WEBVTT mathematics/math-analysis/selhorst-jones
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Hi, these are the trigonometry lectures for educator.com, and we're here to learn about the inverse trigonometric functions.
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We've already learned about sine, cosine and tangent, and today we're going to learn about the inverse sine, inverse cosine, and inverse tangent, probably better known as arc sine, arc cosine, and arc tangent.
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The arc sine function is also known as the inverse sine function, basically, it's kind of the opposite of the sine function.
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The idea is that you're given a value of x, and you want to find an angle whose sine is that value of x.
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In order to make this work, sines only occur between -1 and 1, so you have to be given a value of x between -1 and 1.
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You're going to try to give an angle between -π/2 and π/2.
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You'll try to give an angle between -π/2 and π/2 that has the given value as a sine.
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If arcsin(x)=θ, what that really means is that the sin(θ)=x.
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Now, there's some unfortunate notation in mathematics which is that arcsin is sometimes written as sin to the negative 1 of x.
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This is very unfortunate because people talk about, for example, sin²x means (sin(x))².
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You might think that sin^-1(x), would be (sin(x))^-1, which would be 1/sin(x).
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Now, that's not what it means, the sine inverse of x doesn't mean 1/sin(x), it means arcsin(x).
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This notation really is very ambiguous because this inverse sine notation could mean arcsin(x) or it could mean 1/sin(x), and so this notation is ambiguous because arcsin(x) and 1/sin(x) are not the same.
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The safest thing to do is to not use this notation sin^-1 at all.
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Let me just say, avoid this notation completely because it is ambiguous, it could be interpreted to mean these two different things that are not equal to each other.
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Instead, it's probably safer to use the notation arcsin(x), which definitely means inverse sin(x), and cannot be confused.
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The arccos(x), the arc cosine function is sort of the opposite of the cosine function.
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You're given a value of x, and you have to find an angle whose cosine is that value of x.
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Again, the value of x you must be given would have to be between -1 and 1, because those are the only values that come up as answers for cosine.
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What you try to do is produce an angle between 0 and π, so there's 0, π/2, π.
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You try to produce an angle between 0 and π that has that value as its cosine.
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Just like we have with sine, there's the problem of this misleading notation cos^-1(x), it could be interpreted as 1/cos(x) or it could be interpreted as arccos(x).
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The best thing to do is to avoid using this notation completely, cos^-1(x) is just misleading, it could be interpreted either way.
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Try not to use it at all, instead stick to the notation arccos(x).
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Finally, arc tangent is known as the inverse tangent function.
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You're given a value of x, and you want to find an angle whose tangent is x.
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For arc tangent, we're going to try to find the angles between -π/2 and π/2 because that covers all the possible tangents we could get.
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If arctan(x)=θ, that really means that tan(θ)=x.
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Just like with sine and cosine, we have this potentially misleading notation, tan^-1(x), could be interpreted to mean arctan(x) or 1/tan(x).
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Those are both reasonable interpretations but they mean two different things.
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Again, let's try to avoid this notation completely because it could mean two completely different things.
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We'll just try not to use that at all when we're talking about inverse tangents we'll say arctan instead of tan^-1.
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Let's get started with some examples.
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First off we have to identify the domain and range of the arc sine function, and then graph the function.
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Let's start out with the graph of sin(x), because arcsin is really meant to be an inverse to sin(x).
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I'll start with the graph of sin(x) here.
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Remember that when you're trying to find inverse functions, you take the function and you reflect it across the line y=x.
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In order for something to be a function, it has to pass the vertical line test.
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If you draw a vertical line, you shouldn't cross the graph twice.
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Since we're reflecting across the line y=x, that kind of switches the x's and y's, so we wanted something that will pass the horizontal line test.
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Let me draw this in red.
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We don't want to be able to draw a horizontal line and cross the graph twice.
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As you see, when I've drawn these horizontal lines, we crossed the graph in lots of places.
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We have a problem when we wanted to find the inverse sine function.
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The way we solve that is by not using all of the sine graph, we just cut off a piece of the sine graph that will pass the horizontal line test.
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I'm going to cut off a piece of the sine graph from -π/2 to π/2.
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If you just look at this portion of the sine graph, you see that it passes the horizontal line test.
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That means we can take the inverse just at that part of the sine function.
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Let me draw what that looks like when we reflect it.
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We're just taking this thick piece here, and I'm going to reflect that across that line y=x.
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Now, the notations that I had on the y and x axis are going to switch, it goes from now -1 to 1 on the x-axis, and -π/2 to π/2 on the y-axis.
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I can't keep going with this, I can't draw the rest of the graph because if I do draw any more, I'm going to get something that fails the vertical line test, it won't be a function anymore.
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This is the entire arcsin function.
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The domain here, it's all the numbers that you can plug into the arcsin, that's all values of x with -1 less than or equal to x, less than or equal to +1.
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We can't plug any other values of x into arcsin, and that's really because in the other direction the only values that come out of the sine function are between -1 and 1.
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The only values that you can plug into the arcsin function are between -1 and 1.
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The range, the numbers that come out of the arcsin, all values of x between -π/2 and π/2.
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Let me write that as y because when we think of those are the values coming out of the arcsin function.
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Those are all the y values you see here and you see that the smallest y-value is -π/2 and the biggest value we see is π/2.
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Arcsin takes in a number between -1 or 1, gives you a number between -π/2 and π/2 and its graph looks like a sort of chopped of piece of the sine graph.
