Question 1052574
sinθ= -2/3 and tanθ > 0
<pre>
-2/3 is a negative number, so the sine is negative.
tan&#952; > 0 says the tangent is positive.

That tells us that &#952; is in QIII

So we draw the picture of &#952;.  We know that 
{{{matrix(1,5,sine,""="",(opposite)/(hypotenuse),""="",y/r)}}}

Since {{{sin(theta)=-2/3=(-2)/(""+3)}}}

We draw a right triangle in QIII with opposite side -2
and hypotenuse +3.  We can use the Pythagorean theorem
to find the adjacent side or x,
adjacent = x = {{{-sqrt(3^2-(-2)^2)=-sqrt(9-4)=-sqrt(5)}}}
We take it negative because x, the adjacent side, goes to
the LEFT of the origin.



{{{drawing(300,300,-2.9,2.9,-2.9,2.9,
red(arc(0,0,1.46,-1.46,0,222)), line(-2.236067977,-2,-2.236067977,0),
line(0,0,-2.236067977,0), locate(-2.7,-.7,-2), locate(-1.1,-1,3),
locate(-1.5,.5,-sqrt(5)),red(locate(-.5,.9,theta)),
graph(300,300,-2.9,2.9,-2.9,2.9), line(0,0,-2.236067977,-2) )}}} 

Since {{{matrix(1,3,cosine,""="",adjacent/hypotenuse)}}},

{{{cos(theta)}}}{{{""=""}}}{{{(-sqrt(5))/3}}}{{{""=""}}}{{{-sqrt(5)/3}}}. 

Since {{{matrix(1,3,cotangent,""="",adjacent/opposite)}}},

{{{cot(theta)}}}{{{""=""}}}{{{(-sqrt(5))/(-2)}}}{{{""=""}}}{{{sqrt(5)/2}}}.

Edwin</pre>