Hello, mankauff!

It's easy to get tangled in this problem . . .

Prove, for 0 < x < π/2 and x ≠ π/4:

sin(4x) [1 - cos(2x)]

------------------------ .= .tan(x)

cos(2x) [1 - cos(4x)]

I assume you know the necessary identities:

. . sin2θ .= .2·sinθ·cosθ

. . 1 - cos2θ .= .2·sin²θ

In the numerator: .sin(4x) .= .2·sin(2x)·cos(2x)

. . . . . . . . . and: .1 - cos(2x) .= .2·sin²(x)

In the denominator: .1 - cos(4x) .= .2·sin²(2x)

. . . . . . . . . . . . . . . . . 2·sin(2x)·cos(2x) · 2·sin²(x)

The problem becomes: . --------------------------------

. . . . . . . . . . . . . . . . . . . . cos(2x) · 2·sin²(2x)

. . . . . . . . . . . . . . . 2·sin²(x). . . . . . 2·sin²(x) . . . . . .sin(x)

which simplifies to: . ---------- . = . ----------------- . = . ------- . = . tan(x)

. . . . . . . . . . . . . . . .sin(2x) . . . . 2·sin(x)·cos(x) . - - .cos(x)