Prove \( \frac{\sin 3 x}{\sin x}=3-4 \sin ^{2} x \)

To prove the identity \( \frac{\sin 3x}{\sin x} = 3 - 4 \sin^2 x \), we start with the left-hand side and use the sine triple angle formula, which states that \[ \sin 3x = 3 \sin x - 4 \sin^3 x. \] Thus, we can rewrite the left-hand side: \[ \frac{\sin 3x}{\sin x} = \frac{3 \sin x - 4 \sin^3 x}{\sin x}. \] Assuming \(\sin x \neq 0\), we can simplify this: \[ \frac{3 \sin x - 4 \sin^3 x}{\sin x} = 3 - 4 \sin^2 x. \] Now we have shown that \[ \frac{\sin 3x}{\sin x} = 3 - 4 \sin^2 x. \] This completes the proof. Therefore, the equality holds true, and we have successfully demonstrated the relationship.