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)6x)
cos(sin^(-1)6x)=□ (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 $\theta=\sin^{-1}(6x)$

This means $\sin\theta = 6x=\frac{6x}{1}$. In a right - triangle, if $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$, we can assume the opposite side to angle $\theta$ has length $a = 6x$ and the hypotenuse has length $c = 1$.

Step2: Find the adjacent side $b$ using the Pythagorean theorem

By the Pythagorean theorem $a^{2}+b^{2}=c^{2}$. Substituting $a = 6x$ and $c = 1$, we get $(6x)^{2}+b^{2}=1^{2}$, so $b^{2}=1 - 36x^{2}$ and $b=\sqrt{1 - 36x^{2}}$ (since $x>0$ and we are dealing with the length of a side in a right - triangle, we take the positive square root).

Step3: Find $\cos(\sin^{-1}(6x))$

Since $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$ and $\theta=\sin^{-1}(6x)$, then $\cos(\sin^{-1}(6x))=\cos\theta=\frac{b}{c}=\sqrt{1 - 36x^{2}}$.

Answer:

$\sqrt{1 - 36x^{2}}$