# show sin x+sin3x+sin5x=(1+2cos2x)sin3x

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You should write `sin x = sin(3x - 2x)` and `sin 5x = sin(3x + 2x)` such that:

`sin(3x - 2x) + sin 3x + sin(3x + 2x)= (1+2cos2x)sin3x `

You need to use the following formulas such that:

`sin(a - b) = sin a*cos b - sin b*cos a`

`sin(a + b) = sin a*cos b+ sin b*cos a`

`sin(3x - 2x) = sin 3x*cos2x - sin 2x*cos 3x`

`sin(3x+ 2x) = sin 3x*cos 2x+ sin 2x*cos 3x`

Adding `sin(3x + 2x)` to `sin(3x - 2x)` yields:

`sin(3x + 2x)+sin(3x - 2x) = sin 3x*cos 2x - sin 2x*cos 3x + sin 3x*cos 2x + sin 2x*cos 3x`

Reducing like terms yields:

`sin(3x + 2x)+sin(3x - 2x) = 2sin 3x*cos 2x`

You need to add `sin 3x` to `2sin 3x*cos 2x ` such that:

`sin(3x + 2x)+sin(3x - 2x) + sin 3x = sin 3x + 2sin 3x*cos 2x`

You should factor out `sin 3x ` such that:

`sin(3x + 2x)+sin(3x - 2x) + sin 3x = sin 3x(1 + 2 cos 2x)`

**Notice that the last line checks the given identity such that `sin x + sin 3x + sin 5x = sin 3x(1 + 2 cos 2x).` **