WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi this is Will Murray for educator.com and we're talking about the addition and subtraction formulas for the sine and cosine functions.
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The basic formulas are all listed here.
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We have a formula for cos(a-b), cos(a+b), sin(a-b), and sin(a+b).
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Unfortunately, you really need to memorize these formulas but it is not quite as bad as it looks.
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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.
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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.
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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).
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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.
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It's not as bad as it might sound in terms of memorization here.
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There's a couple of cofunction identities that we're going to be using as we prove and apply the addition and subtraction formulas.
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It's good to remember that cos(π/2 - x) is the same as sin(x).
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The similar identity sin(π/2 - x) is equal to cos(x).
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Those aren't too hard to remember if you kind of keep a graphical picture in your head.
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Let me show you how those work out.
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Let me draw an angle x here.
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Then the cosine and sine, remember the x and y coordinates of that angle.
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That's the cosine, and that's the sine.
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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.
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There's x and then that right there is π/2 - x.
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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.
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When you go from x to π/2 - x, you're just switching the sine and cosine.
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That's kind of how I remember that cos(π/2 - x)=sin(x) and sin(π/2 - x)=cos(x).
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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.
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Let's get some examples here.
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The first example is to derive the formula for cos(a-b) without using the other addition and subtraction formulas.
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There's a key phrase here, it says, without using the other formulas.
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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.
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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.
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You need one of these formulas to get started and we'll have to go a bit of work to prove that.
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Figuring out the other formulas from the first one turns out not to be so difficult.
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What we'll do is we'll work out the formula for cos(a-b).
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Then in our later example, we'll show how you can work out all the others just from knowing the cos(a-b).
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This is a bit of a trick, it's probably not something that you would easily think about.
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It really takes a little bit of ingenuity to prove this.
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We'll start with a unit circle.
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There's my unit circle.
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I'm going to draw an angle a and an angle b.
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I'm going to draw an angle a over here, so there's a, this big arc here.
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I'll draw a b a little bit smaller, so there's b.
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Then (a-b) is the difference between them.
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This arc between them is going to be (a-b).
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That's (a-b) in there.
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Now, I want to write down the coordinates of each of those points.
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The coordinates there, I'll write them in blue, are cos(a), the x-coordinate, and cos(b), the y-coordinate.
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That's the coordinates of endpoint of angle a.
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In red, I'm going to write down the coordinates of the endpoint of angle b, cos(b), sin(b).
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Now, I'm going to connect those two points up with a straight line.
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I want to figure out what the distance of that line is, and I'm going to use the Pythagorean formula.
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Remember, the distance formula that comes from the Pythagorean formula is you look at the differences in the x-coordinates.
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So, (x₂-x₁)²+(y₂-y₁)², you add those together and you take the square root of the whole thing.
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That's the distance formula.
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Here, the x₂ and the x₁ are the cosines, so my distance is cos(b)-cos(a).
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Actually, I think I'm going to write that the other way around, this cos(a)-cos(b).
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It doesn't matter which way I write it because it's going to be squared anyway.
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Plus [sin(a)-sin(b)]², then I'll have to take the square root of the whole thing.
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To get rid of the square root, I'm going to square both sides.
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I get d²=(cos(a)-cos(b))²+(sin(a)-sin(b))².
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That's one way of calculating that distance.
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Now, I'm going to do something a little different.
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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).
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There's that line segment again.
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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)).
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That point has coordinates (cos(a-b),sin(a-b)).
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Now, I'm going to apply the distance formula, again, to the new line segment in the new place.
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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.
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So, d is equal to change in x coordinates, that's cos(a-b).
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Now, the old x-coordinate is just 1 because I'm looking at the point (1,0).
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That quantity squared plus the change in y coordinates, sin(a-b) minus, the old y-coordinate is 0, squared.
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Then I take the square root of the whole thing.
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I'm going to square both sides, d²=(cos(a)-1)²+(sin²(a-b)).
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What I'm going to do is look at these two different expressions here for d².
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Well, they're both describing the same d², they must be equal to each other.
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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.
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From here on, it's just algebra, so we're going to set these two equations equal to each other.
