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The next step in our study of the trig functions is to define the inverse sine, inverse cosine, and inverse tangent
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functions. Let's start by reviewing the graph of the sine function. You can see I've drawn it here. Its domain is
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all real numbers. Its range is from negative 1 to 1, and our goal is to find an inverse for this function. It'll help
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you a great deal to think of these x-values as angles. So your sine function takes an angle, an x-value,
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and spits out, or gives out, a y-value, a real number between negative 1 and 1 that is the sine of that angle.
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What we'd like to do is find an inverse function that takes one of these real number values between negative 1 and 1
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and gives us back a unique angle. But there's a big problem. Notice that if I pick say 1/2 there are a lot of angles
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whose sine is 1/2. It wouldn't be a unique angle. The term for this is that this is not a one-to-one function.
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A [non] one-to-one function doesn't have an inverse, so we have to do a little work to find an inverse function.
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We're going to take just a one-to-one piece of our sine function and work with that. So it might help to remember
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the horizontal line test from algebra. The horizontal line test, if a horizontal line crosses more than once the
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graph's not one-to-one. So we're looking for a little piece of the graph where when we draw that horizontal line it'll
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only cross one time, anywhere we draw it. There are a lot of those pieces. There's this piece. There's a piece here.
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There's a piece back here, and so on. It's customary to take this piece. It starts when x is negative pi over 2
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and ends here when x is pi over 2, and this is the piece that we're going to work on.
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So let's draw this one-to-one piece of the sine function separately. We'll start here. This is (negative pi over 2, -1).
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Here's the point (0, 0) and then the point (pi over 2, 1). And then we'll draw our sine curve. So this is our function
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y equals sine of x for negative pi over 2 less than or equal to x, less than or equal to pi over 2. And we can see this
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is our one-to-one piece we want to use. Each of these angles or x-values corresponds to one y or sine value, and
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furthermore, each y-value corresponds to exactly one angle value. Since we have a one-to-one piece, we're ready
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to define our inverse sine function. So y equals the inverse sine of x is the inverse of our function y equals sine of x
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for negative pi over 2 less than or equal to x, less than or equal to pi over 2. This function, y equals the inverse sine
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of x, has domain closed interval from negative 1 to 1 and range negative pi over 2 to pi over 2. Ok, so our
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inverse sine function is going to take a real number in the closed interval negative 1 to 1 and return a unique angle
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in the closed interval negative pi over 2 to pi over 2. Ok, one word about our notation. I know this looks like an
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exponent, this minus 1, but it's not an exponent. It's just a notational device.
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We read this "y equals the inverse sine of x." It's not an exponent.
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The definition of the inverse sine function tells us that the inverse sine of x is an angle. We're going to call that angle
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theta, in the closed interval negative pi over 2 to pi over 2, whose sine, s-i-n-e, is x. So we can write
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theta equals the inverse sine of x. It's just an angle, and we know that theta is in negative pi over 2 to pi over 2,
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this closed interval. So let's picture this here, and write this as our usual coordinate system where we look at
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the different quadrants. An angle of 0 radians would be drawn here, pi over 2 radians along the positive y-axis,
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and so on. If angle theta is in the closed interval negative pi over 2 to pi over 2, then where could theta lie? Well,
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here's negative pi over 2, there's pi over 2, so it can be anywhere in quadrant one or quadrant four. It could be along
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the positive x-axis, and it could be on either side of the y-axis, the positive or the negative. It will help us a lot
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to make a table. It will help us to sort of track when x is a certain number where does the terminal side of theta lie.
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So what are our choices? X can be negative 1. Remember our inverse sine function has domain negative 1 to 1,
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that closed interval, so the smallest is negative 1. Alright, well where would that be? That would mean it was along
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the negative y-axis, the terminal side there. In that case, we actually know the angle right away. It's going to be
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negative pi over 2. Anywhere in the fourth quadrant, the x must be a negative number, so we do from negative 1 to 0.
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Notice those are not allowing equal to, just an open interval. This is going to be in quadrant four. If x is actually
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equal to 0, it also corresponds to theta equals 0. That's on the positive x-axis. X might be between 0 and 1,
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and that's going to make the angle have its terminal side lie somewhere in quadrant one, and the last choice is
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x could be 1, in which case, the terminal side lies here on the positive y-axis, and we know in that case that the
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angle would be positive pi over 2. I think you'll find this chart very helpful as you look at these next examples.
