you are correct, since (-4,6) is in QII.
Recall that cosine is x/r.
r^2 = 4^2+6^2 = 42
r = √42 = 2√13
so, your cos is -4/2√13 = -2/√13
Or, as they clumsily put it, -2√13/13
The point (-4, 6) is on the terminal side of theta in standard position. Which is the cosine of theta?
1. - 2sqrt13/13
2. -2/13
3. 2/13
4. 2sqrt13/13
I think it is one of the negative ones, but I'm not sure of which.
1 answer
1 answer