Question 231164
In the fourth quadrant, x's are positive and y's are negative. So let's find a point on the line in the fourth quadrant by picking a positive value for x and the using the equation to find the corresponding y. To avoid fractions, I'm going to pick 3 for x:
2(3) + 3y = 0
6 + 3y = 0
3y = -6
y = -2
Here's a drawing of what we have so far:
{{{drawing(300,300, -1, 4, -4, 1, line(0, 0, 3, 0), line(0, 0, 3, -2), line(3, 0, 3, -2), locate(1.5, .2, 3), locate(3.2, -1, -2))}}}
We already have what we need for tan. But we need the hypotenuse for the sin and cos. We'll use the Pythagorean Theorem for this:
{{{3^2 + 2^2 = h^2}}}
{{{9 + 4 = h^2}}}
{{{13 = h^2}}}
{{{sqrt(13) = h}}}
Now we can find our values. (Since Algebra.com's formula software does not "do" theta, for some unknown reason, I'll use x instead):
{{{sin(x) = (-2)/sqrt(13)}}}
{{{cos(x) = 3/sqrt(13)}}}
{{{tan(x) = (-2)/3}}}
We don't usually leave square roots in denominators so we'll rationalize the sin and cos:
{{{sin(x) = (-2)/sqrt(13) = ((-2)/sqrt(13))(sqrt(13)/sqrt(13)) = (-2sqrt(13))/13}}}
{{{cos(x) = 3/sqrt(13) = (3/sqrt(13))(sqrt(13)/sqrt(13)) = (3sqrt(13))/13}}}