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Let's practice finding exact values using our inverse sine function. The first example is to find the inverse sine
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of positive 1/2. So remember we're looking for an angle. We have to decide where that angle might be, and we know
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that its sine is positive 1/2. Ok, so in this problem, this is our x, and we can see it's positive 1/2 which is between
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0 and 1. So we're looking for an angle in quadrant one. In quadrant one, the angles are positive and acute, and
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let's look at this x-value more closely and see if it will help us figure out which special angle is the answer.
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The sine of pi over 6 is 1/2, so there you go. That's your answer. The positive acute angle in quadrant one whose
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sine is 1/2 is pi over 6. Let's try a second one. This time x is negative. It's between negative 1 and 0. So that means
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its terminal side is going to be somewhere in quadrant four. It needs to be a negative angle in quadrant four because,
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if you remember, the inverse sine function's range is negative pi over 2 to pi over 2. So we're looking for a negative
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angle, and I think we might want to use a reference angle to help us figure this out. Remember a reference angle is
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a positive acute angle. We hope that this is one of those special angles we know. The sine of pi over 4 is
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1 over the square root of 2, so our reference angle is pi over 4. So let's just look at the angle and make sure that we
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pick the right one. We want a negative angle in quadrant four. The reference angle is pi over 4, so this is actually
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that angle negative pi over 4. Ok, last one, inverse sine of negative 1. Negative 1 is one of these sort of special
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cases. The terminal side of theta will land on the negative y-axis, and theta has to be negative pi over 2.
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Let's turn our attention to defining the inverse cosine function. As you look at the graph of our cosine function,
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we see that again it is not a one-to-one function, so we need to find a one-to-one piece and define our inverse on
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that piece. Again we have many choices, but the customary choice is to take this first piece from 0 to pi, and it's first
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from the origin so that makes sense. We'll draw that one-to-one piece here and examine it more closely.
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We have the point (0, 1), (pi over 2, 0), and this one is (pi, -1). We're going to draw our cosine curve here. This is the
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function y equals cosine x for 0 less than or equal to x, less than or equal to pi, and we see that it is in fact a
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one-to-one function. We're ready to write the definition of the inverse cosine. So y equals the inverse cosine of x is the
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inverse of the function y equals cosine x on the interval from 0 to pi, the closed interval. The inverse cosine function
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has domain negative 1 to 1, again a closed interval, and range the closed interval from 0 to pi. So this inverse
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cosine function will take in a real number from this interval and give us back an angle in the closed interval 0 to pi.
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Let's use our definition of the inverse cosine function to help us find exact values. The definition tells us that inverse
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cosine of x is an angle, which we'll call theta, in the closed interval from 0 to pi whose cosine is x. So we're going
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to write this again a little bit differently. Theta is, the angle theta is the inverse cosine of x, and we know
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that 0 is less than or equal to theta, is less than or equal to pi. So where will we find the terminal side of such an
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angle theta? Well, it has to be from 0 to pi. Zero to pi, that's this part, quadrant one, quadrant two, the positive x-axis,
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the positive y-axis, or the negative x-axis. Ok, let's organize the information that we have. What are the possible
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values of x? Well, what's the domain of our inverse cosine function? The closed interval from negative 1 to 1.
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So if we have x equal to negative 1, then the terminal side will lie on the negative x-axis, and that is the angle
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theta equals pi. X can be something between negative 1 and 0. So that's a negative number that's going to lie,
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that means the terminal side will lie in quadrant two. X could be 0, in which case the terminal side lies on the
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positive y-axis, and so that means the angle theta is pi over 2. We have two more possibilities. If x is between 0 and 1,
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then the terminal side is somewhere in quadrant one. Or x could be 1, and that means that the angle lies on the
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positive x-axis, and that would be theta equals 0.
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Let's look at some examples of finding exact values of the inverse cosine function. Here's our first one. Find the
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inverse cosine of 1 over the square root of 2. So our x-value is 1 over the square root of 2. This is between 0 and 1,
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so we know that we're looking for an angle in quadrant one. In quadrant one, our angles are acute positive angles,
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and we just need to try to figure out if this is one of our special angles that we know. That's right. Cosine of pi over 4
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is 1 over the square root of 2, so the angle in the first quadrant whose cosine is 1 over the square root of 2 is
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pi over 4. Ok, second one, the inverse cosine of negative 1/2. So in this case, the x-value is negative 1/2. That's
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going to be in this interval, negative 1 to 0. That means the terminal side of our angle theta is in quadrant two.
