WEBVTT mathematics/trigonometry/murray
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Hi, these are the trigonometry lectures for educator.com and today we're going to talk about the half-angle formulas.
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The main formulas that we're going to be using today, we have a formula for sin(1/2 x) and cos(1/2 x).
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They're a little bit cumbersome, sin(1/2 x) is equal to plus or minus the square root of 1/2 of 1-cos(x).
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Cos(1/2 x), same formula, except there's a plus in it.
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They're a little bit cumbersome but we'll practice using them and you'll see that they're not so bad.
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What makes them difficult is the plus or minus and the square root signs.
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That's probably the confusing part.
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Actually, what's inside the square root sign isn't bad at all, the 1-cos(x), 1+cos(x) aren't too bad.
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Let's try them out right away with some examples.
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Our first example is to find the sine and cosine of 15 degrees and then we'll check that our answers satisfy the Pythagorean identity sin²+cos²=1.
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The first thing to notice here is 15 degrees is not a common value, it's not one that where we've memorized the sine and cosine.
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We'll have to use the half-angle formulas here.
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We'll start with 15 is 1/2 of 30, so we're going to use the sine formula.
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Remember, sin(1/2 x) is equal to plus or minus the square root of 1/2(1-cos(x)).
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The x in question here is 30, we're trying to find the sin(15), sin(15) is equal to plus or minus the square root of 1/2(1-cos(30)).
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That's plus or minus the square root of 1/2.
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Now, 30 degrees is a common value, that's π/6, and I've got all the sines and cosines of the common values memorized.
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Cos(π/6), cos(30) is root 3/2.
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I'm just going to do a little bit of algebra with this expression here, plus or minus the square root of 1/2.
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I'm going to put 1 in root 3 over 2 over a common denominator, that will be 2-3/2.
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I'm just writing 1 there as 2/2.
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If I combine these, I get 2 minus root 3 over 4.
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I still have this plus or minus which is not very good because I want to give a single answer, I don't want to give two different answers.
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Let's remember where 15 degrees is.
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Here's 0 degrees, and here's 90 degrees, 15 degrees is way over here right in the first quadrant.
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Since sine and cosine are the x and y values, actually sine is the y-value, and cosine is the x-value.
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They're both positive in the first quadrant.
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15 degrees is in quadrant 1.
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The sine is its y-value, it's positive, sin(15) is greater than 0, it's positive.
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Our answer then, is sin(15) must be the positive square root there, it's the square root of 2 minus root 3 over 4.
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Now, if you've been paying really close attention to the educator.com trigonometry lectures, you'll know that we've actually solved this problem before.
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15 degrees if you convert it to radians, is actually π/12, and we worked out the sin(π/12) before not using the half-angle identities but using the addition and subtraction formulas.
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We worked out the sine and cosine of π/12 by realizing it as π/4-π/6.
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When we worked it out, sin(π/12) using the subtraction formulas, we got the answer square root of 6 minus the square root of 2 over 4.
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There's a little bit of a worry here, because it's seems like we did the same problem using two different sets of formula and we got two quite different looking answers.
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We got the square root of 6 minus the square root of 2 over 4 last time we did it.
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This time we have the square root of 2 minus the square root of 3 over 4.
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Actually, that can be simplified a little bit into, if we just take the square root of the top part, then the square root of 4 is just 2.
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We have this two different answers here, or at least they seem to be different.
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Let me show you that these two answers can actually be reconciled.
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How do these answers agree?
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Let me start with the old answer, the one we did in a previous lecture on educator.com.
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Our old answer was root 6 minus root 2 over 4.
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What I'm going to do is square the numerator, root 6 minus root 2 squared.
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To pay for that, I have to take a square root later.
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Remembering an algebra formula (a-b)² is a²-2ab+b², I remember that algebra formula.
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I've got a quantity that I'm squaring here, root 6 squared is 6 minus 2 root 6 root 2 plus root 2 squared is just 2, all over 4.
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This simplifies down to 8 minus 2 root 12 over 4.
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But root 12, I could pull a 4 out of that, and it turns into a 2 on the outside.
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But I already had a 2 on the outside, you combine those and you get 4 root 3 over 4.
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Now, I can factor 4 out of the numerator.
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When it comes outside it becomes a 2, left on the inside will be, the 8 turns into a 2, and the minus 4 root 3 becomes a minus root 3 over 4.
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That simplifies down to 2/4 cancel into 1/2, square root of 2 minus root 3 over 2.
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Look at that, that's our new answer that we've just arrived using the half-angle formula.
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We could do that problem using the addition and subtraction formulas as we did a couple of lectures ago or we could do it using our new half-angle working it out from what we now about 30 degrees.
