WEBVTT
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Let's take a look at double and half angle formulas. Here I've listed in this box the identities and formulas
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for sine 2 theta, cosine 2 theta, and tangent 2 theta. In order to get sine 2 theta, you could use the sum
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identity for sine alpha plus beta, where alpha and beta are both theta, and you would get this identity.
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The rest of these identities can be found in a similar way. These two in particular can be obtained by using
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the Pythagorean identity. So you need to be familiar with these identities. Let's look at an example where
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we use one. Here we want to evaluate the expression as the sine, cosine, or tangent of a double angle.
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That's our first step. Then we want to find the exact value without using a calculator. So look at this.
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You have cosine squared of 11 pi over 12 minus sine squared of 11 pi over 12. Which one does this look like?
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Well notice it looks like this one, where theta is 11 pi over 12. So it looks like cosine squared theta minus
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sine squared theta. So I can see that theta is 11 pi over 12. Now that I know that, this is equal to the
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cosine of 2 theta, and we just determined that theta was 11 pi over 12. Well if you reduce that,
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what do you get? That gives you 11 pi over 6. So we've taken this expression to values we couldn't find
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without a calculator to something that we know one of the exact values. This is one of the special
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angles that you should be familiar with. So here's the first part. We rewrote it as, in this case, the cosine of
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a double angle. And then do you remember what the cosine of 11 pi over 6 is? Where does the terminal
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side of 11 pi over 6 lie? Exactly, in quadrant four, and cosine in quadrant four is positive or negative?
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Positive, exactly. So this is the same as the cosine of pi over 6, and those should be values you know.
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It's the square root of 3 over 2. So first we wrote it as a double angle, cosine of a double angle,
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and then we found that value.
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Let's look at this example where we're given sine of theta is 5 over 7 and we know the terminal side of
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theta lies in quadrant two. We want to find the exact value of sine 2 theta, cosine 2 theta, and tangent 2 theta.
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Well the first thing we need to do is very important. We need to draw a picture for theta. We know the
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terminal side lies in quadrant two, so let's draw a picture. So here's my angle theta. We know that we
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need to come up with our coordinates. Well we know the sine of theta is 5 over 7, and remember that the
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sine of theta is defined to be y over r. So what I know is y is 5 and r is 7. What we need to figure out is x.
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Well we can easily use the Pythagorean theorem to find it, but we need to pay attention. So this point is
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x comma 5, and r is 7. In quadrant two, remember x is negative, so when we find x it's going to be negative
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the square root of 7 squared minus 5 squared. What does that give me? 49 minus 25 is 24, and remember
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we're always asked to simplify radicals if we can. This is the same thing as negative 2 square roots of 6.
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So now we know x, y, and r. This is very important. Always pay attention to whether x and y are
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positive or negative in your quadrants, and we know the sine of theta is 5 over 7, and we know the cosine
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of theta would be what? Well it's going to be x over r. It's going to be negative 2 square roots of 6 over 7.
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Tangent of theta would be y over x, or negative 5 over 2 square roots of 6, and let's go ahead and write
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our sine here. That was 5 over 7. So it's very important to do this part right because if you make a
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mistake here then the rest of it will be wrong. So now that we've got a picture, let's first find the sine of
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2 theta. Well the identity for that is 2 sine theta cosine theta, and now we just have to plug in our values.
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Sine of theta, 5 over 7, cosine of theta, negative 2 square root of 6 over 7, and let's see. What does that
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give me? Negative 20 square root of 6 over 49, and that can't be reduced. So that's my value for sine 2 theta.
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Now let's find cosine 2 theta. Well notice that there are three identities for cosine 2 theta. Since we were
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given the sine of theta, the easiest one to use is the identity that only involves sine of theta. So I will use
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1 minus 2 sine squared theta, and then we just have to plug in our value. Sine of theta remember was
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5 over 7. We want to square that, and remember your order of operations. So you want to square this.
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We get 25 over 49. That gives me 1 minus 50 over 49, and 1 is the same thing as 49 over 49, minus 50 over 49,
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gives me negative 1 over 49. So cosine 2 theta is negative 1 over 49. Now let's find the tangent of 2 theta.
