WEBVTT
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Let's look at compositions of the trig functions and their inverses. Sometimes a trig function and its inverse
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undo each other. Here are the six relationships we need to know. The sine of the inverse sine of x is x, as long as
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x is in the closed interval negative 1 to 1. Notice I switched the order, so the inverse sine of the sine of theta is theta,
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as long as theta is in the closed interval negative pi over 2 to pi over 2. We have similar relationships for cosine
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and inverse cosine compositions, and the same thing with tangent or inverse tangent compositions. The key here is
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the relationship is true as long as your x is in the proper interval. So you need to memorize and learn all of these
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relationships. Let's try some examples. The first one, the cosine of the inverse cosine of 4/5. So we need to look
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at 4/5 right there. In order to use this relationship, the cosine of the inverse cosine of 4/5, the answer is just 4/5
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as long as 4/5 is in the interval negative 1 to 1. Is it? Yes it is. It's a little bit less than 1.
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So this is just 4/5. Let's try a second one. The inverse cosine of the cosine of negative pi over 4.
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So we find the relationship we want to use. Inverse cosine of the cosine of theta is theta as long as theta is in
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the interval 0 to pi. Well, it's not. Negative pi over 4 is definitely not in this interval. So what does that mean?
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It doesn't mean we can't answer the question. It just means that we have to work a little harder to find that answer.
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So let's start with this part, the cosine of negative pi over 4. We already know how to find this. Negative pi over 4,
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that's in the fourth quadrant. The cosine of an angle in the fourth quadrant is positive. So this inside part is just
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1 over the square root of 2, positive cosine of pi over 4. By evaluating the inside part, it now reduces to a different
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kind of problem, and we know how to solve this as well. The inverse cosine of 1 over the square root of 2.
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This is a positive number between 0 and 1, so in what quadrant will the terminal of this angle lie? Right,
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quadrant one. So 1 over the square root of 2 is positive, and so this is going to lie in quadrant one. What angle is it?
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Well, it's in quadrant one, so it's a positive acute angle, and its cosine is 1 over the square root of 2, so pi over 4.
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We have one more. The inverse cosine of the cosine of 2 pi over 3. Alright, so we go back and we look at our
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relationship. The inverse cosine of the cosine of theta is theta as long as theta is in the closed interval 0 to pi. So
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we need to know is 2 pi over 3 between 0 and pi. Yes, it is. It's somewhere in the second quadrant. That's correct.
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Here are a few more examples. The inverse sine of the sine of 2 pi over 7. So we find the relationship that works
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here, or at least we hope it works. The inverse sine of the sine of theta is theta for any theta in the closed interval
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from negative pi over 2 to pi over 2. So the big question here is, is 2 pi over 7 in that interval, negative pi over 2
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to pi over 2? Ok, so let's think about 2 pi over 7. It's a little bit less than pi over 2, half of pi, so yes, it's right there.
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It's somewhere in this interval. Since the answer is yes, then I'm allowed to use this relationship. The inverse sine
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of the sine of 2 pi over 7 is 2 pi over 7. Alright, let's look at the next one. The sine of the inverse sine of 3/2.
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The sine of the inverse sine of 3/2, and I see a problem already. 3/2 is too big. It's not in negative 1 to 1,
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that interval. So I can't use this relationship. Again, I have to stop and think about whether I can even solve
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the problem. This number is an angle whose sine is 3/2. Right, that's not even possible, so this expression
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is undefined. Let's look at one more. The inverse sine of the sine of 2 pi over 3. So here's our relationship.
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Inverse sine of the sine of 2 pi over 3 would be 2 pi over 3, as long as 2 pi over 3 is in this closed interval
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negative pi over 2 to pi over 2. But correct, it's not in that interval. 2 pi over 3 is too big. Well, when the relationship
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doesn't hold we have to stop and think if we can work it out in some other way using other infomation we know.
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So let's look at the inside part here, the sine of 2 pi over 3. In what quadrant does 2 pi over 3's terminal side lie?
