Question

use a right triangle to write the following expression as an algebraic expression. assume that x is positive and that the given inverse trigonometric function is defined for the expression in x. cos(sin^(-1)2x)
cos(sin^(-1)2x)=□
(simplify your answer, including any radicals. use integers or fractions for any numbers in the expression. rationalize all denominators. do not factor.)

Explanation:

Step1: Let $\theta=\sin^{-1}(2x)$

This means $\sin\theta = 2x=\frac{2x}{1}$. In a right - triangle, if $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, we can assume the opposite side to the angle $\theta$ has length $a = 2x$ and the hypotenuse has length $c = 1$.

Step2: Use the Pythagorean theorem

By the Pythagorean theorem $a^{2}+b^{2}=c^{2}$, where $a = 2x$, $c = 1$. Substitute these values: $(2x)^{2}+b^{2}=1^{2}$, so $b^{2}=1 - 4x^{2}$, and $b=\sqrt{1 - 4x^{2}}$ (since $x>0$ and we are dealing with the length of a side of a triangle, we take the positive square root).

Step3: Find $\cos\theta$

We know that $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. Since the adjacent side to the angle $\theta$ is $b=\sqrt{1 - 4x^{2}}$ and the hypotenuse is $c = 1$, then $\cos(\sin^{-1}(2x))=\cos\theta=\sqrt{1 - 4x^{2}}$.

Answer:

$\sqrt{1 - 4x^{2}}$