How is sin6x/sin2x equal to (3sin2x-4sin(2x)^3)/sin2x

I am taking limes of sin6x/sin2x and I can't see how the next step would be (3sin2x-4sin(2x)^3)/sin2x

Please provide the formula!

I don't think if your question is correctly worded.

I think I should show that sin6x = 3sin2x- [4sin(2x)]^3

Recall that sin2x = 2sinxcosx, and sin(a+b) = sinacosb + sinbcosa, cos(2x) = 1-2(sinx)^2; cos^x = 1-sin^x

So, sin6x = sin(4x+2x) = sin4xcos2x + sin2xcos4x

=> (2sin2xcos2x)cos2x + sin2x[(cos2x)^2 - (sin2x)^2]

=> 2sin2x(cos2x)^2 + sin2x[1-2(sin2x)^2]

=> 2sin2x(1-(sin2x)^2) + 2sin2x - 2(sin2x)^3

=> 2sin2x - 2(sin2x)^3 + 2sin2x - 2(sin2x)^3

=> 3sin2x- [4sin(2x)]^3

But if you want to take the limits lim sin6x/sin2x = lim sin6x/6x * 6x *2x/(2xsin2x)

so lim sin6x/sin2x = lim sin6x/6x * 3 * 2x/sin2x = 3

Please gimme best answer