Question

The point (-2,3) is on the terminal side of angel theta, in standard position. What are the valus of sine, cosine, and tangent of theta?

Please show work

Answer 1

The x value is the adjacent side and the y is the opposite side and to find the hypotenuse you use the Pythagorean theorem a^2 + b^2 = c^2.

x = -2 and y = 3

-2^2 + 3^3 = c^2

4 + 9 = c^2

13 = c^2

c = square root of 13

Then plug in the values for sine, cosine, and tangent:

sine = opp / hyp = 3 / square root of 13

cosine = adj / hyp = -2 / square root of 13

tangent = opp / adj = 3 / -2

Answer 2

Angels don't help with math homework.

We have x=-2 and y=3 and magnitude aka hypotenuse

`h = √(x^2 + y^2) =√((-2)^2+(3)^2)=√(13)`

Now the trig functions are straightforward;

`\cos \theta = \frac x h = - (2)/(√(13))`

`\sin \theta = \frac y h = (3)/(√(13))`

`\tan \theta = \frac y x = - \frac 3 2`

I won't bother to move the radicals to the numerator because I think that's a waste of time, but your teacher may feel differently.