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

`√(9x^2+36) = 6 \csc(\theta)`

Explanation:

Given

`(x)/(2)=\cot(\theta)`

Required

Express
`√(9x^2+36)` as trigonometry expression

`√(9x^2+36)`

Factorize

`√(9x^2+36) = √(9(x^2+4))`

Split

`√(9x^2+36) = √(9) * √((x^2+4))`

`√(9x^2+36) = 3 * √((x^2+4))`

`√(9x^2+36) = 3√((x^2+4))`

We have:

`(x)/(2)=\cot(\theta)`

Make x the subject

`x = 2 \cot(\theta)`

So:

`√(9x^2+36) = 3√((x^2+4))`

`√(9x^2+36) = 3√(((2 \cot(\theta))^2+4))`

Evaluate all squares

`√(9x^2+36) = 3√(4\cot^2(\theta)+4)`

Factorize

`√(9x^2+36) = 3√(4(\cot^2(\theta)+1))`

Split

`√(9x^2+36) = 3√(4) * √(\cot^2(\theta)+1)`

`√(9x^2+36) = 3*2 * √(\cot^2(\theta)+1)`

`√(9x^2+36) = 6 * √(\cot^2(\theta)+1)`

In trigonometry

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

So, we have:

`√(9x^2+36) = 6 * √(\csc^2(\theta))`

Evaluate the square root

`√(9x^2+36) = 6 * \csc(\theta)`

`√(9x^2+36) = 6 \csc(\theta)`