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Well the identity is 2 tangent theta over 1 minus tangent squared theta. Well we need to plug in our
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values. Our tangent was, let's see. Where did we write it down? Negative 5 over 2 square roots of 6 over
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1 minus negative 5 over 2 square roots of 6 and then that's squared. So the first thing we need to do is
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we can reduce this to negative 5 over the square root of 6, and then we've got to square this. When you
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square this, remember you're going to square the numerator and the denominator. So you're going to get
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25 in the numerator. In the denominator, you're going to square 2 and get 4, square root of 6 [squared] is 6,
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4 times 6 is 24, and 1 minus 25 over 24, that's 24 over 24, so that's negative 1 over 24. We can't leave
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complex fractions. We're asked to simplify that. Remember that's the same thing as negative 5 over the
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square root of 6 times the reciprocal, times negative 24, which gives me a positive number. I have a
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negative times a negative. That's 120 over the square root of 6. So that's my value for tangent 2 theta.
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One comment about tangent 2 theta. Since we found the sine of 2 theta and the cosine of 2 theta, if you
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knew both of those you could have found the tangent 2 theta by doing sine 2 theta over cosine 2 theta.
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You already had those values. So if you took negative 20 square roots of 6 over 49 divided by
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cosine, which was negative 1 over 49, this would simplify to positive 20 square roots of 6. The 49's will
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divide out. These two answers are equivalent. So if you don't know the sine 2 theta and cosine 2 theta,
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then you would just go ahead and use this formula. If you know them then of course you have a choice.
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Let's take a look at the half-angle formulas. Here I've listed on the board all the half-angles for sine alpha
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over 2, cosine alpha over 2, tangent alpha over 2. Notice for sine alpha over 2 there are two formulas, one
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if alpha over 2, the terminal side, lies in quadrant one or two and negative square root of 1 minus cosine
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alpha over 2 when alpha over 2, the terminal side, lies in quadrant three or four. So it's very important
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to first know where alpha over 2 is and pick the correct formula. The same thing for cosine and tangent.
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Although for tangent, notice the last two. Here, same thing, you'd have to know where alpha over 2 lies,
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but the last two, alpha over 2 can be in any quadrant. Let's look at an example. We want to use a
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half-angle formula to evaluate cosine 7 pi over 8 without using a calculator. Notice 7 pi over 8 is not a
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special angle, so it says use a half-angle formula. That means that 7 pi over 8 is the half-angle, so alpha
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over 2, and from this we can determine what alpha is. If we multiply both sides by 2, then we see that alpha
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is 7 pi over 4. So we want to find the exact value of [cosine] 7 pi over 8. So we've determined alpha. We have
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to choose the formula. So where does 7 pi over 8 lie? Think about that. 7 pi over 8 is a little less than pi,
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so it's in quadrant two. Cosine in quadrant two is negative, so we would choose this formula. So you write
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down the formula. It's [negative square root of] 1 plus cosine alpha over 2. A lot of students make the
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mistake of choosing the sign based on alpha. Don't do that. You always choose the formula based on where
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alpha over 2 is. Always write down the formula, and then plug in your value. We found alpha to be
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7 pi over 4, so we want the cosine of 7 pi over 4. Don't skip any steps, and then you have to find the cosine
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of 7 pi over 4. Well 7 pi over 4 lies in what quadrant? 7 pi over 4 is in quadrant four. Cosine in quadrant four
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is positive. So this means this is going to be positive. Cosine of 7 pi over 4 is the same thing as cosine of
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pi over 4, which is 1 over the square root of 2. Now we can't leave complex fractions. We need to simplify them.
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In order to do that, what you're going to do is you're going to multiply underneath the radical by the
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square root of 2 over the square root of 2, and so what does that give me? Well, 1 times the square root
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of 2 you get the square root of 2, plus 1, in the numerator, and in the denominator you get 2 square roots
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of 2. So the cosine of 7 pi over 8 is negative square root of square root of 2 plus 1 over 2 square roots of 2.
