Sin 3A Identity 5Ed1B5

1. **State the problem:** We want to prove the trigonometric identity $$\sin 3a = 3 \sin a - 4 \sin^3 a.$$\n\n2. **Recall the formula for sine of a sum:** $$\sin(x+y) = \sin x \cos y + \cos x \sin y.$$\n\n3. **Express $\sin 3a$ as $\sin(2a + a)$:** $$\sin 3a = \sin(2a + a) = \sin 2a \cos a + \cos 2a \sin a.$$\n\n4. **Use double angle formulas:** $$\sin 2a = 2 \sin a \cos a,$$ $$\cos 2a = 1 - 2 \sin^2 a.$$\n\n5. **Substitute these into the expression:** $$\sin 3a = (2 \sin a \cos a) \cos a + (1 - 2 \sin^2 a) \sin a = 2 \sin a \cos^2 a + \sin a - 2 \sin^3 a.$$\n\n6. **Rewrite $\cos^2 a$ using $\cos^2 a = 1 - \sin^2 a$:** $$\sin 3a = 2 \sin a (1 - \sin^2 a) + \sin a - 2 \sin^3 a = 2 \sin a - 2 \sin^3 a + \sin a - 2 \sin^3 a.$$\n\n7. **Combine like terms:** $$\sin 3a = (2 \sin a + \sin a) - (2 \sin^3 a + 2 \sin^3 a) = 3 \sin a - 4 \sin^3 a.$$\n\n8. **Conclusion:** We have shown $$\sin 3a = 3 \sin a - 4 \sin^3 a,$$ which proves the identity.