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Either way, we get down to the same answer.
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We still have to find the cos(15).
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We're going to use the half-angle formula for cosine, cos(1/2 x) is equal to plus or minus the square root of 1/2 times 1 plus cos(x).
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The cos(15), 15 is 1/2 of 30, this is square root of 1/2 times 1 plus cos(30).
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Cos(30) is a common value that I remember very well, 1 plus root 3 over 2.
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If I put those, combine those over a common denominator, I get 1/2 2 plus root 3 over 2.
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That simplifies down to 2 plus root 3 over 4, or if I take the square root of the bottom 2 plus root 3 over 2.
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I still have that plus or minus but remember 15 degrees is safely there in the first quadrant.
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It's sine and cosine, it's x and y values.
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They're both positive, 15 degrees is in quadrant 1, cos(15) is positive.
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The cos(15) must be the positive square root 2 plus root 3 over 2.
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There's my answer.
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Last thing we were supposed to check there was that the answer satisfies the Pythagorean identity sin²+cos²=1.
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Sin²(15)+cos²(15), sin²(15) was square root of 2 minus root 3/2.
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We're going to square that.
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Plus cos²(15) is 2 plus root 3, square of that, over 2.
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We'll square that one out.
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Now, in the top, the square root and the square will cancel each other away.
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We get 2 minus root 3.
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In the bottom, we have 2, squared is 4, plus 2 plus root 3/4.
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When we add those together, the root 3s cancel.
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We just get 4 over 4 which is 1.
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When we worked out sin²(15)+cos²(15), we did indeed get 1, showing that it does confirm the Pythagorean identity.
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The key to that problem was really just recognizing that 15 is 1/2 of 30, and then invoking the sine and cosine half-angle formulas plugging in x=30 working them through doing a little bit of algebra, and getting our answers there.
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The only other step that was a little bit tricky was recognizing whether we wanted to use the positive or the negative square root.
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That's the matter of recognizing that 15 degrees is in the first quadrant, in both cases sine and cosine are both positive.
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For our second example here, we're asked to use, to prove a trigonometric identity, (cos(1/2 x)+sin(1/2 x))/(cos(1/2 x)-sin(1/2 x))=sec(x)+tan(x).
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That's a pretty complicated identity.
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It's not really obvious where to start.
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You might want to jump into the half-angle formulas because you see cos(1/2 x), sin(1/2)x.
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I'm going to say, let's try to avoid the half-angle formulas here if we can.
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Here's why.
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Remember that cos(1/2 x) is equal to plus or minus the square root of something or other, so is sin(1/2 x) is equal to plus or minus the square root of something or other.
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If we start putting those in, we're going to have plus or minuses, or lots of square roots, it's going to get complicated.
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I'm going to try to avoid those.
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Instead, I have another strategy which we've seen before in proving trigonometric identities.
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If you have (a+b)×(a-b), remember from algebra, that's the difference of squares formula.
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That's a²-b².
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That can be really useful if you have an (a+b) in the denominator or an (a-b) in the denominator.
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You multiply both sides by the conjugate, by the other one, and then you get the difference of squares.
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Let's try that out on this one.
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The left-hand side, I'm going to work with the left-hand side because I see that (a-b) in the denominator.
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That's cos(1/2 x)+sin(1/2 x)/(cos(1/2 x)-sin(1/2 x)).
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Now, I'm going to multiply top and bottom by the conjugate of the denominator.
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That means where I saw a minus before, I'm going to multiply by the same expression with a plus in it, sin(1/2 x).
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Of course, I have to multiply the top by the same thing, (cos(1/2 x)+sin(1/2 x).
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Let's see where we'll go with that.
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In the numerator, we actually have cos((1/2 x)+sin(1/2 x))², so that's cos²(1/2 x)+2sin(1/2 x)×cos(1/2 x)+ sin²(1/2 x).
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We can invoke this difference of squares formula.
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We get cos²(1/2 x)-sin²(1/2 x).
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There's several good things that are going to happen right now but they will only happen if you remember the double angle identities.
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Let me write those down for you.
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I'm going to write them in θ instead of x.
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Remember that sin(2θ)=2sin(θ)×cos(θ), cos(2θ)=cos²(θ)-sin²(θ).
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Now, look at what we have here.
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There's several good things that are going to happen.
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First of all, cos² and sin², those combined, and those give me a 1.
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Now, we have 2 sine of something, cosine of something, and the something is 1/2 x.
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If you look back at our sin(2θ), 2 sine of something and cosine of something is equal to sine of 2 times that thing.
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We have sin(2×1/2 x).
