Hi. I think I've reversed a sign somewhere, but for the life of me I can't seem to find it.

I have these parametric equations (using t to represent theta):

$\displaystyle x = 2 - 3\,cos\,t$

$\displaystyle y = 3 + 2\,sin\,t$

I've got to find the equation of the tangent line at various given points on the curve. So, I want to create a general equation first that I can plug the points into. First, I solve for t:

$\displaystyle y = 3 + 2\,sin\,t$

$\displaystyle t = \arcsin{\frac{y-3}{2}}$

Then I get the slope:

$\displaystyle \frac{dy/dt}{dx/dt} = \frac{3\,sin\,t}{2\,cos\,t} = \frac{3}{2}\,tan\,t$

Now, the intercept in the tangent line equation will be

$\displaystyle B = y_{1} - (\frac{3}{2}\,tan\,(\arcsin{\frac{y_{1}-3}{2}}))\,x_{1}$

Giving me

$\displaystyle y = (\frac{3}{2}\,tan\,(\arcsin{\frac{y_{1}-3}{2}}))\,x + y_{1} - (\frac{3}{2}\,tan\,(\arcsin{\frac{y_{1}-3}{2}}))\,x_{1}$

or

$\displaystyle y = \frac{3}{2} (tan \circ arcsin) (\frac{y_{1}-3}{2}}) (x - x_{1}) + y_{1}$

Great. But if I plug in the first coordinate I'm given

$\displaystyle (x_{1}, y_{1}) = (\frac{4 + 3\sqrt{3}}{2}, 2)$

I eventually get

$\displaystyle \frac{-\sqrt{3}}{2}\,x + \sqrt{3} + \frac{17}{4}$

A tangent line with a negative slope! Yet when I look diagram provided in the book itself, the tangent line at this point clearly has a positive slope.