For more information, please see full course syllabus of Trigonometry
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Addition and Subtraction Formulas
Main formulas:
 Addition and subtraction formulas
cos(A −B) = cosA cosB + sinA sinB cos(A + B) = cosA cosB − sinA sinB sin(A −B) = sinA cosB − cosA sinB sin(A + B) = sinA cosB + cosA sinB  Cofunction identities
cos ( π 2−x ) = sinx sin ( π 2−x ) = cosx
Example 1:
Derive the formula for cos(A−B) without using the other addition and subtraction formulas.Example 2:
Use the addition and subtraction formulas to find the cosines and sines of (π/12)^{R} and 105^{°} .Example 3:
Use the addition and subtraction formulas to prove the following identity:

Example 4:
Use the formula for cos(A−B) and the cofunction identities to derive the other three addition and subtraction formulas.Example 5:
Convert 75^{°} and 15^{°} to radians and use the addition and subtraction formulas to find their cosines and sines.Addition and Subtraction Formulas
 210^{°} = 30^{°} + 180^{°}
 cos(A + B) = cosAcosB  sin AsinB, sin(A + B) = sinAcosB + cosAsinB
 cos 210^{°} = cos30^{°}cos180^{°}  sin30^{°}sin180^{°}
 cos210^{°} = ( [(√3 )/2] ) ( − 1 ) − ( [1/2] )(0) ⇒ cos210^{°} = − [(√3 )/2]
 sin210^{°} = sin30^{°}cos180^{°} + cos30^{°}sin180^{°}
 sin210^{°} = ( [1/2] ) ( − 1 ) + ( [(√3 )/2] )(0) ⇒ sin210^{°} = − [1/2]
 [(3π)/4] = π− [(π)/4]
 cos(A  B) = cosAcosB + sin AsinB, sin(A  B) = sinAcosB  cosAsinB
 cos[(3π)/4] = cosπcos[(π)/4] + sinπsin[(π)/4]
 cos[(3π)/4] = ( − 1 ) ( [(√2 )/2] ) + (0)( [(√2 )/2] ) ⇒ cos[(3π)/4] = − [(√2 )/2]
 sin[(3π)/4] = sinπcos[(π)/4]  cosπsin[(π)/4]
 sin[(3π)/4] = ( 0 ) ( [(√2 )/2] ) − ( − 1 ) ( [(√2 )/2] ) ⇒ sin[(3π)/4] = [(√2 )/2]
 285^{°} = 135^{°} + 150^{°}
 sin(A + B) = sinAcosB + cosAsinB
 sin285^{°} = sin135^{°}cos150^{°} + cos135^{°}sin150^{°}
 sin285^{°} = ( [(√2 )/2] ) ( − [(√3 )/2] ) + ( − [(√2 )/2] ) ( [1/2] )
 sin285^{°} = − [(√6 )/4] + ( − [(√2 )/4] ) = [( − √6 − √2 )/4]
 [(7π)/12] = [(π)/4] + [(π)/3]
 cos(A + B) = cosAcosB  sin AsinB
 cos[(7π)/12] = cos[(π)/4]cos[(π)/3] − sin[(π)/4]sin[(π)/3]
 cos[(7π)/12] = ( [(√2 )/2] ) ( [1/2] ) − ( [(√2 )/2] ) ( [(√3 )/2] )
 cos[(7π)/12] = [(√2 )/4] − [(√6 )/4]
 sin(A  B) = sinAcosB  cosAsinB, where A = [(7π)/6] and B = [(π)/3]
 sin( [(7π)/6] − [(π)/3] ) = sin[(7π)/6]cos[(π)/3]  cos[(7π)/6]sin[(π)/3]
 sin( [(7π)/6] − [(π)/3] ) = ( − [1/2] ) ( [1/2] ) − ( − [(√3 )/2] ) ( [(√3 )/2] )
 sin( [(7π)/6] − [(π)/3] ) = − [1/4] + [3/4] = [2/4] = [1/2]
 cos(A  B) = cosAcosB + sinAsinB, where A = 120^{°} and B = 30^{°}
 cos( 120^{°} − 30^{°} ) = cos120^{°}cos30^{°} + sin120^{°}sin30^{°}
 cos( 120^{°} − 30^{°} ) = ( − [1/2] ) ( [(√3 )/2] ) + ( [(√3 )/2] ) ( [1/2] )
 cos( 120^{°} − 30^{°} ) = − [(√3 )/4] + [(√3 )/4] = 0
 sin(A  B) = sinAcosB  cosAsinB
 195^{°} = 225^{°} − 30^{°}
 sin 195^{°} = sin225^{°}cos30^{°} − cos225^{°}sin30^{°}
 sin 195^{°} = − sin45^{°}cos30^{°} + cos45^{°}sin30^{°}
 sin 195^{°} = ( − [(√2 )/2] ) ( [(√3 )/2] ) + ( [(√2 )/2] ) ( [1/2] )
 sin 195^{°} = − [(√6 )/4] + [(√2 )/4]
 [(11π)/12] = [(3π)/4] + [(π)/6]
 cos(A + B) = cosAcosB  sin AsinB
 cos[(11π)/12] = cos[(3π)/4]cos[(π)/6]  sin[(3π)/4]sin[(π)/6]
 cos[(11π)/12] = ( − [(√2 )/2] ) ( [(√3 )/2] ) − ( [(√2 )/2] ) ( [1/2] )
 cos[(11π)/12] = [( − √6 )/4] − [(√2 )/4]
 255^{°} = 225^{°} + 30^{°}
 sin(A + B) = sinAcosB + cosAsinB
 sin255^{°} = sin225^{°}cos30^{°} + cos225^{°}sin30^{°}
 sin255^{°} = ( − [(√2 )/2] ) ( [(√3 )/2] ) + ( − [(√2 )/2] ) ( [1/2] )
 sin255^{°} = − [(√6 )/4] + ( − [(√2 )/4] ) = [( − √6 − √2 )/4]
 [(17π)/12] = [(3π)/4] − [(π)/6]
 cos(A  B) = cosAcosB + sin AsinB
 cos[(17π)/12] = cos[(9π)/4]cos[(5π)/6] + sin[(9π)/4]sin[(5π)/6]
 cos[(17π)/12] = ( [(√2 )/2] ) ( − [(√3 )/2] ) + ( [(√2 )/2] ) ( [1/2] )
 cos[(17π)/12] = [( − √6 )/4] + [(√2 )/4]
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Answer
Addition and Subtraction Formulas
Lecture Slides are screencaptured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
 Intro