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Now, I have cos² of something minus sin² of something, and I know that cos² of something minus sin² of something is equal to cosine of 2 times that something, so cos(2×1/2 x).
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You can simplify this a little bit, this is (1+sin(x))/cos(x).
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I'll split that up into 1/cos(x) + sin(x)/cos(x).
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Those are expressions that I recognize, 1/cos(x) is sec(x), sin(x)/cos(x) is tan(x).
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Look, now we've got the right-hand side of the original trigonometric identity.
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That was the right-hand side right there.
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That was a pretty tricky one.
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There were several key steps involved there.
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The first is looking at the left-hand side and noticing that we have something minus something in the bottom, so we're going to use this difference of squares formula.
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We're going to multiply top and bottom by the conjugate.
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Once we multiply top and bottom by the conjugate, we get something that looks pretty messy, but we start invoking these identities all over the place.
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First of all, sin²+cos² gives you 1.
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Secondly, 2 sine of something cosine of something, that's the double angle formula for sine.
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Then, cos² of something minus sin² of something, that's the double angle formula for cosine.
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That simplifies it down to (1+sin(x))/cos(x).
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Those split apart and convert easily into secant and tangent, and all of a sudden we have the right-hand side.
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You may have to experiment a bit with different techniques when you're proving these trigonometric identities.
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The ones that I'm using for example, these are ones I've worked out ahead of time, so I know right away which technique I'm going to use.
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Even when I'm working on this, I'll try multiplying a few different things together, maybe splitting up things differently and invoking different half-angle formulas, double angle formulas, and finally I find the sequence that works.
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When you're asked to prove this trigonometric identities, go ahead and experiment a little.
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If it seems like it's getting really complicated, maybe go back and try something else.
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Eventually, you'll find something that converts to one side of the equation into the other.
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For our next example, we have to prove a half-angle formula for tangents, and we're told to be careful about removing plus or minus signs.
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The reason for that is we're going to be using the sine and cosine half-angle formulas, and those both have plus or minus in them.
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Let me remind you what those are.
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The cos(1/2 x) is plus or minus the square root of 1/2 times 1+cos(x).
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The sin(1/2 x) is plus or minusthe square root of 1/2 times 1-cos(x).
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Those are the formulas we're going to be using.
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We're given the tan(1/2 x).
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Let me start with that, I'll call that the left-hand side.
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Left-hand side is tan(1/2 x).
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Now, I don't have a formula yet for tan(1/2 x), I'm going to split that up into sin(1/2 x)/cos(1/2 x) because I do have formulas for those.
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Those are my half-angle formulas.
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In the numerator, I get plus or minus the square root of 1/2.
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Sin(1/2 x) is 1-cos(x).
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In the denominator, cos(1/2 x), the same thing except that I get 1/2 (1+cos(x)).
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Here's the thing, you might think that you can cancel plus or minus signs, but you really can't.
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The reason is that plus or minus signs means that both the top and bottom could be positive or could be negative.
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When you divide them together, you don't know if the answer's going to be positive or negative.
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I'm going to put one big plus or minus sign on the outside but I can't just cancel those away.
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I'm also going to combine everything under the square root here.
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I get (1/2 (1-cos(x)))/(1/2 (1+cos(x))).
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The obvious thing to do there is to cancel the (1/2)'s, so you get (1-cos(x))/(1+cos(x)).
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It's not so obvious where to go from here but remember that we've been practicing this rule (a-b)×(a+b)=a²-b².
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That comes up all over the place for trigonometric identities and with other algebraic formulas as well.
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The trick is when you have either one of those in the denominator, you multiply by the conjugate.
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Here, we have 1-cos(x) in the denominator, I'm going to multiply by 1+cos(x).
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Of course, I have to multiply the numerator by the same thing.
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Sorry, I have 1+cos(x) in the denominator, so the conjugate would be 1-cos(x).
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Multiply top and bottom by 1-cos(x).
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The point of that is to invoke this difference of squares formula in the denominator.
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This is all taking place under a big square root.
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1-cos(x), I'll just write that as (1-cos(x))².
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I don't need to multiply that out.
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In the bottom, I got 1-cos(x) times 1+cos(x), that's the difference of squares formula, that's 1-cos²(x).
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Now, I'm going to separate out the top and the bottom part here, because in the top, I've got a square root of a perfect square.
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On the top, I'm just going to write it as 1-cos(x), because I had the square root of (1-cos(x))².
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In the bottom, I still have a square root, 1-cos²(x), that's something that should set off some warning bells in your brain.
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Certainly doesn't mind, because I remember that sin²+cos²=1, the Pythagorean identity.
