Find the exact value of each real number y if it exists. Do no...

Question

Find the exact value of each real number y if it exists. Do not use a calculator.

y = sec⁻¹ 1

Accepted Answer

Welcome back. I am so glad you're here. We are asked to determine the exact value of the given inverse trigonometric function without using a calculator. Our given trigonometric function is Y equals the inverse C can of the square rut of two. Our answer choices are answer choice A Y equals pi divided by four answer choice by equals negative pi divided by four answer choice cy equals three pi divided by four and answer choice dy equals negative three pi divided by four. All right. The first step is to figure out whether this square two is within the domain of our inverse C can't function. So I'm going to draw a few different functions yet so that we can think about what is the domain of inverse C can't. So I'm going to draw a vertical Y axis, a horizontal X axis. They come together at the origin in the middle. I'm going to graph some functions very quickly. So if you need to review graphing functions, please definitely check out previous lesson videos. CC A is the reciprocal function to cosine. So I'm going to start off with a cosine function. And now keep in mind that all of these inverse functions need to be 1 to 1. So that means that they have to pass a horizontal line test. And therefore, we're only going to draw about half of a period for each of these functions. So for this first cosine function, only half of the period, it's going to start off at the point 01, then it's going to pass through the point pi divided by 20 And then it's going to end for our purposes at the point pi negative one. And we can draw a smooth line connecting those. And there is our cosine function. Now we're talking about C can't, which is going to be the reciprocal function of that. So C can't is going to again begin at that 0.01. But then instead of heading down for Y values, it's going to head up and then it's going to have an asom tote at X equals pi divided by two. And then to the right of that vertical Asymptote, it's coming up from negative infinity curving and then meeting at that point pi negative one, there's our C can't function. So now we're talking about the inverse C A, which means that all of our Y values are exchanged for X values and all of our X values are exchanged for Y values. And what's important is that we think about the domain and range for our regular C can't function. Well, the domain for our inverse C can is going to be the range of our regular C can't. And the range for our inverse is going to be the domain of our regular C can. So just looking at our graph talking about the regular C can function, we can see that the domain is going to be from zero all the way to pi divided by two, but not including pi divided by two because there's an Asymptote there. So we'll close that with parentheses in union with. And then we'll start again right after pi divided by two. So that gets a parentheses all the way up to pi inclusive of pie. So that's the domain for our regular C can function the range. It's coming all the way up from negative infinity up to negative one inclusive of negative one. And then that's in union with and it starts again at positive one and heads up to positive infinity. So now for our inverse C can't domain, that's going to be the range of our C can't function. So the inverse C cans domain is oops, it's going to be from negative infinity to negative one inclusive of negative one in union with one to infinity. So does the square root of two fall within our domain? And yes, it does. The square root of two is close to 1.4 a little bit more specific than that, but it does fall in our domain from one to infinity. So we can continue all that. And now, what we're going to do is we take a look at our Y equals the inverse C can of the square root of two. And it can help to put that phrase it in a different way. We know that we're also saying that Y is the number whose C can't is the square rut of two. Or in other words, the C can't of Y equals the square root of two. That is the same thing and that's a little bit easier to solve. So thinking of the C can of Y equals the square root of two, we're looking for an angle here. The C can of what angle equals the square root of two? All right, we recall from very previous lessons that the C can of an angle. If we're talking about the unit circle that would be equal to one divided by X. And then we take a look at our unit circle going to draw a quick sketch of a unit circle still have a vertical Y axis, a horizontal X axis. They come together the origin, the unit circle lies over that and every point on the unit circle is one unit away from the origin. And if we think of our first quadrant, so up into the right of the origin, we have that um angle at pi divided by four with the X and Y values of square root two divided by two square it two divided by two. And if you take one, divided by X equals what our C can of Y is equal to the square root of two. Here, our X value is going to be if we solve for X. So let's say we multiply both sides by X. Then on the right, we've got square root two, multiplied by X is equal to, on the left one, then we can divide both sides by the square root of two. So we know that X is equal to one divided by the square root of two. But we want to rationalize that denominator. So we're multiplying both the numerator and the denominator by the square root of two. And we are going to come up with an X value of square root two divided by two. And there, there's our X value for that angle pi divided by four. So if we have the C can of pi divided by four, it is absolutely going to equal the square root of two. So that's where we find our Y value Y here is equal to pi divided by four. We look at our answer choices and this matches with answer choice. A well done. We'll catch you on the next one.

Find the exact value of each real number y if it exists. Do not use a calculator.
y = sec⁻¹ 1

Key Concepts

Inverse Secant Function (sec⁻¹ x)

Secant Function Definition

Exact Values of Trigonometric Functions

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