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56. sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

0 4 Views | Posted 4 months ago
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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    56. L.H.S = sin x + sin 3x + sin 5x + sin 7x.

    = (sin x + sin 7x) + (sin 3x + sin 5x)

    Using,

    sin A + sin B = 2 sin A+B2 cos A−B2.

    L.H.S. = 2. Sin x+7x2 cos x−7x2 + 2 sin 3x+5x2 cos 3x−5x2

    = 2 sin 8x2 cos (−6x2) + 2 sin 8x2 cos (−2x2)

    = 2 sin 4x cos 3x + 2 sin 4x cosx.[ ? cos (-x) = cosx]

    = 2 sin 4x[cos 3x + cosx]

    Using cos A + cos B = 2 cos A + B2 cos A−B2.

    So, L.H.S. = 2 sin 4x [2·cos3x+x2cos3x−x2].

    = 2 sin 4x [2·cos4x2·cos2x2]

    = 4 sin 4x. cos 2x cosx = R.H.S.

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