Write each trigonometric expression as an algebraic expression...

Question

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

tan (sin⁻¹ u/(√u² + 2))

Accepted Answer

Hello, today we are going to be rewriting the trigonometric expression as an algebraic expression with the terms X, we will also be assuming that the value of X is greater than zero. The trigonometric expression given to us is tangent of two multiplied by sine inverse of X divided by the square root of X squared plus six. Now how can we simplify this trigonometric expression? We recall that whenever we solve for a trigonometric function, whether it be an inverse or standard function, that value will always give us some angle theta. So using this, we can set sign inverse of X divided by the square root of X squared plus six equal to some angle theater. Using this, we can simplify the trigonometric expression to be tangent of two theta. Now using the Pythagorean identity recall that tangent of two theta is equal to two multiplied by tangent of theta divided by one minus tangent squared theta. We can use this identity to rewrite the expression to be two multiplied by tangent of theta divided by one minus tangent squared theta. Now currently we have no way to further simplify the expression or rewrite it as an algebraic expression. How can we continue from the step? Well, we can continue by using the substitution that we created earlier. If we were to take the sine function of both sides of this equation, we can rewrite the equation to be X divided by the square root of X squared plus six equal to sine of theta. Now, you may be wondering why I just sold for sign of data in the given substitution. The reason for this is because we can use sine theta is equal to X divided by the square root of X squared plus six. And we can use these values to create a right triangle. We're going to place the angle theta to the far left of the right triangle. Now recall that the value sine theta is equal to the opposite side with respect to the angle theta divided by the hypotenuse of the right triangle using X divided by the square root of X squared plus six. The opposite side of the right triangle will equal to X. So we're going to label that side as X, the denominator will be the value of the hypotenuse which is the square root of X squared plus six. So the hypotenuse will equal to the square root of X squared plus six. We are missing one side of this right triangle and we can label that side as a. Now, if we were to solve for a, we need to use the Pythagorean theorem using the Pythagorean theorem will leave us with the square root of X squared plus six squared is equal to X squared plus A squared. On the left hand side, the squared will allow us to, to cancel out the square root leaving us with X squared plus six equal to X squared plus A squared. So tracking X squared to both sides of this equation will leave us with sixes equal to A squared. And taking the square root of both sides of this equation will give us the value A is equal to the square root of six. That will allow us to rewrite the missing side of the right triangle as the square root of six. Now we have all three sides of this right triangle recall that the trigonometric value for tangent of data is equal to the opposite side with respect to the divided by the adjacent side with respect to the, we know that the opposite side is the value X and we just saw for the adjacent side which is the square root of six. Using this value, we can rewrite the current trigonometric expression as an algebraic expression that will allow us to rewrite the expression as two multiplied by X divided by the square root of six, divided by one minus the quantity X divided by the square root of six quantity squared. If we were to multiply the two values in the numerator together, that will allow us to rewrite the numerator as two X divided by the square root of six. Now, for the numerator, we cannot allow a square root to be within the denominator. So we're going to have to rationalize the numerator by multiplying the numerator and denominator by the square root of six. This will allow us to simplify the expression to be two X multiplied by the square root of six divided by six. Now, we want to simplify the denominator in the denominator, we have one minus the quantity X divided by the square root of six squared X divided by the square root of six quantity squared will allow us to simplify the denominator to be one minus X squared divided by six, we can rewrite one to be six divided by six. And using this, we can combine both of the fractions in the denominator to be six minus X squared divided by six. This will allow us to simplify the denominator of the expression to be six minus X squared divided by six. The expression is in the form of a complex fraction. So we're going to multiply the numerator and denominator by the reciprocal of the denominator. This value will be six divided by six minus X squared. This will allow us to reduce the denominator to be one. Now, in the numerator, we can reduce both of the sixes to be one and that will allow us to reduce the expression to be two X multiplied by the square root of six divided by six minus X squared. 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 a. 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.
tan (sin⁻¹ u/(√u² + 2))

Key Concepts

Inverse Trigonometric Functions

Trigonometric Identities

Pythagorean Theorem in Trigonometry

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