How do I write tan(2 sin^-1(x)) as an algebraic expression?

Question

An open algebra book with blank pages. On the right page, draw a large protractor and a compass to illustrate the importance of geometry in algebra. On the left page, gently sketch the sine and tangent curves to demonstrate their relevance to trigonometry and algebra. Do not use any text within the image itself.

How do I write tan(2 sin^-1(x)) as an algebraic expression?

Answer 1

Consider a right-angled triangle with hypotenuse = 1, and opposite = x

sinθ = opposite/hypotenuse = x/1
then by Pythagoras the adjacent side = √(1-x^2)
and cosθ = √(1-x^2)

sin^-1(x) is the angle θ so that sinθ = x
then tan(2 sin^-1 (x) )
= tan (2θ)
= sin 2θ / cos 2θ
= 2sinθcosθ/(cos^2 θ - sin^2 θ)
= 2x√(1-x^2) / (1-x^2 - x^2)
= 2x√(1-x^2) / (1- 2x^2)

Answer 2

To write tan(2 sin^-1(x)) as an algebraic expression, we'll start by using the double angle identity for tangent. The double angle identity for tangent states that tan(2θ) = 2tan(θ) / (1 - tan^2(θ)).

Step 1: Substitute θ with sin^-1(x).
tan(2 sin^-1(x)) = 2tan(sin^-1(x)) / (1 - tan^2(sin^-1(x)))

Step 2: Let's find the value of tan(sin^-1(x)).
tan(sin^-1(x)) is equal to the ratio of the opposite side to the adjacent side of a right triangle with angle sin^-1(x). Since sin^-1(x) represents an angle that satisfies sin(sin^-1(x)) = x, we can choose a right triangle where the opposite side is x and the hypotenuse is 1. Then, by using Pythagoras' theorem, the adjacent side can be found as √(1 - x^2).
Thus,
tan(sin^-1(x)) = x / √(1 - x^2)

Step 3: Substitute this value back into the original expression.
tan(2 sin^-1(x)) = 2 * (x / √(1 - x^2)) / (1 - (x / √(1 - x^2))^2)

Simplifying the expression gives:
tan(2 sin^-1(x)) = 2x / (1 - x^2)

Answer 3

Well, it looks like you're tangled up in a web of trigonometry! Let me see if I can untangle it for you. First, let's break it down step by step.

We have tan(2 sin^(-1)(x)). To simplify this expression, we can use the double-angle formula for tangent, which states that tan(2θ) = (2tan(θ))/(1-tan^2(θ)).

Now, let's substitute θ with sin^(-1)(x). tan(2 sin^(-1)(x)) becomes (2tan(sin^(-1)(x)))/(1-tan^2(sin^(-1)(x))).

But hang on a minute! We can substitute tan(sin^(-1)(x)) with x as well, because tan(sin^(-1)(x)) is just equal to x. So, our expression becomes (2x)/(1-x^2).

And voila! We have written tan(2 sin^(-1)(x)) as an algebraic expression, which is (2x)/(1-x^2). Keep calm and keep solving those equations!

Answer 4

To write tan(2 sin^-1(x)) as an algebraic expression, we can use the double-angle trigonometric formula:

tan(2θ) = 2tan(θ) / (1 - tan^2(θ))

First, let's substitute θ with sin^-1(x). So, θ = sin^-1(x).

Now, substitute θ into the formula to get:

tan(2 sin^-1(x)) = 2tan(sin^-1(x)) / (1 - tan^2(sin^-1(x)))

Next, we need to express tan(sin^-1(x)) in terms of x. To do this, we can make use of the Pythagorean identity for trigonometric functions:

sin^2(θ) + cos^2(θ) = 1

Since sin(θ) = x (from sin^-1(x)), we can substitute it into the Pythagorean identity:

x^2 + cos^2(θ) = 1

Solving for cos(θ), we get:

cos(θ) = √(1 - x^2)

Now, substitute the values back into the expression:

tan(2 sin^-1(x)) = 2tan(sin^-1(x)) / (1 - tan^2(sin^-1(x)))
= 2x / (1 - (√(1 - x^2))^2)
= 2x / (1 - (1 - x^2))
= 2x / x^2
= 2/x

Therefore, the algebraic expression for tan(2 sin^-1(x)) is 2/x.