How does sqrt(9-9(sin theta)^2 = sqrt(9(cos theta)^2) ?
Thanks!
$\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) $