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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)).
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Now, I'm just going to manipulate this expression expanded out, cancel few things and it should give us the identity that we want.
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Remember, the square formula (a-b)²=a²-2ab+b².
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We're going to be using that a lot because we have a lot of squares of differences.
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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).
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Now, there's a lot of nice ways to invoke the Pythagorean identity here.
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If you look at this term, and this term, cos²(a) and sin²(a), that gives me 1-2cos(a)cos(b).
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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.
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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.
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There's another 1 in there, -2cos(a-b).
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That's it because we already took care of the sin²(a-b) that got absorbed with the cos(a-b).
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There's a lot of terms that will cancel now.
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The 1s will cancel, 1, 1, 1 and 1, those will cancel.
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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).
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Now, if we cancel the -2s, look what we have.
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We have exactly cos(a)×cos(b)+sin(a)×sin(b)=cos(a-b).
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That's the formula for cos(a-b).
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That was really quite tricky.
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The key element to that is that we did not use the other addition and subtraction formulas.
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We really derived this from scratch, which means that we can use this as our starting point.
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Later on, we'll derive the other addition and subtraction formulas but we'll be able to use this one to get started.
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The others will be a lot easier.
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This one was trickier because we really had to later on geometric ideas from scratch.
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What we did was we graphed this angle a and angle b.
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We connected them up with this line segment d, and we found the length of that line segment using the Pythagorean distance formula.
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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).
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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.
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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.
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Then we got this sort of big algebraic and trigonometric mess, but there was no more real geometric insight after that.
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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.
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It all reduced down into the formula for cos(a-b).
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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.
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In particular, we're going to find the cosine and sine of π/12 radians and 105 degrees.
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Let's start out with cosine of π/12 radians.
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Cos(π/12), that's not one of the common values.
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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.
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Here's the trick, remember π/12=π/4 - π/6, that's because π/4 is 3π/12 and π6 is 2π/12.
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You subtract them, and you get π/12.
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The reason I do it like that is that I know the sines and cosines for π/4 and π/6.
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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.
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I'm going to use my subtraction formula cos(a-b)=cos(a)×cos(b)+sin(a)×sin(b).
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Here, the (a-b) is π/12, so a and b are π/4 and π/6.
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This is cos(π/4-π/6), which is cos(π/4)×cos(π/6)+sin(π/4)×sin(π/6).
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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.
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The cos(π/4) is square root of 2 over 2.
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The cos(π/6) is square root of 3 over 2.
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The sin(π/4) is root 2 over 2.
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The sin(π/6) is 1/2.
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Those are values that I have memorized, you should have them memorized too.
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Now, we simply combine these, root 2 times root 3 is root 6.
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I see I'm going to have a common denominator of 4 here, and root 2 times 1 is just root 2.
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That gives me the cos(π/12) as root 2 plus root 6 over 4.
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I'm going to work out sin(π/12) very much the same way, it's the sine of (π/4) - (π/6).
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I remember my formula for the sin(a-b), it's sin(a)×cos(b)-cos(a)×sin(b).
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I'll just plug that in as sin(π/4), cos(b) is π/6, minus cos(π/4)×sin(π/6).
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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.
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Again, I have a common denominator of 4, and I get root 2 times root 3 is root 6, minus root 2.
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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.
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I can invoke my cosine and sine formulas to figure out what the cosine and sine are of (π/12).
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Now, let's do the same thing with a 105 degrees.
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We'll do everything in terms of degrees now.
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I know that 105, well, to break that up to some common values that I recognize, that's 45+60.
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I'm going to be using my addition formulas now.
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I'll write those down to review them.
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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).
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The cos(105), that's the same as cos(45+60).
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Using the formulas with a as 45 and b as 60, I get cos(45)×cos(60)-sin(45)×sin(60).
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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.
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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.
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I'll put those together.
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Common denominator is 4, and I get square root of 2 minus the square root of 6, as my cos(105).
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Sin(105) works very much the same way.
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We'll write that as sin(45+60), which is sin(45)×cos(60)+cos(45)×sin(60).
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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.
