Question 614437
One way to do this is by using the coordinates directly. In order to do it this way, you have to know how the "opposite", "adjacent" and "hypotenuse" of the various ratios translate into coordinates:<ul><li>Opposite = y</li><li>Adjacent = x</li><li>Hypotenuse = r. We can use the Pythagorean Theorem to express r in terms of x and y:
{{{x^2 + y^2 = r^2}}}
{{{sqrt(x^2 + y^2) = r}}} (Always use a positive r.) So:
Hypotenuse = {{{sqrt(x^2+y^2)}}}</li></ul>We are no ready to requested functions:
{{{csc(theta) = hypotenuse/opposite = sqrt(x^2+y^2)/y = sqrt((-7)^2+(5)^2)/5 = sqrt(49+25)/5 = sqrt(74)/5}}}
{{{sin(theta) = 1/csc(theta) = 5/sqrt(74)}}}
{{{tan(theta) = opposite/adjacent = y/x = 5/(-7) = -5/7}}}<br>
P.S. You may want/need to rationalize the denominator of {{{sin(theta)}}}.