Since [tex]P[/tex] lies in the fourth quadrant, you know the sine is negative.
Also since the point lies on the unit circle, the coordinates satisfy
[tex]x^2+y^2=1\implies \left(\dfrac{33}{64}\right)^2+y^2=1[/tex]
When you solve for [tex]y[/tex], you have two possible solutions, which are
[tex]y=\pm\sqrt{1-\left(\dfrac{33}{64}\right)^2}=\pm\sqrt{\dfrac{3007}{4096}}=\pm\dfrac{\sqrt{3007}}{64}[/tex]
but since [tex]y[/tex] is negative, you take the negative root.
So,
[tex]\sin\theta=\dfrac yx=\dfrac{-\frac{\sqrt{3007}}{64}}{\frac{33}{64}}=-\dfrac{\sqrt{3007}}{33}[/tex]
