Write each trigonometric expression as an algebraic expression...

Question

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

sin (2 sec⁻¹ u/2)

Accepted Answer

Hello. Today we will be transforming the given trigonometric expression into an algebraic expression. We will be transforming it into an algebraic expression in terms of X. And we will be assuming that the value of X is greater than zero. So the trigonometric expression given to us is sign of two multiplied by seeking inverse of X divided by four. In order to transform this trigonometric expression, we want to first simplify the angle of the sine function. Now recall that whenever we evaluate a trigonometric function, whether it be an inverse or a standard function, its output is always going to be the angle theta. So using this, we can go ahead and represent see inverse of X divided by four as some angle theta. This will allow us to simplify the expression to be sign of two theta. Now recall that there is a trigonometric identity that states that sine of two theta is equal to two multiplied by sine theta multiplied by cosine of theta. Using this identity, we can further simplify the expression to be two multiplied by sine theta multiplied by cosine of the. Now, at this point, we don't have anything that we can use to further simplify the expression or to rewrite it as an algebraic expression in terms of X. So how can we continue from this step? Well earlier, we stated that seeking inverse of X divided by four will equal to the angle theta because when we evaluate this function, its output would be some angle. So what we can do is we can set seek an inverse of X divided by four equal to some angle theta. Now, if we want to simplify this equation, we can take the seeking function of both sides of the equation. Doing so will allow us to rewrite the equation as X divided by four is equal to second theta. And if we reorder the terms, we can rewrite this equation to be second of theta is equal to X divided by four. Now you may be wondering why are we doing all of this? Well, the reason why we want to do this is because we can use this current equation to create a positive right triangle. And we can use that right triangle to solve for the values of sine theta and cosine theta. Now recall that the trigonometric identity for seek and theta is given to us as seek of theta is equal to the value of the hypotenuse of the right triangle divided by the X value of the right triangle. So if we were to create a positive right triangle and we can allow the angle theta to be located on the far side of the triangle, we can use the value of sin to label the missing sides. Since sin theta is equal to the hypotenuse divided by the X value. And we know that the hypotenuse is equal to X and the adjacent side to the angle is equal to four. We can label the hypotenuse to be X and the adjacent site to theta to be four. Now we are missing one side of this right triangle. We can label this side to be a, we can further solve for this side by using the Pythagorean theorem. And the Pythagorean theorem states that X squared will equal to four squared plus the value A squared evaluating this will leave us with X squared is equal to 16 plus A squared. If we subtract 16 to both sides of this equation, we are left with A squared is equal to X squared minus 16. And if we were to take the square root of both sides of this equation, we get A is equal to the square root of X squared minus 16. Now that we have the value for A, we can rewrite A in the right triangle to B the square root of X squared minus 16. Now that we have all three missing sides, we can solve for sine theta and cosine theta with respect to the current angle theta in the right triangle recall that sign the is equal to the opposite side from the angle, theta divided by the hypotenuse of the right triangle. And cosine theta is equal to the adjacent side from the angle theta divided by the hypotenuse. Currently, we know that the hypotenuse is the value X, the opposite side is the square root of X squared minus 16. And the adjacent side to the angle theta is equal to four using these identities, we can rewrite sign data and cosine data using the respective values from the right triangle. Thus, we can rewrite the algebraic expression to be two multiplied by the opposite side from the angle theta which is the square root of X squared minus 16 divided by the hotpot news which is X multiplied by the adjacent side from the angle theta which is four divided by the hypotenuse which is X. Now, what we need to do is we need to simplify the algebra multiplying the two fractions together will allow us to rewrite the expression as two multiplied by four multiplied by the square root of X squared minus 16 divided by X squared. And multiplying the two into the fraction will leave us with eight multiplied by the square root of X squared minus 16 divided by X squared. This will be the simplest algebraic form of the given trigonometric 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 type of problem and I'll go ahead and see you all in the next video.

Struggling with Trigonometry?

Write each trigonometric expression as an algebraic expression in u, for u > 0.
sin (2 sec⁻¹ u/2)

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Inverse Sine, Cosine, & Tangent practice set

Struggling with Trigonometry?

Write each trigonometric expression as an algebraic expression in u, for u > 0.sin (2 sec⁻¹ u/2)

Watch next

Inverse Sine, Cosine, & Tangent practice set

Write each trigonometric expression as an algebraic expression in u, for u > 0.
sin (2 sec⁻¹ u/2)