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In this example, we're given the tangent of alpha is 5 over 12. We're told that alpha is between pi and
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3 pi over 2, and we want to determine the exact values of sine of alpha over 2, cosine alpha over 2,
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tangent alpha over 2. Well the first thing we need to do is we need to draw a picture for alpha. We know
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the terminal side lies in quadrant? If we're between pi and 3 pi over 2 that means it lies in quadrant three.
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So let's draw a picture for alpha. We know the tangent of alpha is 5 over 12, and remember tangent is
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defined to be y over x. You might be tempted to think both x and y are positive, but they're not. Remember,
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you're in quadrant three, and in quadrant three x and y are both negative. So that means that x is negative 12
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and y is negative 5. We use the Pythagorean theorem. We get r to be 13. So now that we know our x, y, and r,
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we can find, for example, what's the cosine of alpha? Well it would be negative 12 over 13. Sine of alpha is
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negative 5 over 13, and tangent we already know. Tangent is 5 over 12. Ok, so now that we know this, let's
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start with the sine of alpha over 2. Well the first thing we have to do is we have to pick a formula, and
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remember, we pick the formula based on where alpha over 2 is. This is a picture for alpha. Where's alpha over 2?
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Well what we can do is take this expression. We know that alpha is between pi and 3 pi over 2. If we
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divide this inequality by 2, that will tell us where the terminal side of alpha over 2 lies, so we can see
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that the terminal side of alpha over 2 is between pi over 2 and 3 pi over 4. What quadrant is that?
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Quadrant two. So that's how we decide. We know that the terminal side of alpha over 2 lies in quadrant two.
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We want a sine, and sine in quadrant two is positive, so we choose this formula. Write down your formula.
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Don't skip any steps, and then plug in your value. Cosine of alpha is negative 12 over 13, so underneath the
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radical you have 1 plus 12/13. Well 1 is 13/13 plus 12/13 gives you 25/13, and that's the same thing as
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25 over 26, which can be simplified because the square root of 25 is 5. So the sine of alpha over 2 I get to be
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5 over the square root of 26. Now let's find the cosine of alpha over 2. Well we've already determined, remember
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that alpha over 2, the terminal side, lies in quadrant two, and cosine in quadrant two, is that positive or negative?
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Negative, exactly, so you're going to use the negative square root of 1 plus cosine alpha over 2,
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and then you plug in your value for cosine. Let's see I'll go this way. Cosine of alpha was negative 12 over 13,
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and so what is that? That's 13 over 13 minus 12, so that's 1 over 13 over 2, which is the same thing as
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1 over 26, and the square root of 1 of course is 1. So this is negative 1 over the square root of 26.
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So now we've found the cosine of alpha over 2. Last, let's find the tangent of alpha over 2. Well since
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we have a choice on this identity, let's use one where we don't have to choose whether it's positive
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or negative. Let's use this one, 1 minus cosine alpha over sine alpha, and then plug in our values.
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Cosine of alpha was negative 12 over 13. Sine of alpha was negative 5 over 13. So again this is the same
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thing as 1 plus 12 /13 which is 13 over 13 plus 12, so that give me 25 over 13 over negative 5 over 13.
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That's the same thing as 25 over 13 times negative 13 over 5, and that gives you what? Negative 5
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for tangent alpha over 2. So the tangent of alpha over 2 is negative 5. Let me point out, it's very important,
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this part of the step is very important. If you draw the picture wrong, if you don't pay attention to x, y,
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whether they're positive or negative, you'll make a mistake, and of course all this will be wrong.
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Pay special attention when you're drawing your picture. Make sure you choose the right formula,
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based on where alpha over 2 lies. Make sure you plug in the values.
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In this example, we want to find the exact value without using a calculator. So look at this first example.
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We have the sine of 2 inverse cosine of negative 1/3. First we want to look at this, the inverse cosine of
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negative 1/3, and we're going to think of that as an angle. We're going to call it theta, so we're going to
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let theta be the inverse cosine of negative 1/3. Which means if you rewrite that, that's the same thing as the cosine of
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theta is negative 1/3. But remember your range for inverse cosine. Where does this angle lie? What interval?
