Okay, let's solve this step-by-step.
Given the function:
[tex]\[ r(\theta) = \frac{1}{16} v^2 \sin \theta \cos \theta \][/tex]
We need to use a double-angle identity to rewrite this formula.
Recall the double-angle identity for sine:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \][/tex]
We can rearrange this identity to express [tex]\( \sin(\theta) \cos(\theta) \)[/tex]:
[tex]\[ \sin(\theta) \cos(\theta) = \frac{1}{2} \sin(2\theta) \][/tex]
Let's substitute [tex]\( \sin(\theta) \cos(\theta) \)[/tex] in the original function with [tex]\( \frac{1}{2} \sin(2\theta) \)[/tex]:
[tex]\[ r(\theta) = \frac{1}{16} v^2 \left( \frac{1}{2} \sin(2\theta) \right) \][/tex]
When we simplify this, it becomes:
[tex]\[ r(\theta) = \frac{1}{16} \cdot \frac{1}{2} v^2 \sin(2\theta) \][/tex]
[tex]\[ r(\theta) = \frac{1}{32} v^2 \sin(2\theta) \][/tex]
Thus, the formula rewritten using a double-angle identity is:
[tex]\[ r(\theta) = \frac{1}{32} v^2 \sin(2\theta) \][/tex]
This is the required transformation of the original function using the double-angle identity.
