Hello, bobbooey!
$\displaystyle \text{I recently read somewhere that: }\:\sin(\theta)\:=\:\dfrac{y}{\sqrt{x^2+y^2}}$
$\displaystyle \text{Does anyone know the identities for the other trig functions?}$
Take a look at where that equation comes from . . .
Code:

 *
 * 
 h * 
 *  y
 * 
 * @ 
  *      +  
 x
We have angle $\displaystyle \,\theta$ in a right triangle.
The adjacent side is $\displaystyle \,x$; the opposite side is $\displaystyle \,y.$
Using Pythagorus, we find that: .$\displaystyle h \:=\:\sqrt{x^2+y^2}$
Therefore, we have:
. . $\displaystyle \begin{array}{cccccc}
\sin\theta &=& \dfrac{opp}{hyp} &=& \dfrac{y}{\sqrt{x^2+y^2}} \\ \\[3mm]
\cos\theta &=& \dfrac{adj}{hyp} &=& \dfrac{x}{\sqrt{x^2+y^2}} \\ \\[3mm]
\tan\theta &=& \dfrac{opp}{adj} &=& \dfrac{y}{x} \\ \\[3mm]
\cot\theta &=& \dfrac{adj}{opp} &=& \dfrac{x}{y} \\ \\[3mm]
\sec\theta &=& \dfrac{hyp}{adj} &=& \dfrac{\sqrt{x^2+y^2}}{x} \\ \\[3mm]
\csc\theta &=& \dfrac{hyp}{opp} &=& \dfrac{\sqrt{x^2+y^2}}{y} \end{array}$