Prove that: \( \sin 3 x=3 \sin x-4 \sin ^{3} x \)

To prove the identity \( \sin 3x = 3 \sin x - 4 \sin^3 x \), we can use the angle addition formulas. Start by using the formula for \( \sin(3x) \): \[ \sin(3x) = \sin(2x + x) = \sin(2x)\cos(x) + \cos(2x)\sin(x) \] Now, we can express \( \sin(2x) \) and \( \cos(2x) \) using the double angle formulas: \[ \sin(2x) = 2\sin(x)\cos(x) \] \[ \cos(2x) = 1 - 2\sin^2(x) \] Substituting these into the original expression gives: \[ \sin(3x) = (2\sin(x)\cos(x))\cos(x) + (1 - 2\sin^2(x))\sin(x) \] \[ = 2\sin(x)\cos^2(x) + \sin(x) - 2\sin^3(x) \] From the Pythagorean identity, we know \( \cos^2(x) = 1 - \sin^2(x) \). So, we can substitute for \( \cos^2(x) \): \[ \sin(3x) = 2\sin(x)(1 - \sin^2(x)) + \sin(x) - 2\sin^3(x) \] \[ = 2\sin(x) - 2\sin^3(x) + \sin(x) - 2\sin^3(x) \] \[ = 3\sin(x) - 4\sin^3(x) \] Thus, we have proven the identity: \[ \sin(3x) = 3\sin(x) - 4\sin^3(x) \]