Write each trigonometric expression as an algebraic expression...

Question

Write each trigonometric expression as an algebraic expression in u, for u > 0.

sec (arccot (√4―u² )/ u)

Accepted Answer

Hello, today we are going to be transforming the given trigonometric expression into an algebraic expression. In terms of X, we will also be assuming that the value of X is greater than zero. So we are given sequent of cotangent inverse of the square root of 25 minus X squared divided by X. The first thing we want to do is we want to simplify the trigonometric expression. Now recall that any time we solve for a trigonometric function, whether it be a standard function or an inverse function, the output is always going to be some angle theta. Using this, we can set cotangent inverse of the square root of 25 minus X squared divided by X equal to some angle theta. This will allow us to rewrite the trigonometric expression as sequent of theta. Now, currently, we have no way to further simplify seeking of data and we currently don't have a way to rewrite it as an algebraic expression. So how can we continue from this step? Well, take a look at the substitution we made earlier. If we were to take the code tangent function of both sides of this equation, we can rewrite the equation to be the square root of 25 minus X squared divided by X is equal to cotangent of theta. Now, you may be wondering why it just solved for cotangent of the. The reason for this is because we currently know that a code tangent of the is equal to the square root of 25 minus X squared divided by X. We can use this value to label the sides of a right triangle. We will place the angle theta to the far left of the right triangle. And if we solve for all the sides of the right triangle, we can use this right triangle to rewrite seek. And the as an algebraic expression now recall that the trigonometric value for co tangent of data is equal to the adjacent side with respect to theta divided by the opposite side with respect to theta. That means that the numerator, the square root of 25 minus X squared will be the value of the adjacent side with respect to theta. That's going to be the bottom side of this right triangle. And the denominator for the value of cotangent the is going to be the value of the opposite side with respect to theta. So we will label the opposite side as X, we are missing the side of the right triangle, which is the hypotenuse, we will label that side as a, we can solve for a by using the Pythagorean theorem. The Pythagorean theorem will allow us to write a squared is equal to the quantity the square root of 25 minus X squared. That quantity squared plus X squared squaring a square root will allow us to eliminate the square root. So we can simplify this equation to be A squared is equal to 25 minus X squared plus X squared, negative X squared plus X squared will cancel each other out. So we are left with A squared is equal to 25. And if we take the square root of both sides of this equation, we get the value of A is equal to five. So we know that the value of the hypotenuse is five. Now that we have the values of all three sides of this right triangle, we can rewrite seek. And the as an algebraic expression now recall that the trigonometric value for seeking of theta is equal to the hypotenuse divided by the adjacent side with respect to theta. We know that the adjacent side with respect to theta is the square root of 25 minus X squared. And we know the value of the hypotenuse is five using these values we can rewrite seek and the to be five divided by the square root of 25 minus X squared. Now recall that we cannot leave a square root in the denominator of our answer. So we're going to need to rationalize our expression. We're going to multiply the numerator and denominator by the square root of 25 minus X squared. This will allow us to simplify the expression to be five multiplied by the square root of 25 minus X squared divided by the quantity 25 minus X squared. And this is going to be the simplest form of the algebraic expression. And with that being said, the answer to this problem is going to be C so I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

Write each trigonometric expression as an algebraic expression in u, for u > 0.
sec (arccot (√4―u² )/ u)

Key Concepts

Trigonometric Functions

Inverse Trigonometric Functions

Algebraic Manipulation

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