If cos x = 2/9, and tan x < 0, what is sin x?

sin x = -√(1-(2/9)²)  It has to be negative because cos x is positive and tan x is negative only in the fourth quadrant, where sin x is negative.

$${\mathtt{sinx}} = {\mathtt{\,-\,}}{\sqrt{{\mathtt{1}}{\mathtt{\,-\,}}{\left({\frac{{\mathtt{2}}}{{\mathtt{9}}}}\right)}^{{\mathtt{2}}}}} \Rightarrow {\mathtt{sinx}} = -{\mathtt{0.974\: \!996\: \!043\: \!043\: \!569\: \!1}}$$

I always draw pictures for these. 

Now cos is positive (1st and 4th quad)  and tan is negative (2nd and 4th quadrant)

so x has to be in the 4th quadrant.  So Sin x is negative

$$Sin(x)=\frac{-\sqrt{77}}{9}$$