\( \frac { \cos ( 90 ^ { \circ } - 2 \theta ) \cdot \sin \theta } { \sin ^ { 2 } \theta ( 180 ^ { \circ } + \theta ) \cdot \cos ( 210 ^ { \circ } + \theta } \)

To simplify the expression \( \frac { \cos ( 90 ^ { \circ } - 2 \theta ) \cdot \sin \theta } { \sin ^ { 2 } \theta ( 180 ^ { \circ } + \theta ) \cdot \cos ( 210 ^ { \circ } + \theta ) } \), we can start by using the co-function identity for cosine: \[ \cos (90^\circ - x) = \sin x. \] This means that \( \cos (90^\circ - 2\theta) = \sin (2\theta) \). Therefore, we can rewrite the numerator: \[ \sin (2 \theta) \cdot \sin \theta. \] Next, we can apply the double angle identity for sine, which states that \( \sin (2\theta) = 2 \sin \theta \cos \theta \). Thus, the numerator becomes: \[ 2 \sin^2 \theta \cos \theta. \] Now our expression looks like: \[ \frac{2 \sin^2 \theta \cos \theta}{\sin^2 \theta (180^\circ + \theta) \cdot \cos (210^\circ + \theta)}. \] Next, we can simplify by canceling the \( \sin^2 \theta \) in the numerator and denominator (assuming \( \theta \neq n\pi \) for \( n \in \mathbb{Z} \)): \[ \frac{2 \cos \theta}{(180^\circ + \theta) \cdot \cos (210^\circ + \theta)}. \] Now, taking into account that \( \cos(210^\circ + \theta) = \cos(210^\circ) \cos(\theta) - \sin(210^\circ) \sin(\theta) \), where \( \cos(210^\circ) = -\frac{\sqrt{3}}{2} \) and \( \sin(210^\circ) = -\frac{1}{2} \): \[ \cos(210^\circ + \theta) = -\frac{\sqrt{3}}{2} \cos(\theta) + \frac{1}{2} \sin(\theta). \] Substituting this into the expression gives: \[ \frac{2 \cos \theta}{(180^\circ + \theta) \left(-\frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta\right)}. \] The final form depends on the specific values of \( \theta \), so be careful with your substitutions and simplifications!