Write each trigonometric expression as an algebraic expression...

Question

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

sin (arccos u)

Accepted Answer

Hello, today we are going to be transforming the given trigonometric expression into an algebraic expression with the terms X. We will also assume that the value of X will be positive. So what we are given is sign of cosine inverse of two X. So how can we start this type of problem? Well, what we want to do is we want to first simplify the trigonometric expression. Now recall that every time we solve for a trigonometric function, whether it be an inverse function or a standard function, the output of that function will always be some angle theta. So using this, we can state that cosine inverse of two X will equal to some unknown angle theta. This will allow us to simplify the trigonometric expression to be sign of theta. Now, we have a problem here because we can't simplify this any further. And we currently have no way to rewrite this as an algebraic expression. So what can we do? Well earlier, we stated that if we were to evaluate any inverse trigonometric function or any standard trigonometric function, its output would be some angle theta. So what we said earlier was that cosine inverse of two X is going to equal to some angle theta. If we were to simplify this equation, we can start by taking the cosine function of both sides of the equation. This will allow us to rewrite the equation as two X is equal to cosine of theta. And reordering the terms will allow us to rewrite the equation as cosine data is equal to two X. Now you may be wondering why did I just solve for cosine data with the given substitution? While using the statement, we can create a right triangle with some angle theta in it and use that right triangle to find the algebraic value for sign of theta. Recall that cosine theta is equal to the adjacent side from the angle theta divided by the hypotenuse of the right triangle. Let's go ahead and draw a right triangle using this right triangle. We're going to set the angle theta at the far left corner of the triangle. Now again, cosine theta is equal to the adjacent side divided by the hypotenuse. Going back to the equation we solved cosine theta is equal to two X. But we can represent two X as a fraction. If we divide it by one, this now matches the form of the adjacent side divided by the hypotenuse the numerator that cosine is equal to will be the adjacent side to theta. Since the adjacent side is equal to two X, we are going to label the adjacent side as two X. The denominator for cosine data will be the hypotenuse of the right triangle. The denominator for cosine theta is one. So its hypotenuse of the right triangle will be one, we are only missing one side to this right triangle. So we're going to label that side A we can solve for this side by using the Pythagorean theorem. And the Pythagorean theorem states that the hypotenuse squared is equal to the sum of the two sides squared, which can be labeled as A squared plus B squared. We know the values for the two sides. And the hypotenuse, we labeled the hypotenuse as the value one squared and that will equal to one of the side links, which in this case is two X squared plus the second side length which we set to be A. So we can rewrite the Pythagorean theorem as one squared is equal to two X squared plus A squared. Simplifying the values will allow us to rewrite the theorem as one is equal to four X squared plus A squared. Since we want to solve for A, we're going to subtract four X squared to both sides of the equation. This will allow us to rewrite the equation as A squared is equal to one minus four X squared. And if we take the square root of both sides of this equation, we will be left with A is equal to the square root of one minus four X squared. This will allow us to rewrite the missing side as the square root of one minus four X squared. Now that we have the missing value, how can we use these values to solve for sine theta? Well, recall that the trigonometric identity for sine theta can be written as the opposite side to the angle theta divided by the hypotenuse of the right triangle. We just solved for the opposite side to the angle theta. It is the square root of one minus four X squared. So we can let sine theta equal to the opposite side of theta, which is the square root of one minus four X squared divided by the hypotenuse of the triangle, which is one. And since the denominator of this fraction is just one, we can rewrite the trigonometric expression to be the square root of one minus four X squared. This is going to be the simplest form of the algebraic expression that we want. And with that being said, the answer to this problem is going to be B. 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.

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Write each trigonometric expression as an algebraic expression in u, for u > 0.
sin (arccos u)