First of all, use the fundamental trigonometric equation to get the magnitude of the sine function. We'll discuss the sign later. From [tex] \sin^2(x)+\cos^2(x)=1 [/tex], we deduce [tex] \sin(x) = \pm\sqrt{1-\cos^2(x)} [/tex]. In your case, the equation becomes
[tex] \sin(x) = \pm\sqrt{1-\cfrac{9}{25}} = \pm\sqrt{\cfrac{16}{25}} = \pm\cfrac{4}{5} [/tex].
To choose the sign, remember that sine and cosine are, respectively, the x and y coordinates of points on the unit circle. Since we are in the second quadrant, the x coordinates are negative (and in fact the given cosine value is negative), while the y coordinates are positive. So, we choose the positive value of 4/5
