Evaluate cot(Cos^-1(-15/17)) The answer needs to be in fraction form. I drew out a right triangle and labeled the adjacent side -15 and the hypotenuse 17. So I get -15^2+b^2=17^2. -- b^2=17^2+15^2

b=sqrt17^2+15^2

What do I do know brahs ;~;?

Question

Evaluate cot(Cos^-1(-15/17)) The answer needs to be in fraction form. I drew out a right triangle and labeled the adjacent side -15 and the hypotenuse 17. So I get -15^2+b^2=17^2. --> b^2=17^2+15^2

b=sqrt17^2+15^2

What do I do know brahs ;~;?

anonymous · Answer

I think you almost had it.  Bear with me, because my trig is a little bit ... not great.  Anyway, if we look at the problem in two parts it might make it easier to figure out.  So, I broke it up as 1. cotangent of some theta   2.  theta = arccosine (-15/17).

We can then, instead of bring the arccosine over the equal sign to get cosine theta = (-15/17).  From here, looking at your work above, I think you know how to find the missing side length (if you draw a triangle).

Once you have that, I think you can now look at that triangle, and pick the sides you need for cotangent.

anonymous · Answer

Thanks man :)

anonymous · Answer

b shouldn't be the sum of the squares of the other two because it isn't the hypotenuse.  Your b should be the square root of the difference between the square of the hypotenuse and the square of the given side.