Question

Use the suggested substitution to write the expression as a trigonometric expression. Simplify your answer as much as possible. Assume 0≤θ≤π2.
$\sqrt{9x^2+36$ , x/2=cot(∅)

Answer 1

Answer:

$\sqrt{9x^2+36} = 6 \csc(\theta)$

Step-by-step explanation:

Given

$\frac{x}{2}=\cot(\theta)$

Required

Express $\sqrt{9x^2+36}$ as trigonometry expression

$\sqrt{9x^2+36}$

Factorize

$\sqrt{9x^2+36} = \sqrt{9(x^2+4)}$

Split

$\sqrt{9x^2+36} = \sqrt{9} * \sqrt{(x^2+4)}$

$\sqrt{9x^2+36} = 3 * \sqrt{(x^2+4)}$

$\sqrt{9x^2+36} = 3\sqrt{(x^2+4)}$

We have:

$\frac{x}{2}=\cot(\theta)$

Make x the subject

$x = 2 \cot(\theta)$

So:

$\sqrt{9x^2+36} = 3\sqrt{(x^2+4)}$

$\sqrt{9x^2+36} = 3\sqrt{((2 \cot(\theta))^2+4)}$

Evaluate all squares

$\sqrt{9x^2+36} = 3\sqrt{4\cot^2(\theta)+4}$

Factorize

$\sqrt{9x^2+36} = 3\sqrt{4(\cot^2(\theta)+1)}$

Split

$\sqrt{9x^2+36} = 3\sqrt{4} * \sqrt{\cot^2(\theta)+1}$

$\sqrt{9x^2+36} = 3*2 * \sqrt{\cot^2(\theta)+1}$

$\sqrt{9x^2+36} = 6 * \sqrt{\cot^2(\theta)+1}$

In trigonometry

$\cot^2(\theta)+1 = \csc^2(\theta)$

So, we have:

$\sqrt{9x^2+36} = 6 * \sqrt{\csc^2(\theta)}$

Evaluate the square root

$\sqrt{9x^2+36} = 6 * \csc(\theta)$

$\sqrt{9x^2+36} = 6 \csc(\theta)$