula.0897

Then we did this very clever idea of translating and moving that line segment d over, so that it had a base of one endpoint at (1,0).0905

We found another expression for the length of that line segment or that distance, also using the Pythagorean distance formula but starting and ending at different places.0916

We get these two expressions for the length of that line segment d, and then we set them equal to each other in this line.0929

Then we got this sort of big algebraic and trigonometric mess, but there was no more real geometric insight after that.0937

It was just a matter of sort of expanding out algebraically using the Pythagorean identity to cancel some things that kind of collapse together, sin²+cos²=1.0944

It all reduced down into the formula for cos(a-b).0957

Now, let's try applying the addition and subtraction formulas to actually find the cosines and sines of some values that we wouldn't have been able to do without these formulas.0964

In particular, we're going to find the cosine and sine of π/12 radians and 105 degrees.0975

Let's start out with cosine of π/12 radians.0982

Cos(π/12), that's not one of the common values.0987

I don't have that memorized, instead I'm going to write π/12 as a combination of angles that I do have the common values memorized for.0992

Here's the trick, remember π/12=π/4 - π/6, that's because π/4 is 3π/12 and π6 is 2π/12.1003

You subtract them, and you get π/12.1020

The reason I do it like that is that I know the sines and cosines for π/4 and π/6.1022

I can use my subtraction formulas to figure out what the cosine and sine of π/12 are in terms of π/4 and π/6.1030

I'm going to use my subtraction formula cos(a-b)=cos(a)×cos(b)+sin(a)×sin(b).1042

Here, the (a-b) is π/12, so a and b are π/4 and π/6.1060

This is cos(π/4-π/6), which is cos(π/4)×cos(π/6)+sin(π/4)×sin(π/6).1066

Now, π/4, π/6, those are common angles that I have those sines and cosines absolutely memorized so I can just come up with those very quickly.1095

The cos(π/4) is square root of 2 over 2.1105

The cos(π/6) is square root of 3 over 2.1109

The sin(π/4) is root 2 over 2.1113

The sin(π/6) is 1/2.1115

Those are values that I have memorized, you should have them memorized too.1117

Now, we simply combine these, root 2 times root 3 is root 6.1123

I see I'm going to have a common denominator of 4 here, and root 2 times 1 is just root 2.1130

That gives me the cos(π/12) as root 2 plus root 6 over 4.1139

I'm going to work out sin(π/12) very much the same way, it's the sine of (π/4) - (π/6).1145

I remember my formula for the sin(a-b), it's sin(a)×cos(b)-cos(a)×sin(b).1160

I'll just plug that in as sin(π/4), cos(b) is π/6, minus cos(π/4)×sin(π/6).1177

So, sin(π/4) is root 2 over 2, cos(π/6) is root 3 over 2, minus cos(π/4) is root 2 over 2, and sin(π/6) is 1/2.1196

Again, I have a common denominator of 4, and I get root 2 times root 3 is root 6, minus root 2.1212

What we did there was we just recognized that (π/12) is (π/4)-(π/6), and those are both common values that I know the sine and cosine of.1227

I can invoke my cosine and sine formulas to figure out what the cosine and sine are of (π/12).1238

Now, let's do the same thing with a 105 degrees.1246

We'll do everything in terms of degrees now.1250

I know that 105, well, to break that up to some common values that I recognize, that's 45+60.1254

I'm going to be using my addition formulas now.1264

I'll write those down to review them.1267

cos(a+b)=cos(a)×cos(b)-sin(a)×sin(b), and when I'm at it, I'll remember that the sin(a+b)=sin(a)×cos(b)+cos(a)×sin(b).1269

The cos(105), that's the same as cos(45+60).1300

Using the formulas with a as 45 and b as 60, I get cos(45)×cos(60)-sin(45)×sin(60).1310

Again, 45 and 60 are both common values, I've got the sines and cosines absolutely committed to memory, and hopefully you do too by the time you've gotten this far in trigonometry.1328

Cos(45) is square root of 2 over 2, cos(60) is 1/2, sin(45) is square root of 2 over 2, and the sin(60) is root 3 over 2.1338

I'll put those together.1353

Common denominator is 4, and I get square root of 2 minus the square root of 6, as my cos(105).1355

Sin(105) works very much the same way.1366

We'll write that as sin(45+60), which is sin(45)×cos(60)+cos(45)×sin(60).1371

Now, I'll just plug in the common values that I have committed to memory, root 2 over 2, cos(60) is 1/2, plus cos(45) is root 2 over 2, and sin(60) is root 3 over 2.1391

Common denominator there is 4.1408

This is root 2 over 2 plus root 6 over 4.1413

That was a matter of recognizing that 105 degrees.1422

It's not a common value itself but we can get it from the common values as 45+60.1426

Those both are common values, so I know the sines and cosines, so I can figure out what the sine and cos of 105 is, by using my addition and subtraction formulas.1434