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Common denominator there is 4.
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This is root 2 over 2 plus root 6 over 4.
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That was a matter of recognizing that 105 degrees.
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It's not a common value itself but we can get it from the common values as 45+60.
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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.
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I'll mention one more thing there which is that we could write 105.
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If we convert that into radians, that's 7π/12 radians.
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Remember, the way to convert back and forth is you just multiply by π/180.
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Then, 7π/12, well that's the same as 6π/12, otherwise known as π/2 + 1π/12.
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We figured out what the sine and cosine of π/12 were on the previous page.
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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.
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This is really an alternate way we could have solved this problem.
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Given that we had already figured out the sine and cosine of π/12.
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Let's try another example there.
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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).
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It really may not be obvious how to start with something like this.
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The trick here is to write 5x, to realize 5x is 3x+2x, and x itself is 3x-2x.
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If we start with a=3x and b=2x, then 5x=a+b, and x itself is a-b.
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That's what the connection between this identity and the addition and subtraction formulas is.
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We're going to use the addition and subtraction formulas to prove this identity.
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Let me write them down now and show how we can combine them in clever ways.
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I'm going to write down the formula for sin(a-b).
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Remember, that's sin(a)×cos(b)-cos(a)×sin(b).
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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).
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Now, I'm going to do something clever here.
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I'm going to add these two equations together.
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The point of that is to make the cos(a)×sin(b) cancel.
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If we add these equations together, on the left-hand side we get sin(a-b)+sin(a+b).
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Remember, you're thinking in the back of your head, a is going to be 3x and b is going to be 2x.
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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.
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On the right side, we get 2sin(a)×cos(b), and then the cos(a)×sin(b), they cancel.
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That was the cleverness of adding these equations together.
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We get 2sin(a)×cos(b).
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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.
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On the right-hand side, I'll get 2sin(a) is 3x, cos(b) is x.
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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.
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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.
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Let me write those down.
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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).
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I wanted to figure out cos(a+b).
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It's just the same thing changing the positives and negatives, so cos(a)×cos(b)-sin(a)×sin(b).
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I'm going to do the same thing here, I'm going to add them together in order to make them cancel nicely.
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On the left-hand side, I get cos(a-b)+cos(a+b)=2cos(a)×cos(b).
00:29:45.000 --> 00:29:51.000
That's it, because the sin(a) and sin(b) cancel with each other.
00:29:51.000 --> 00:30:10.000
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.
00:30:10.000 --> 00:30:23.000
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).
00:30:23.000 --> 00:30:32.000
I'm going to combine those and see if I can prove the identity.
00:30:32.000 --> 00:30:36.000
I'll start with the left-hand side of the identity.
00:30:36.000 --> 00:30:41.000
I'll see if I can transform it into the right-hand side.
00:30:41.000 --> 00:31:00.000
The left-hand side is sin(5x)+sin(x) over cos(5x)+cos(x).
00:31:00.000 --> 00:31:21.000
Now, by what we did on the previous page, I have an expression for sin(5x)+sin(x), that's sin(3x)×cos(2x).
00:31:21.000 --> 00:31:30.000
That's by the work we did on the previous page.
00:31:30.000 --> 00:31:41.000
Also on the previous page, cos(5x)+cos(x)=2cos(3x)×cos(2x).
00:31:41.000 --> 00:31:46.000
That was also what we did on the previous page.
00:31:46.000 --> 00:32:00.000
But now look at this, the cos(2x) is cancelled, and what we get is 2sin(3x) over 2cos(3x).
00:32:00.000 --> 00:32:09.000
The 2s cancel as well and we get just tan(3x), which is equal to the right-hand side.
00:32:09.000 --> 00:32:13.000
We finished proving it.
00:32:13.000 --> 00:32:31.000
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.
00:32:31.000 --> 00:32:47.000
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.
00:32:47.000 --> 00:32:54.000
That gives us the expressions that we had in the identity here.
00:32:54.000 --> 00:33:17.000
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.
00:33:17.000 --> 00:33:24.000
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.
00:33:24.000 --> 00:52:52.000
We'll try some more examples of that later