For some reason or another every time I come at this I get tangled:
[1-cos(2x)] sin(4x)
________________ = tanx
cos(2x) [1-cos(4x)]
I have to prove it when {0,pi/2} and so x can't be 45 degree
For some reason or another every time I come at this I get tangled:
[1-cos(2x)] sin(4x)
________________ = tanx
cos(2x) [1-cos(4x)]
I have to prove it when {0,pi/2} and so x can't be 45 degree
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)
It turns out that on about half an hour before you posted the answer, when I was playing around with the two identities that you checked to see if I knew, I ended up figuring it out on my own. But your explanation was still helpful! Thanks again!