Prove:Sinx-2sin3x+sin5x=2sinx (cos4x-cos2x)

A = sinx - sin3x,

B = - sin3x + sin5x

First A:

The average of x and 3x is 2x, and they (x and 3x, that is) are each a distance of x from this average. That's fancy talk for:

x = 2x-x,

3x = 2x+x.

So, A = sin (2x-x) - sin (2x+x)

Using angle sum formulas:

A = (sin2x cosx - cos2x sinx) - (sin2x cosx + cos2x sinx)

A = - 2 cos2x sinx

Similarly,

B = - sin (4x-x) + sin (4x+x)

= - (sin4x cosx - cos4x sinx) + (sin4x cosx + cos4x sinx)

B = 2 cos4x sinx

Now,

sinx - 2sin3x + sin5x = A+B = - 2 cos2x sinx + 2 cos4x sinx

= 2 sinx (cos4x - cos2x).

Sinx-2sin3x+sin5x=2sinx (cos4x-cos2x)

Step-by-step explanation:

sinx - 2sin3x + sin5x = sinx - sin (3x) + sin (5x) - sin (3x)

= 2· cos[ (x+3x) / 2] · sin[ (x-3x) / 2] + 2·cos[ (5x+3x) / 2]· sin[ (5x-3x) / 2]

= 2· cos (2x) ·sin (-x) + 2· cos (4x) · sin (x)

= - 2·cos (2x) ·sinx + 2· cos (4x) ·sinx

= 2·sinx · [ cos (4x) - cos (2x) ]