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Ok, so we need to figure out which angle that is in quadrant two. We're going to use our reference angle to help us
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figure that out. The reference angle is a positive acute angle whose cosine is 1/2. Right, the reference angle should
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be pi over 3. Let's draw a picture of the angle that is going to be our answer. It's in quadrant two, the reference angle
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is pi over 3, and here's the angle we need. How do we figure this out? It's pi minus our reference angle,
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or 2 pi over 3. Ok, one more, the inverse cosine of 0. When x is 0, then we know that the terminal side of theta
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lies on the positive y-axis, and that makes our angle theta pi over 2.
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Let's now try to define an inverse tangent function. So the graph of the tangent function is shown on the board
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and we see, again, that it is not a one-to-one function, so in order to find an inverse we need to choose
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a one-to-one piece. Actually each of the pieces is one-to-one. Any will do, but the customary choice is this piece
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that runs from negative pi over 2 to pi over 2. Notice that at the ends those are vertical asymptotes, so we're going to
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choose the open interval from negative pi over 2 to pi over 2 to restrict the x's to. We'll highlight it here, and now let's
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draw it separately so we can look at it carefully. We have vertical asymptotes at pi over 2 and at negative pi over 2,
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and then we just have this one piece. This is definitely a one-to-one piece of the tangent function, and we've
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restricted the domain to this open interval to get a one-to-one piece. We're ready to write the definition of the
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inverse tangent function. Ok, y equals the inverse tangent of x is the inverse of y equals tangent x on the open interval
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negative pi over 2 to pi over 2. This inverse tangent function has domain all real numbers and range the open
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interval from negative pi over 2 to pi over 2.
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Let's talk about using the inverse tangent function to find exact values. From our definition, the inverse tangent of
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x is an angle, which we'll call theta, that is in the interval from negative pi over 2 to pi over 2, an open interval, whose
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tangent is x. We can write this a different way. The angle theta is inverse tangent of x. So again, this is just an angle,
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and we know that negative pi over 2 is less than theta which is also less than pi over 2. So let's look at where we find
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angles, theta, that meet this criteria. So negative pi over 2 to pi over 2 is here on the right side of this. It includes
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the first quadrant and the fourth quadrant. It includes the positive x-axis, and that's it because remember this is an
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open interval and the tangent is actually undefined at both pi over 2 and negative pi over 2. So we'll organize our
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information in a chart. If x is a negative number, then the terminal side of theta will lie in quadrant four, here.
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If x is equal to 0, then the terminal side of theta will lie on the positive x-axis, and that means that theta is 0. The
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only other possibility is that x is a positive number, and that would place the terminal side of theta in quadrant one.
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Let's look at some specific examples of finding exact values of the inverse tangent function. So our first one,
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inverse tangent of negative 1 over the square root of 3. So x here is a negative number, and whenever x is a negative
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number, the terminal side of the angle theta we're trying to find will be in quadrant four. Ok. Right, that's the tangent
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there of one of our special angles. So let's see if we can find a reference angle. Ok, the tangent of pi over 6 is
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1 over the square root of 3. So our reference angle is pi over 6, and we'll sketch a picture to make sure that we give
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the right name of our angle. Our angle, theta, is a negative angle in quadrant four, and the reference angle is pi over 6.
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So here is the angle we need. That's right. It's negative pi over 6. Ok, the inverse tangent of 1. So we look at our
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chart to help us out. 1 is a positive number, so that tells me the terminal side of theta is going to land somewhere in
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quadrant one. In quadrant one, what's the special angle whose tangent is 1? Right, pi over 4, and we have one
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more, inverse tangent of 0. Whenever x is 0, the terminal side of theta lies on the positive x-axis, so the angle is 0.
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Let's look at the instructions for this last group of examples. Use a calculator to approximate each of the following.
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Give your answer in radians. Round to four decimal places as needed. Well, I've already worked these out with
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my calculator, but I'd like to talk to you about some helpful tips. The first thing is notice that you don't really have any
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other way to work this problem except for the calculator approximation because this doesn't remind me of any
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of the special angles that we know. Ok, so how do we get this number? Well, this is important. Our answer should be
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in radians, so make sure your calculator is in radian mode. Where's the inverse cosine button? Usually it's the
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second button on the cosine key. If you can't find it, be sure to look in your calculator instruction manual. Type in
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inverse cosine of 4/5 and then look at the answer on the screen. Round it off to four decimal places if needed,
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and that's your answer. I've given you a couple of other ones to try. Make sure that you practice, that you learn
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how to use your calculator, and that you can get these rounded answers.