# Math Help - couple problems i have

1. ## couple problems i have

I need help with a few problems I have left.

1. Given : r = csc²Θ
how do I find the range of all possible values of r and also how do I find a Cartesian equation without fractions

2. Given : √(1+3cos²Θ(2-cos²Θ)) dΘ
how do i show that the expression under the radical is positive for all angle

2. ## solved

To solve the first one, it's best to change the problem to Cartesian first, just to see it much easier. In order to do that, you'll want to do this:
$r=1/sin^2\Theta$
$r \times sin^2\Theta=1$
$r^2 \times sin^2\Theta = r$ *multiply both sides by r*
$y^2 = r$ *substitute $y^2$ for $r^2 sin^2\Theta$ *
$y^2 = \pm\sqrt{x^2 + y^2}$ *substitute in $\pm\sqrt{x^2 + y^2}$ for r*

Now you have your Cartesian. So, now let's solve for x, much easier than to solve for y. You should get $\pm y \sqrt{y^2 - 1} = x$
Now, since we want the inverse, just look at the Domain of the above function and you have the Range you wanted originally, which is $\Re/(-1,1)$

to answer part 2, we look at first $cos^2\Theta$ and see that for $cos\Theta$ is negative when $\Theta\pi/2,3\pi/4)" alt="\Theta\pi/2,3\pi/4)" />, but since it's squared, then we don't have to worry about that term. So, since $cos^2\theta$ is positive for all $\Theta$ and since the largest value of $cos^2\Theta$ is 1, then 2-1 is 1, hence positive.

Hope this helps.

3. ## Thanks!!!!!

Thanks soooooo much! You helped me out a ton!! Thanks also for explaining how it works.