How do I show work for this

Question

-4sin³x + 3sinx

I know the final answer is sin(3x) but I'm not sure how to show how to get there

Answer 1

Sin(3x) = -4Sin³(x) + 3Sin(x)

Sin(x + 2x) Sin(x)Cos(2x) + Cos(x)Sin(2x) from Sin(a+b) = Sin(a)Cos(b) + Cos(a)Sin(b)

and Cos(2a) = 1 - 2Sin²(a)

and Sin(2a) = 2Sin(a)Cos(a)

then Sin(x)[1-2Sin²(x)] + Cos(x)[2Sin(x)Cos(x)]

Sin(x) - 2Sin³(x) + 2Sin(x)Cos²(x)

and Cos²(a) = 1 - Sin²(a)

then Sin(x)[1-2Sin²(x) + 2 - 2Sin²(x)]

so Sin(x)[ 1-4Sin²(x) + 2) = Sin(x)( 3 - 4Sin²(x)] = Sin(x)[3-4Sin²(x)] = 3Sin(x) - 4Sin²(x)] = Sin(3X) !!!!!