sorry I wrote this one wrong previously. It should read as below. thanks for any help!?

Question

cos4x-c0s2x
___________ = -sinx
2sin3x

Answer 1

Anami's answer was correct and very simple. You should give him credit for it.

Here it is:

ID: cos(A+B)-cos(A-B)= -2sinAsinB
so RHS: A=3x and B=x
=> -2sin3xsinx = cos4x -cos 2x
proved.

What he did is this (only working with the left side):

cos4x - cos2x
------------------
2sin3x

cos(3x + x) - cos(3x - x)
------------------------------
2sin3x

cos3x cosx - sin3x sinx - (cos3x cosx - sin3x sinx)
-----------------------------------------------------------------
2sin3x

-2sin3x sinx
---------------
2sin3x

-sinx

is that more clear?

Answer 2

cos4x-cos2x
----------------- = -sinx
2sin3x

LHS =

cos4x-cos2x
----------------- =
2sin3x

(2cos²2x - 1) - cos2x
---------------------------- =
2sin(2x + x)

2cos²2x - cos2x - 1
------------------ ------------------ =
2(2sin2xcosx + cos2x sinx)

(2cos2x + 1)(cos2x - 1)
-------------------- ---------------------- =
2(2sinxcos²x + (1 - 2 sin²x) sinx)

(2(1 - 2sin²x) + 1)((1 - 2sin²x) - 1)
--------------------- ---------------------- =
2(2sinx(1 - sin²x) + sinx - 2sin³x)

(2(1 - 2sin²x) + 1)((1 - 2sin²x) - 1)
---------------------- --------------------- =
2(3sinx - 4sin³x)

(3 - 4sin²x)( - 2sin²x)
--------------------------- =
2sinx(3 - 4sinx)

-sinx

= RHS ......................................... QED!!!!!!

Answer 3

are you sure it's correct cause i'm not coming up with the proof

Answer 4

already given answer previously check my answer again