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Right, quadrant two. We know that in quadrant two the sine of an angle is positive or negative? Right, positive.
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We could use a reference angle to help a little bit. The reference angle for 2 pi over 3 would be pi over 3,
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and we know the sine of pi over 3 is the square root of 3 over 2. Ok, so we fill in just this inside part. We've
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discovered that the sine of 2 pi over 3 is positive square root of 3 over 2, using this information and our
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reference angle. We have one more step to finish the problem, the inverse sine of square root of 3 over 2.
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Ok, inverse sine of this number is an angle. Where does the terminal side of this angle lie? Correct, this is positive,
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so it's going to lie in quadrant one, and the angle in quadrant one whose sine is square root of 3 over 2 is pi over 3.
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Let's look at another set of examples. We'd like to find the tangent of the inverse tangent of 15. So the tangent of
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the inverse tangent of x equals x for any x. This is my favorite one because it's going to have to just be 15.
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The second one, the inverse tangent of the tangent of negative pi over 3. This relationship, the inverse tangent
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of the tangent of an angle is that angle as long as the angle is in this open interval, negative pi over 2 to pi over 2.
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So where will I find negative pi over 3? Inside this interval? Yes, so this one is negative pi over 3.
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The inverse tangent of the tangent of 7 pi over 12. So this would just be 7 pi over 12, if 7 pi over 12 is in this
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interval, negative pi over 2 to pi over 2. Is it? Unfortunately no, 7 pi over 12 is a little bit bigger than 6 pi over 12,
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so it's a little bit bigger than pi over 2. This isn't going to work. I have to do something else to find the answer.
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Well, there's another problem with this one. It's not one of your special angles that you know and love, right? Ok,
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so we're going to draw some pictures and see if we can figure it out. We said 7 pi over 12 is a little bit bigger
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than pi over 2, so 7 pi over 12 has its terminal side in quadrant two. Well, I know a couple of things about angles
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in quadrant two. For instance, would the tangent of 7 pi over 12 be a positive number or a negative number?
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A negative number, so that might be useful information in a few minutes. I'm just going to make a note of that
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for now. What about the reference angle? Can we figure out the reference angle? We'll call that theta r. Do you
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remember how to find a reference angle? All the way to here would be pi, so we're going to find that theta r is
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pi minus 7 pi over 12, and that means theta r is 5 pi over 12. Now what? We're going to try to find the
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inverse tangent of whatever this number is, and we know that it's a negative number. So let's think about our
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inverse tangent function. If you recall, the inverse tangent function has to give you an angle over here, between
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negative pi over 2 and pi over 2. If this is a negative number, then it must be in quadrant four. So our answer
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is somewhere down here in quadrant four. I'm going to draw that in purple on this one. We want an angle whose
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terminal side is in quadrant four. The reference angle is going to have to match this one, 5 pi over 12.
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So if I draw that negative angle in the fourth quadrant, the same reference angle as this one, then I find out
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that my angle is negative 5 pi over 12.
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In this next example, notice that we're looking for the tangent of the inverse sine of 1 over the square root of 2.
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So I've sort of mixed up our trig functions and inverse trig functions. None of those special relationships will help us,
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and we simply need to work this out starting with the inside part. So let's start with the inverse sine of
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1 over the square root of 2. This is an angle. In what quadrant does this angle's terminal side lie? Well, since this is
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a positive number and this is the inverse sine function, it must be in quadrant one. This is an angle in quadrant one
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whose sine is 1 over the square root of 2. You already know what that angle is, don't you? It's pi over 4.
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If we make this substitution, this is just pi over 4. We've simplified this to the question just asking what's the
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tangent of pi over 4, and that's 1.
