Parametrics tangent line with point given

[LBK]

Homework Statement

I understand the setup for finding the slope, but always get confused whether I've fully simplified when trig identities get involved. [/B]

Homework Equations

My dy/dx is [/B]
4sin(θ)cos(θ)
-2csc²(θ) which I simplified to just (-2sin(θ)cos(θ))/(csc²(θ)

Does that simpify to anything better than the way I have done it?

The Attempt at a Solution

Once I simplify the above expression fully, I know how to plug in the point to find the equation. Thanks so much!

 

[LCKurtz]

$-2\sin^3\theta \cos\theta$