Proving trig identities?

Question

By writing 3A as (A+2A) prove that SIN3A = 3SINA - 4sin^3A

Answer 1

sin 3A = sin (2A + A)

= sin 2A cos A + sinA cos 2A

= (2 sin A cos A) cos A + sin A (1 - 2 sin^2A)

- 2 sin A cos^2 A + sin A - 2 sin^3 A

= 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

= 3 sin A - 4 sin^3 A

Answer 2

you should know the formula for sin(A+B) = sin A cos B + cos A sin B, and cos(A+B) = cos A cos B - sin A sin B, so

sin(A+2A) = sin A cos 2A + cos A sin 2A
sin(3A) = sin A [ cos A cos A - sin A sin A] + cos A [sin A cos A + cos A sin A]
sin(3A) = sin A [cos² A - sin² A] + cos A [ 2sin A cos A]
sin(3A) = sin A cos² A - sin^3 A + 2sin A cos² A
sin(3A) = 3 sin A cos²A - sin^3 A
sin(3A) = 3 sin A [1 - sin² A] - sin^3 A
sin(3A) = 3 sin A - 3sin^3 A - sin^3 A
sin(3A) = 3 sin A - 4sin^3 A

Answer 3

Use you angle-addition formulas, your double-angle formulas, and a bit of alegebra.

sin(3a) = sin(a + 2a)
= sin(a)cos(2a) + sin(2a)cos(a)
= sin(a)[2cos²(a) - 1] + [2sin(a)cos(a)]cos(a)
= 2sin(a)cos²(a) - sin(a) + 2sin(a)cos²(a)
= 4sin(a)cos²(a) - sin(a)
= 4sin(a)[1 - sin²(a)] - sin(a)
= 4sin(a) - 4sin³(a) - sin(a)
= 3sin(a) - 4sin³(a)