I'll mention one more thing there which is that we could write 105.1443

If we convert that into radians, that's 7π/12 radians.1450

Remember, the way to convert back and forth is you just multiply by π/180.1455

Then, 7π/12, well that's the same as 6π/12, otherwise known as π/2 + 1π/12.1462

We figured out what the sine and cosine of π/12 were on the previous page.1473

Once you know the sine and cosine of π/12, you could work out the sine and cosine of 7π/12 by doing an addition formula on π/2 + π/12.1482

This is really an alternate way we could have solved this problem.1494

Given that we had already figured out the sine and cosine of π/12.1504

Let's try another example there.1509

We're going to use the addition and subtraction formulas to prove a trigonometric identity sin(5x)+sin(x) over cos(5x)+cos(x) is equal to tan(3x).1512

It really may not be obvious how to start with something like this.1525

The trick here is to write 5x, to realize 5x is 3x+2x, and x itself is 3x-2x.1530

If we start with a=3x and b=2x, then 5x=a+b, and x itself is a-b.1544

That's what the connection between this identity and the addition and subtraction formulas is.1561

We're going to use the addition and subtraction formulas to prove this identity.1566

Let me write them down now and show how we can combine them in clever ways.1571

I'm going to write down the formula for sin(a-b).1574

Remember, that's sin(a)×cos(b)-cos(a)×sin(b).1580

Right underneath it, I'll write the formula for sin(a+b) which is the same formula sin(a)×cos(b)+cos(a)×sin(b).1590

Now, I'm going to do something clever here.1608

I'm going to add these two equations together.1613

The point of that is to make the cos(a)×sin(b) cancel.1617

If we add these equations together, on the left-hand side we get sin(a-b)+sin(a+b).1623

Remember, you're thinking in the back of your head, a is going to be 3x and b is going to be 2x.1638

On the left side, we really got now sin(x)+sin(5x), which is looking good because that's what we have in the identity.1644

On the right side, we get 2sin(a)×cos(b), and then the cos(a)×sin(b), they cancel.1650

That was the cleverness of adding these equations together.1667

We get 2sin(a)×cos(b).1669

If I plug in a=3x and b=2x, I will get sin(a-b) is just sin(x), plus sin(a+b) which is 5x.1674

On the right-hand side, I'll get 2sin(a) is 3x, cos(b) is x.1692

That seems kind of hopeful because that's something I can plug in to the left-hand side of my identity and see what happens with it.1703

Before we do that though, I'm going to try and work out a similar kind of formula with the addition and subtraction formulas for cosine.1710

Let me write those down.1718

Cos(a-b) is equal to cos(a)×cos(b) plus, cosine is the one that switches the positive and the negative, plus sin(a)×sin(b).1720

I wanted to figure out cos(a+b).1743

It's just the same thing changing the positives and negatives, so cos(a)×cos(b)-sin(a)×sin(b).1749

I'm going to do the same thing here, I'm going to add them together in order to make them cancel nicely.1760

On the left-hand side, I get cos(a-b)+cos(a+b)=2cos(a)×cos(b).1767

That's it, because the sin(a) and sin(b) cancel with each other.1785

I'm going to plug in a=3x and b=2x, so I get cos(x) plus cos(a+b) is 5x, is equal to 2 cosine, a is 3x, and b is 2x.1791

Let's keep this in mind, I've got an expression for sin(x)+sin(5x), and I've got an expression for cos(x)+cos(5x).1810

I'm going to combine those and see if I can prove the identity.1823

I'll start with the left-hand side of the identity.1832

I'll see if I can transform it into the right-hand side.1836

The left-hand side is sin(5x)+sin(x) over cos(5x)+cos(x).1841

Now, by what we did on the previous page, I have an expression for sin(5x)+sin(x), that's sin(3x)×cos(2x).1860

That's by the work we did on the previous page.1881

Also on the previous page, cos(5x)+cos(x)=2cos(3x)×cos(2x).1890

That was also what we did on the previous page.1901

But now look at this, the cos(2x) is cancelled, and what we get is 2sin(3x) over 2cos(3x).1906

The 2s cancel as well and we get just tan(3x), which is equal to the right-hand side.1920

We finished proving it.1929

The trick there and it really was quite a bit of cleverness that might not be obvious the first time you try one of these problems, but you'll practice more and more and you'll get the hang of it, is to look at this 5x and x, and figure out how to use those in the context in the addition and subtraction formulas.1933

The trick is to let a=3x and b=2x, and the point of that is that (a-b), will then be x, and a+b will be 5x.1951

That gives us the expressions that we had in the identity here.1967

Once we see (a-b) and (a+b), it's worthwhile writing down the sine and the cosine each one of (a-b) and (a+b), and kind of looking at those formulas and kind of mixing and matching them, and finding something that gives us something that shows up in the identity.1974

Once we get that, we start with the left-hand side of the identity, we work it down until we get to the right-hand side of the identity.1997

We'll try some more examples of that later2004