In Exercises 52–53, use a right triangle to write each expression...

Question

In Exercises 52–53, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x.sec(sin⁻¹ 1/x)

Accepted Answer

Welcome back in this problem, we want to transform the trigonometric expression of the core. Second of the inverse of the cosine of seven divided by X. It's definitely a mouthful into an algebraic expression. And we want to use a right triangle to help us to write that algebraic expression. So first, let's ask ourselves, what do we already know? We know that everything inside this bracket represents an angle. So our angle for or right angle triangle is equal to the inverse cosine of seven divided by X. We also know that all of this is a trigonometric ratio. So it represents the part of a right angled triangle. So if we can figure out what the sides of that right angle triangle are, then we'll be able to find the second. So let's see if we can use this to help us figure that out. Now, if the anger is equal to the inverse cosine of seven divided by X, that means the core sign of the anger would be equal to seven divided by X because that's the ratio, what do you know about the co sign? Remember that the cosine of an angle is equal to the adjacent side divided by the hypotenuse. So that means seven is our adjacent side and X is our hypotenuse. And if I were to do a little sketch of that right triangle over here with my anger right here and my 90 degree angle in this corner, then it means that the site adjacent to the angle would be seven and the hypotenuse would be X since I have both these sides, I can figure out the third side by using the Pythagorean theorem because remember the Pythagorean theorem tells us that the square of the two shorter sides is equal to the square of the longest side. So let's see if we can figure that out. So by the Pythagorean theorem, this tells us that seven squared plus or unknown side, let's call it a. So seven squared plus A squared equals X squared. The square of the hypo continues, we can solve for that unknown side because that means A squared is going to be equal to X squared minus whatever seven squared is, which would be 49. And that means A would be equal to the square root of all of that. So that's X squared minus 49 all under the square root. So now we've figured out that unknown side, we know that this side is the square root of X squared minus 49. So let's replace it right here. Let's put it in a different color. Oops. Yeah, let's change it to orange. I think that's a great contrast. So remember, what do we want to know? We want to know the of this expression and we've got a far away because now we have all the three sides. What do you know about the? Well, recall that the of an anger is equal to the reciprocal of its sign? OK. So that means that our co is going to be equal to one divided by sin X. No, the remember sine is equal to opposite over the hypotenuse. So it's reciprocal would be equal to the hypotenuse divided by the opposite side. Do we know the hypo? Yes, we know that it's X and do we know the opposite side to the anger? Yes, we know that it's X, the square root of X squared minus 49. We're almost there. The only thing we need now is to get rid of the square root in the denominator. So we can do that by rationalizing. So, rationalizing our expression means that we'd multiply it by its radical. So that's X squared minus 49 all under the square root times the radical of X squared minus 49 divided by the radical of X squared minus 49. No. Guess what? When we multiply both of these, it will get rid of the square root and that will leave us with just X squared minus 49. But in the numerator, we're going to have X times the square root of X squared minus 49. And that would be our result for the second of the inverse cosine of seven divided by X, which would be in our answers. Answer B.

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In Exercises 52–53, use a right triangle to write each expression as an algebraic expression. Assume that x is positive and that the given inverse trigonometric function is defined for the expression in x.sec(sin⁻¹ 1/x)

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