Trig Help

Hi, I need help finding the sin of the angle alpha in standard postion whose terminal side contains the given point. (-4,-6).

Here's what I have so far.

r = sqrt{(-4)^2 + (-6)^2} = sqrt{52}
x = -4, y= -6, r= sqrt{52}

sin(alpha) = (-6/sqrt(52))
= (-6/sqrt(52)*sqrt(52)/sqrt(52)
=-6(sqrt(13*2*2) / 52)


The answer ends up being sin(alpha)= {3(sqrt(13)/13}

And that is where I am getting lost... It's simpler algebra that I should already know, but I just don't remember. Here's an image that might be easier on the eyes...



The box is around the answer I should get to. Thanks for any help.

You're good.

Let us continue,
sin(alpha) = (-6/sqrt(52))
Rationalize the denominator,
= (-6/sqrt(52)*sqrt(52)/sqrt(52)
= -6(sqrt(13*2*2) / 52)
Get the 2*2 out of the sqrt sign,
= -6[2sqrt(13)] / 52
= -12[sqrt(13)] / 52
Reduce it to its lowest term, divide both numerator and denominator by 4,
= -3[sqrt(13)] / 13 ----------answer.

:) Thank you very much! Your help is much appreciated