 Addition and Subtraction Formulas
 Cofunction Identities
 Example 1: Derive the Formula for cos(AB)
 Example 2: Use Addition and Subtraction Formulas
 Example 3: Use Addition and Subtraction Formulas to Prove Identity
 Extra Example 1: Use cos(AB) and Cofunction Identities
 Extra Example 2: Convert to Radians and use Formulas
 Intro 0:00
 Addition and Subtraction Formulas 0:09
 How to Remember
 Cofunction Identities 1:31
 How to Remember Graphically
 Where to Use Cofunction Identities
 Example 1: Derive the Formula for cos(AB) 3:08
 Example 2: Use Addition and Subtraction Formulas 16:03
 Example 3: Use Addition and Subtraction Formulas to Prove Identity 25:11
 Extra Example 1: Use cos(AB) and Cofunction Identities
 Extra Example 2: Convert to Radians and use Formulas
Trigonometry Online Course
I. Trigonometric Functions  

Angles  39:05  
Sine and Cosine Functions  43:16  
Sine and Cosine Values of Special Angles  33:05  
Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D  52:03  
Tangent and Cotangent Functions  36:04  
Secant and Cosecant Functions  27:18  
Inverse Trigonometric Functions  32:58  
Computations of Inverse Trigonometric Functions  31:08  
II. Trigonometric Identities  
Pythagorean Identity  19:11  
Identity Tan(squared)x+1=Sec(squared)x  23:16  
Addition and Subtraction Formulas  52:52  
Double Angle Formulas  29:05  
HalfAngle Formulas  43:55  
III. Applications of Trigonometry  
Trigonometry in Right Angles  25:43  
Law of Sines  56:40  
Law of Cosines  49:05  
Finding the Area of a Triangle  27:37  
Word Problems and Applications of Trigonometry  34:25  
Vectors  46:42  
IV. Complex Numbers and Polar Coordinates  
Polar Coordinates  1:07:35  
Complex Numbers  35:59  
Polar Form of Complex Numbers  40:43  
DeMoivre's Theorem  57:37 
Transcription: 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(AB) 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(AB).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(AB) 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(AB) 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(AB).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/2A). 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 AB, I have (pi2)(AB).0263
So, this is cos of the first term, cos(pi/2A), cos of the second term is B + sin of the first term x sin of the second term.0273
But now, cos(pi/2A) again using the co function identity is just sin A, sin A cos B.0293
Now, sin(pi/2A) 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(AB) 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(AB) 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(AB) and then we made some clever substitutions to figure out cos(A+B), sin(A+B), sin(AB).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(AB).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(AB) is(cos A cos B) + (sin A sin B) and sin(AB) 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/12pi/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(ab), cos(a+b), sin(ab), 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(ab) 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(ab).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 90degree 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(ab) 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(ab).0233
Then in our later example, we'll show how you can work out all the others just from knowing the cos(ab).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 (ab) is the difference between them.0293
This arc between them is going to be (ab).0296
That's (ab) 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 xcoordinate, and cos(b), the ycoordinate.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 xcoordinates.0350
So, (x_{2}x_{1})^{2}+(y_{2}y_{1})^{2}, you add those together and you take the square root of the whole thing.0365
That's the distance formula.0379
