Question 628908
We often learn about the Trig functions in terms of opposite, adjacent and hypotenuse. We don't always learn about them in terms of coordinates. Here's a translation guide:
opposite = y
adjacent = x
hypotenuse = {{{sqrt(x^2+y^2)}}} (Note: this is <i>always</i> positive!)<br>
So
{{{sin(theta) = opposite/hypotenuse = y/sqrt(x^2+y^2)}}}
{{{cos(theta) = adjacent/hypotenuse = x/sqrt(x^2+y^2)}}}
{{{tan(theta) = opposite/adjacent = y/x}}}
{{{csc(theta) = 1/sin(theta) = sqrt(x^2+y^2)/y}}}
{{{sec(theta) = 1/cos(theta) = sqrt(x^2+y^2)/x}}}
{{{cot(theta) = 1/tan(theta) = x/y}}}<br>
To find the answers to your question, just take your x, -3, and your y, -5, and plug them into the appropriate formulas above. Since cos has a square root in the denominator you will need to rationalize the denominator.