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    RamanBS's Avatar
    RamanBS Posts: 1, Reputation: 1
    New Member
     
    #1

    Dec 11, 2007, 05:02 PM
    Solve this
    sinx + sin2x + sin3x
    _________________ = tan2x

    cosx + cos2x + cos3x


    and

    sin^3x + cos^3x = (1-sinxcosx)(sinx + cosx)
    jiten55's Avatar
    jiten55 Posts: 105, Reputation: 8
    Junior Member
     
    #2

    Dec 11, 2007, 07:57 PM
    sin A + sin B = 2 sin ½(A + B) cos ½(A - B)

    Use the above formula to get

    sin3x + sinx = 2 sin 2x cos x

    Also: cos A + cos B = 2 cos ½(A + B) cos ½(A - B)

    Cos 3x + cos x = 2 cos 2x cos x

    Hence

    LHS = (2 sin 2x cos x + sin 2x)/(2cos 2x cos x + cos 2x)

    = sin 2x(2cos x +1)/cos 2x (2cos x + 1)

    = sin 2x/cos 2x

    = tan 2x


    For the second part, use factorizing for x^3 + y^3 = (x + y) (x^2 + y^2 - xy)
    itsme_vipsdude's Avatar
    itsme_vipsdude Posts: 29, Reputation: 0
    New Member
     
    #3

    Dec 12, 2007, 01:59 AM
    Hey Raman Its M Vipul... u Can Use The Formula Sin3x = 3sinx - 4sinx^3

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