Find the alternate form of this trig function?

Hello,

sin²(3x) /[1 - cos(2x)] =

let's write the numerator, splitting the argument, as:

[sin(2x + x)]² /[1 - cos(2x)] =

let's apply the addition formula sin(α + β) = sin α cos β + cos α sin β:

[sin(2x) cosx + cos(2x) sinx]² /[1 - cos(2x)] =

let's apply the double-angle formula sin(2x) = 2sinx cosx:

[(2sinx cosx) cosx + cos(2x) sinx]² /[1 - cos(2x)] =

[2sinx cos²x + cos(2x) sinx]² /[1 - cos(2x)] =

let's factor out sinx in the numerator:

{sinx [2cos²x + cos(2x)]}² /[1 - cos(2x)] =

{sin²x [2cos²x + cos(2x)]²} /[1 - cos(2x)] =

let's apply (in both the numerator and the denominator) the double-angle formula

cos(2x) = cos²x - sin²x:

{sin²x [2cos²x + (cos²x - sin²x)]²} /[1 - (cos²x - sin²x)] =

[sin²x (2cos²x + cos²x - sin²x)²] /(1 - cos²x + sin²x) =

letìs apply (in the denominator) the basic identity 1 = sin²x + cos²x:

[sin²x (3cos²x - sin²x)²] /[(sin²x + cos²x) - cos²x + sin²x] =

[sin²x (3cos²x - sin²x)²] /(sin²x + cos²x - cos²x + sin²x) =

[sin²x (3cos²x - sin²x)²] /(2sin²x) =

simplifying to:

(3cos²x - sin²x)² /2 =

let's apply the identity sin²x = 1 - cos²x:

[3cos²x - (1 - cos²x)]² /2 =

(3cos²x - 1 + cos²x)² /2 =

(4cos²x - 1)² /2

I hope it's helpful