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Do you remember? Yes, 0 to pi, so theta must be between 0 and pi, and we know the cosine is negative.
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So where is it? That's either quadrant one or quadrant two, so we know theta lies, the terminal side, in
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quadrant two. So we can draw a picture for this angle. Well since we know the cosine is negative 1 over 3,
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we know cosine is x over r. So I know x is negative 1. I know r is 3. We need to find y. Now notice in the
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second quadrant the y-coordinate is positive, so it's going to be positive square root of 9 minus 1, which is
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square root of 8, which simplifies to 2 square roots of 2. So now we've found x, y, and r. Knowing that,
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let's go back to this problem. We have the sine of 2 theta. Well we can use an identity. Remember the
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identity for sine 2 theta? It's 2 sine theta cosine theta. So we need to know the sine of theta and the cosine
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of theta. Well, we already know the cosine of theta is negative 1/3. Since we found y, what would the
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sine of theta be? It would be 2 square roots of 2 over 3. So we can plug in those values, 2 square roots of 2
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over 3 for sine theta, cosine negative 1/3, and so I get a negative 4 square roots of 2 over 9 for sine 2 theta.
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Let's scoot this over. So once again, it's important to draw the picture. You have to remember your range
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for inverse cosine and draw it in the correct quadrant. So now we know sine 2 theta is negative 4 square roots
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of 2 over 9. Let's look at another example. We want to find the exact value here as well. We see the sine of
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1/2 inverse tangent of negative 1. Well once again, look at this, inverse tangent of negative 1. Think of it as
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as an angle. We're going to call it alpha. We're going to let alpha be the inverse tangent of negative 1, and
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again, what does that mean? It means the tangent of alpha is negative 1, and again, you have to remember
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the range for inverse tangent. It's very important. Do you remember what the range is? Alpha has to be
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between where? Negative pi over 2 and pi over 2, which means the terminal side either lies in quadrant
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four or quadrant one. Well, we know the tangent is negative, so we know that the terminal side lies in
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quadrant four, and remember the tangent is y over x, so we're going to have to figure out our coordinates
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again. Well, let's see, in quadrant four, which value is negative? The y, right. So we know y is negative.
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Let y be negative 1 and x be 1, so the tangent is negative 1. Use the Pythagorean theorem. R is going to be
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the square root of 2. So now we know cosine of alpha is 1 over the square root of 2, and if we need the sine,
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it would be negative 1 over the square root of 2. So we've got all our values. We've paid attention to what
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quadrant, and now we're ready to go back to here. So we want to know the sine of 1/2 alpha. That's the
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same thing as the sine of alpha over 2. Remember there are two formulas for the sine of alpha over 2.
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There's this one, and then there's this one. Oops, negative. And remember, you choose this based on
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where alpha over 2 is. Well we've drawn a picture for alpha, not alpha over 2. But notice, where's alpha?
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We said it's in quadrant four. So we actually know that alpha lies between negative pi over 2 and 0 because
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we've narrowed it down to just quadrant four. We can divide this inequality by 2 to figure out where the
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terminal side of alpha over 2 lies. So let's see. Alpha over 2 is between negative pi over 4 and 0, so that
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means that the terminal side of alpha over 2 lies also in quadrant four. Remember, sine in quadrant
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four is, which one do I use, positive or negative? Quadrant four sine is negative, so I would use this one.
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Write it down here. So it's going to be [negative square root of] 1 minus cosine alpha over 2, and again
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plug in your value for cosine alpha, which is 1 over the square root of 2. Once again, since we have
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a complex fraction, we need to simplify by multiplying by the square root of 2 over the square root of 2
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underneath the radical. So the sine of this alpha over 2 is what? Let's see. It would be the square root of 2
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minus 1 over 2 square roots of 2. So again, it's very important that you know the range for the inverse
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functions, that you draw the picture, you pay attention to whether x and y are positive or negative, and
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don't skip any steps. I see students skipping steps, and they make a mistake. So make sure you write down
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the formula and then plug in your values. Practice working these examples until you've mastered the
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skills and concepts involved.