prove that cos squarex- cos square 6x=sin4x.sin8x
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L.H.S. = cos^2 2x – cos^2 6x
= (cos 2x + cos 6x) (cos 2x – 6x)
= [2cos(2x+6x/2)cos(2x-6x/2)][-2sin(2x+6x/...
= [2 cos 4x cos 2x] [–2 sin 4x (–sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.
= (cos 2x + cos 6x) (cos 2x – 6x)
= [2cos(2x+6x/2)cos(2x-6x/2)][-2sin(2x+6x/...
= [2 cos 4x cos 2x] [–2 sin 4x (–sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.
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