Here, the x_{2} and the x_{1} 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 d^{2}=(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 (ab), exactly equal to an angle of the size (ab), which means that that point right there has coordinates (cos(ab),sin(ab)).0481
That point has coordinates (cos(ab),sin(ab)).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(ab).0535
Now, the old xcoordinate is just 1 because I'm looking at the point (1,0).0543
That quantity squared plus the change in y coordinates, sin(ab) minus, the old ycoordinate is 0, squared.0547
Then I take the square root of the whole thing.0562
I'm going to square both sides, d^{2}=(cos(a)1)^{2}+(sin^{2}(ab)).0566
What I'm going to do is look at these two different expressions here for d^{2}.0590
Well, they're both describing the same d^{2}, 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(ab)1)^{2}+(sin^{2}(ab)).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 (ab)^{2}=a^{2}2ab+b^{2}.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 cos^{2}(a)2cos(a)cos(b)+cos^{2}(b)+sin^{2}(a)2sin(a)sin(b)+sin^{2}(b)=cos^{2}(ab)2×1×cos(ab)+sin^{2}(ab).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, cos^{2}(a) and sin^{2}(a), that gives me 12cos(a)cos(b).0735
Now I have a cos^{2}(b) and a sin^{2}(b), so that's another 12sin(a)sin(b), is equal to, now look, cos^{2}(ab) and sin^{2}(ab), that's another 1.0749
It looks like I forgot one term on the line above when I was squaring out cos(ab)1^{2}, I got cos^{2}(ab)2cos(ab), then there should be +1^{2}, there's another 1 in there.0772
There's another 1 in there, 2cos(ab).0793
That's it because we already took care of the sin^{2}(ab) that got absorbed with the cos(ab).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(ab).0814
Now, if we cancel the 2s, look what we have.0838
We have exactly cos(a)×cos(b)+sin(a)×sin(b)=cos(ab).0840
That's the formula for cos(ab).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, sin^{2}+cos^{2}=1.0944
It all reduced down into the formula for cos(ab).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(ab)=cos(a)×cos(b)+sin(a)×sin(b).1042
Here, the (ab) is π/12, so a and b are π/4 and π/6.1060
This is cos(π/4π/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(ab), 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 3x2x.1530
If we start with a=3x and b=2x, then 5x=a+b, and x itself is ab.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(ab).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 lefthand side we get sin(ab)+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(ab) is just sin(x), plus sin(a+b) which is 5x.1674
On the righthand 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 lefthand 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(ab) 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 lefthand side, I get cos(ab)+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 lefthand side of the identity.1832
I'll see if I can transform it into the righthand side.1836
The lefthand 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 righthand 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 (ab), 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 (ab) and (a+b), it's worthwhile writing down the sine and the cosine each one of (ab) 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 lefthand side of the identity, we work it down until we get to the righthand side of the identity.1997
We'll try some more examples of that later2004
1 answer
Last reply by: Dr. William Murray
Sat Aug 13, 2016 10:54 AM
Post by tae Sin on August 12, 2016
I know this question is frivolous  I don't mind anyone answering this question if they know this is possible, but I used a calculator when I was bored. And I spammed sincostansincostan() with some random value  I'm pretty sure I used a real number that didn't create undefined value. And I actually got a value, so can you somehow explain how this is possible? if it does relate to the addition and subtraction formulas, please explain them as well?