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If you move that around, if you see 1-cos², that's equal to sin².
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So, 1-cos²(x)=sin²(x), that cancels with the square root.
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I've already got a plus or minus outside, I don't need that another one, (1-cos(x))/sin(x).
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I almost got what I want, I've almost got the right-hand side (1-cos(x))/sin(x).
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The problem is this plus or minus.
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The directions of this exercise said we have to be very careful about why we can remove any plus or minus signs.
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Let me write the big question here, "Why can we remove this?"
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That actually takes a bit of explanation.
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I'm going to go on a new slide to explain that.
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From the last slide, we figured out that the left-hand side is equal to plus or minus (1-cos(x))/sin(x), but we aren't sure if we can remove the plus or minus from the right-hand side.
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Let's think about that.
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First of all, I know that cos(x) is always less than 1.
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That's because cos(x), remember, is the x values on the unit circle so it's always between -1 and 1.
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1-cos(x) is always greater than 0, the numerator here, the 1-cos(x), the numerator 1-cos(x) is always positive.
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That part isn't really affected by the plus or minus.
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What about the sin(x)?
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I know that that is not always positive.
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What about sin(x)?
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Let me draw a unit circle, because this really depends on where x lies on the unit circle.
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There are several different cases depending on where x lies on the unit circle.
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Let me write down the four quadrants, 1, 2, 3, and 4.
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Let's try and figure out where x could lie on the unit circle.
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There's sort of 4 cases.
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If x is in quadrant 1, then remember, sin(x) is its y-value, so sin(x) is positive.
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And x/2, if x is in quadrant 1, if that's x right there, then x/2 will also be in quadrant 1.
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Tan(x) will also be positive.
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Remember All Students Take Calculus, they're all positive in the first quadrant, second quadrant, only sine is, third quadrant only tangent is, and fourth quadrant only cosine is.
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If x is in quadrant 1, then they're both positive.
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Both sides here, the tan(1/2 x) and the sin(x), Oops, I said tan(x) and I should have said the tan(1/2 x) or x/2, they're both positive.
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Let's check the second quadrant, if x is in quadrant 2, then sin(x) will still be positive, and x/2 ...
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Well if x is over here in quadrant 2, then x/2 will be in quadrant 1 because it's half of x.
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It's tangent, will still be positive.
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Again, both the sin(x) and the tan(x) will both be positive.
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Third case is if x is in quadrant 3, then sin(x) is less than 0 because x is down here in quadrant 3.
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Where will x/2 be?
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If x is in quadrant 3, that means x is bigger than π, x/2 is bigger than π/2.
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x/2 will be over here in quadrant 2.
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Tan(x/2), in quadrant 2, only the sine is positive, the tangent is negative.
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Sine is negative because x is in quadrant 3, and tan(x/2) is also negative.
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Finally, if x is in quadrant 4, then sin(x) is less than 0 because it's still below the x-axis, its y-coordinate is negative.
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If x is somewhere over here in the fourth quadrant, x is between π and 2π, x/2 will be between π/2 and π.
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So x/2 is still in quadrant 2, tan(x/2) is still negative.
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Now, there's 4 cases there.
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In the first 2 cases, sine was positive and tangent was positive, tan(x/2) is positive.
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In the second 2 cases, in the last 2 cases, sine was negative and tan(x/2) was also negative.
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Sin(x) and tan(x/2), they're either both negative or both positive, that means they have the same plus or minus sign, have the same sign in terms of positive and negative not as any have the same sign.
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We can drop the plus or minus, and finally say tan(x/2) or tan(1/2 x)=(1-cos(x))/sin(x).
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In all 4 cases, the tan(x/2) has the same plus or minus as sin(x).
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Then remember the 1-cos(x) is always positive, the left-hand and the right-hand side will always have the same plus or minus, we don't need to attach another plus or minus.
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That was a pretty tricky one.
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The secret to that was starting with the tan(x/2), expanding it out using the formulas for cos(1/2 x) and sin(1/2 x) or x/2.
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We worked it down, we did some algebra, simplifying a square root.
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Then we still have that plus or minus at the end.
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What we had to do was this sort of case by case study of each of the four quadrants to say when is sine positive or negative, when is tan(x/2) is positive or negative.
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The last term, 1-cos(x) was always positive.
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Finally, we figured out that sin(x) and tan(x/2) could be positive or negative, but they'll always be the same so we don't need to put in a plus or minus there.
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We'll try some more examples later on.
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The examples we're going to be using later, we're going to be using this formula for tan(x/2), so it's worth remembering this formula for tan(x/2).
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Later on, we'll be using that to solve some more trigonometric identities.