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(sin 3x + sin x) sin x + (cos 3x - cos ...

(sin 3x + sin x) sin x + (cos 3x - cos x) cos x =0

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LHS `=(sin3x+sinx)sinx+(cos3x-cosx)cosx`
`=[2sin(3x+x)/(2)cos(3x-x)/(2)]sinx+[2sin(3x+x)/(2)sin(x-3x)/(2)]cosx`
`=[2sin2xcosx]sinx+[2sin2xsini(-x)]cosx`
`=2sin2xsinx.cosx-2sin2x sinx cosx`
=0 = RHS Hence Proved.
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