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Remember, the reason we had to chop it off is to get a piece of the sine graph that would satisfy the horizontal line test, so that the arcsin graph satisfies the vertical line test and really is a function.
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In our second example here, we have to find the arcsin of the sin(2π/3) and then some similar values for our cosine and our tangent.
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The first thing to do here is really to figure out where we are on the unit circle.
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There's 0, π/2, π, 3π/2 and 2π.
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Let's figure out where each of this angle is on the unit circle.
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2π/3 is over here, there's 2π/3.
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Remember, arcsin is always between -π/2 and π/2.
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Arcsin is between -π/2 and π/2, so there's -π/2 down here.
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We want to find an angle between -π/2 and π/2.
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On the right-hand side that has the same sine as 2π/3.
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Well, sine is the y-value so we want something that has the same y-value as 2π/3.
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That angle right there has the same y-coordinate as 2π/3, that's π/3.
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From the graph, we see that π/3 has the same sine as sin(2π/3).
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So, arcsin(2π/3) is π/3.
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It's an angle whose sine is the sin(2π/3).
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Let's try the next one, -5π/6.
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-π is over there, so 5π/6 in the negative direction puts you over there, there's -5π/6.
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We want an angle whose cosine is the same as the cosine(-5π/6).
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Cosine is the x-value, so we want an angle that has the same x-value as the one we just found.
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There it is right there, 5π/6.
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That's between 0 and π, which is the range for the arccos function, it's always between 0 an π.
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That means that arccos of the cos(-5π/6) is 5π/6, that's an angle between 0 and π that has the right cosine.
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Finally, we want to find the arctan of tan(3π/4).
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3π/4 is over there.
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There's 3π/4.
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We want to find an angle that has the same tangent as that one.
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Remember, arctan is forced to be between π/2 and -π/2.
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We want to find an angle between π/2 and -π/2 that has the same tangent as 3π/4.
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If we go straight across the origin there, we started with an angle whose cosine is negative and sine is positive.
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This angle right here has a positive cosine and negative sine, so it'll end up having the same tangent, and that angle is -π/4.
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That's an angle inside the range we want that has the same tangent.
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Arctan of tan(3π/4) is -π/4.
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Those are really quite tricky.
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What we're being asked to do here in each case is we're given an angle, for example 2π/3.
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We want to find arcsin of sin(2π/3).
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We look at the sine of 2π/3, and then we want to find another angle that has that sine but it has to be in the specified range for arcsin.
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We're really trying to find what's an angle between -π/2 and π/2 whose sine is the same as sin(2π/3).
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That's why we did this reversal to find the angle π/3 that has the same sine as the sin(2π/3).
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That was kind of the same process of all three of them, finding angles that have the same cosine as -5π/6, but is in the specified range, finding an angle that has the same tangent as 3π/4 but is in the specified range.
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Our third example here, we're asked to identify the domain and range of the arctan function, and graph the function.
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Remember when we graphed arcsin, we started out with a graph of sine, so let me start out here with a graph of tangent.
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I'll graph it in blue.
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Has asymptotes of π/2, and then it starts repeating itself.
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It has period π.
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This is 0, π/2, -π/2, and that's π, 3π/2.
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What I graphed in blue there is tan(θ).
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Now, I want to take this graph and I want to flip it around the line y=x.
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Let me graph the line y=x.
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I want to get something that it will be a function, it has to pass the vertical line test after I flip it.
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That means it has to pass the horizontal line test before I flip it.
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Just like with sine, the tangent function fails the horizontal line test badly.
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You can draw a horizontal lines and intersect it in lots of places.
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Just like with sine, we're going to cut off part of the tangent function and use that to form the arctan function.
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We're going to cut off just one of these curves.
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I'll cut it off these asymptotes at -π/2 and π/2.
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I'll just take this part of the tangent function and I'll flip it around to make the arctan function.
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I'll do this in red.
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Arctan(x).
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I'm going to flip this around just that one main branch of the tangent function.
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Since the tangent function had vertical asymptotes, the arctan function is going to have horizontal asymptotes at -π/2 and π/2.
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Those are my asymptotes right there at -π/2 and π/2.
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We're also asked to identify the domain and range.
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The domain means what numbers can we plug in to arctan(x).
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Domain is all values of x because we can plug any number in on the x-axis here and we'll get an arctan value.
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-infinity less than x less than infinity.
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The range is all x with, okay, let me write this as y because these are values on the y-axis.
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All y with -π/2 less than y less than π/2.
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Let me emphasize here that π/2 and -π/2 are not included, π/2 themselves are not in the range.
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These inequalities, they're not less than or equal to they're strictly less than.
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The reason for that is that the arctan function never quite gets to -π/2 or π/2, it goes down close -π/2 and it goes up close to π/2 but it never quite gets there.
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The reason for that, if you kind of look back at the tangent function, was that these asymptotes never quite get to -π/2 or π/2.
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You can't take the tangent of -π/2 or π/2, because you're trying to divide by zero back in the tangent function.
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To understand this problem really, you have to remember what the graph of tan(θ) looks like.
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Then you take its one main branch and you flip it over, and you get the graph of arctan(x), and the domain and range just come back to any things in the domain.
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The range can go from π/2 to -π/2, but you don't include those end points.
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The asymptotes are the horizontal lines at -π/2 and π/2.