# Thread: Concerning one step in Trig substitution

1. ## Concerning one step in Trig substitution

How does sqrt(9-9(sin theta)^2 = sqrt(9(cos theta)^2) ?

Thanks!

2. Originally Posted by C.C.
How does sqrt(9-9(sin theta)^2 = sqrt(9(cos theta)^2) ?

Thanks!
You have some mismatched delimiters as WolframAlpha would say. I believe the question at hand is $\displaystyle 9-9\sin^2(\theta)=9\left(1-\sin^2(\theta)\right)=9\cos^2(\theta)$

3. $\displaystyle \sqrt{9 - 9 sin^2(\theta) } = \sqrt{9 * (1 - sin^2(\theta) }$

Recall that

$\displaystyle cos^2 (\theta) + sin^2(\theta) = 1$

Manipulating,

$\displaystyle cos^2(\theta) = 1 - sin^2 (\theta)$

Using this property, we see that

$\displaystyle \sqrt{9 * ( 1 - sin^2(\theta)} = \sqrt{9 * cos^2(\theta)}$

Which neatly reduces to

$\displaystyle +/- 3 * cos(\theta)$

4. Originally Posted by Math Major
[tex] $\displaystyle \sqrt{9 * ( 1 - sin^2(\theta)} = \sqrt{9 * cos^2(\theta)}$

Which neatly reduces to

$\displaystyle 3 * cos(\theta)$
No it does not. That is false.
What about $\displaystyle \theta=\frac{3\pi}{4}$.