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  • trig -

    are we solving ?

    (cosx/sinx)(cos2x/sin2x) - (cos2x/sin2x)(cos3x/sin3x) - (cos3x/sin3x)(cosx/sinx) = 1
    common denominator is sinx(sin2x)(sin3x), so

    [ cosxcos2xsin3x - sinxcos2xcos3x - sin2xcos3xcosx ] / (sinxsin2xsin3x) = 1
    cross-multiply, then bring sinxsin2xsin3x back to left side

    cosxcos2xsin3x - sinxcos2xcos3x - sin2xcos3xcosx - sinxsin2xsin3x = 0
    factor sin3x from 1st and 4th, cos3 from 2nd and 3rd
    sin3x(cosxcos2x - sinxsin2x) - cos3x(sinxcos2x + cosxsin2x) = 0
    sin3x(cos(x+2x)) - cos3x(sin(x+2x)) = 0
    sin3xcos3x - cos3xsin3x = 0
    0 = 0

    Since this is a true statement, the original equation must have been an identity.

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