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