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Here we'd like to find the secant of the inverse tangent of negative square root of 3, and again we're going to start
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on the inside part of this and work our way through. Inverse tangent of negative square root of 3, so this is going
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to be an angle whose terminal side lies in which quadrant? Ok, well this is negative and this is an inverse tangent,
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so quadrant four, and it's the angle in quadrant four whose tangent is negative square root of 3. So have you
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figured out which one it is? Tangent of pi over 3 is the square root of 3, so we want an angle in quadrant four,
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a negative angle, whose tangent is negative square root of 3. We're using pi over 3 as our reference angle,
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so this must be the angle negative pi over 3. Next question, what's the secant of negative pi over 3?
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I think it'll be easier for us to find the cosine of negative pi over 3 and use that to get our answer. Where is
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negative pi over 3's terminal side? Quadrant four. In quadrant four, is the cosine positive or negative? Positive.
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Ok, so it's positive, and we need to know the value. What's the cosine of pi over 3? 1/2, ok and we've already
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decided that it's positive so this is 1/2. Ok, that wasn't quite my answer though. Cosine of negative pi over 3 is 1/2,
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so what's the secant of negative pi over 3? 2 and there we go.
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Let's find the inverse sine of the cosine of 7 pi over 4. So again we'll start on the inside and work our way through
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the problem. 7 pi over 4 is an angle whose terminal side lies in what quadrant? The fourth. Let's just draw a little
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picture of that. It's going to be down here. 7 pi over 4 is all the way around. Is the cosine of an angle positive or
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negative in quadrant four? Positive, so we're going to be looking for the inverse sine of a positive number. You need
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to fill that in. What's the reference angle? Pi over 4, ok, so what's the cosine of pi over 4? Right,
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1 over the square root of 2. We've already decided that should be a positive number. Inverse sine of
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1 over the square root of 2. What quadrant can this angle lie in, its terminal side? The first, right. In the first
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quadrant what's the angle whose sine is 1 over the square root of 2? Correct, it's pi over 4.
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Let's find the inverse cosine of the sine of 5 pi over 3. We'll start with the inside part, the sine of 5 pi over 3.
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Tell me about the angle 5 pi over 3. It lands in, it's terminal side lands in, quadrant four. What would be the reference
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angle? Right, pi over 3. In the fourth quadrant, is the sine of an angle positive or negative? Negative. So the sine
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of 5 pi over 3 is negative. Let's figure that number out. We said the reference angle is pi over 3. We know that the
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sine of pi over 3 is square root of 3 over 2, so this is equal to negative square root of 3 over 2. So we really need
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to find now the inverse cosine of negative square root of 3 over 2. So let's think about our inverse cosine function.
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The answer here is an angle somewhere in the interval 0 to pi. Since this is a negative number, that means I'm looking for an
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angle whose terminal side is in quadrant two and whose cosine is negative square root of 3 over 2. A reference angle
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will probably help us here. What positive acute angle's cosine is square root of 3 over 2?
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Pi over 6. We're going to draw a picture. We know our angle is in quadrant two, that our reference angle is
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pi over 6, and so we know that the angle we're looking for, whose terminal side lies in quadrant two, must be 5 pi over 6.
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Here we'd like to find the sine of the inverse cosine of negative 7/8. We'll start on the inside. We're looking for an
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angle whose cosine is negative 7/8, and where does the terminal side of this angle lie? Well, it's an inverse cosine
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question and this number is negative. In the interval 0 to pi, that must mean we're looking in quadrant number two.
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So we're going to sketch the angle in quadrant two. The terminal side lies here, and we're going to label a point on
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the terminal side of the angle, and we're going to write down the x, the y, and the r that correspond to our picture.
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Since we're given the cosine of the angle is negative 7/8 and cosine is x over r, we know that x is negative 7. R has
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to be positive, so it's just 8, and we'll need to find y. Well, look at our picture. Y is up here, and that means that y is a
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positive number since it lands in quadrant two here. So y has to be the positive square root of r squared minus
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x squared, and when we substitute our known values and work this out we're going to find out that y is the square
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root of 15. Now, unfortunately, I don't know exactly what the value of this angle is, but I have everything I need to
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know about it in order to find the sine of that angle. Sine, remember the definition, it's y over r, so the
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y is square root of 15, and r was positive 8.