Question

Prove:Sinx-2sin3x+sin5x=2sinx(cos4x-cos2x)
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Answer 1

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).

Answer 2

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

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)]