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Question

Mathematics Question on Trigonometric Functions of Sum and Difference of Two Angles

Prove that cos2 2x-cos2 6x = sin 4x sin 8x

Answer

It is known that

2 cosA+cosB=2cos(A+B2)cos(A−B2), cosA−cosB=2sin(A+B2)sin(A−B2)2\,cosA+cosB=2cos(\frac{A+B}{2})cos(\frac{A-B}{2}),\,cosA-cosB=2sin(\frac{A+B}{2})sin(\frac{A-B}{2})

∴L.H.S. = cos2 2x-cos2 6x

= (cos 2x +cos 6x) (cos 2x-6x)

=[2 cos(2x+6x2)cos(2x−6x2)][2 cos(2x+6x2)sin(2x−6x2)]=[2\,cos(\frac{2x+6x}{2})cos(\frac{2x-6x}{2})][2\,cos(\frac{2x+6x}{2})sin(\frac{2x-6x}{2})]

= [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