1. The problem is to find the value of $\frac{1}{60^\circ}$ and verify the given trigonometric identity $\sin 60^\circ = 2 \sin 30^\circ \cos 30^\circ$. 2. First, find the exact values of $\sin 30^\circ$ and $\cos 30^\circ$: - $\sin 30^\circ = \frac{1}{2}$ - $\cos 30^\circ = \frac{\sqrt{3}}{2}$ 3. Substitute these values into the expression $2 \sin 30^\circ \cos 30^\circ$: $$2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$ 4. Note that $\sin 60^\circ$ is known to be $\frac{\sqrt{3}}{2}$, so the identity holds. 5. Now, to find $\frac{1}{60^\circ}$, assuming the question asks for the reciprocal of $60^\circ$ in degrees: $$\frac{1}{60^\circ} = \frac{1}{60}$$ This is simply a numeric value, approximately $0.0167$ per degree. 6. If the question involves $\sin 60^\circ$, the value is $\frac{\sqrt{3}}{2} \approx 0.866$. Final answers: - $\sin 60^\circ = \frac{\sqrt{3}}{2}$ - $2 \sin 30^\circ \cos 30^\circ = \frac{\sqrt{3}}{2}$ (identity confirmed) - $\frac{1}{60^\circ} = \frac{1}{60} \approx 0.0167$ per degree