2 answers
Last reply by: Dr. William Murray
Wed Apr 27, 2016 4:51 PM
Post by Tania Torres on April 26, 2016
Regarding Iris Kim's question, "At 12:14, you wrote that (cos(AB)1)^2 equals cos(AB)^22cos(AB)... shouldn't it be cos(AB)^22cos(AB)+1?" and your response, why is it not '+ 1'?
2 answers
Last reply by: Dr. William Murray
Wed Jul 1, 2015 8:52 AM
Post by Iris Kim on June 30, 2015
At 12:14, you wrote that (cos(AB)1)^2 equals cos(AB)^22cos(AB)... shouldn't it be cos(AB)^22cos(AB)+1?
1 answer
Last reply by: Dr. William Murray
Mon Nov 17, 2014 8:26 PM
Post by olga shevchuk on November 16, 2014
THERE WAS A MISTAKE. IT WAS WRITTEN 2SIN(3X)COS(X) WHEN IT SHOULD HAVE BEEN 2SIN(3X)COS(2X) STARTING @28:10
1 answer
Last reply by: Dr. William Murray
Tue Aug 5, 2014 3:49 PM
Post by Jamal Tischler on July 23, 2014
Very good lesson. I apreciate you derived the formulas ! It helped me.
3 answers
Last reply by: Dr. William Murray
Mon Jun 23, 2014 7:44 PM
Post by Jeffrey Tao on June 21, 2014
In your response to Manfred Berger's question, you stated how it is possible to use the Euler's formula, e^ix=cosx+isinx, to prove the identities,as a way that did not use calculus. But from what I've learned, the derivation of the formula e^ix=cosx+isinx comes from power series, so doesn't this method of proving the identities still use calculus?
1 answer
Last reply by: Dr. William Murray
Tue Dec 10, 2013 11:32 PM
Post by Monis Mirza on December 7, 2013
Write an equivalent expression for sin(2m)cos(n)+ cos(2m) sin(n)
3 answers
Last reply by: Dr. William Murray
Thu Jul 18, 2013 8:20 AM
Post by Manfred Berger on June 28, 2013
Are you going to prove any of the addition formulas in Calc 2?
1 answer
Last reply by: Dr. William Murray
Fri Aug 31, 2012 5:26 PM
Post by Su Jung Leem on August 2, 2012
I know it's a irrelevant question but i wasn't sure where to ask this question. Does anyeone know how to add y+2 over y squared  y 2 and one over 3y+3 ? I keep on getting different answers every time I trying to answer this question. please help!!!
1 answer
Last reply by: Dr. William Murray
Sun May 12, 2013 5:21 PM
Post by Nathan Thomas on January 8, 2012
He didn't include the tangent sum difference formulas which is very important and shouldn't be skipped over.
tan(a + b) =
(tan a + tan b) / (1  (tan a)(tan b)
tan(a  b) =
(tan a  tan b) / (1 + (tan a)(tan b)
1 answer
Last reply by: Dr. William Murray
Sun May 12, 2013 5:19 PM
Post by Elina Bugar on August 23, 2011
how did he get the coordinates of angle a to be cosA,cosB
and for angle B (SinA, CosB)
3 answers
Last reply by: Dr. William Murray
Sun May 12, 2013 5:15 PM
Post by Marco Zendejo on June 22, 2011
Im kinda confuse in Example II.
How did Pie/12 turn into pie/4  pie/6
If anyone could explain this I'll be grateful.
2 answers
Last reply by: Dr. William Murray
Sun May 12, 2013 5:11 PM
Post by Judith Gleco on June 11, 2011
Hi,
I was wandering if anyone is having any problems with the recording glitching, or stopping and going back to the begining of the lesson. Help so I know if it is my computer.
7 answers
Last reply by: Dr. William Murray
Sun May 12, 2013 5:07 PM
Post by Mark Mccraney on January 15, 2010
Lecture 3, ex 1: shouldn't the coords written in blue be A=(cosA, sinB) vs A=(cosA, cosB)