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Trigonometry - Identities and proof

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Show that cot((x+y)/2) = - (sin x - sin y)/(cos x - cos y) for all values of x and y for which both sides are defined.

I tried manipulating both sides in terms of trig identities but I don't really have a solution....help would be appreciated, thanks.

  • Trigonometry - Identities and proof - ,

    Using the sum-to-product formulas,

    sinx-siny = 2cos((x+y)/2)sin((x-y)/2)
    cosx-cosy = -2sin((x+y)/2)sin((x-y)/2)

    now just divide to get

    cos((x+y)/2) / sin((x+y)/2) = cot((x+y)/2)

  • Trigonometry - Identities and proof - ,

    Two of the standard conversion formulas are:

    sinA - sinB = 2sin((A-B)/2) cos( (A+B)/2)
    and
    cosA - cosB = - 2sin( (A-B)/2) sin( (A+B)/2)

    see:
    near bottom of page under:
    Sum-to-Product Formulas
    http://www.sosmath.com/trig/Trig5/trig5/trig5.html

    RS
    = -(sinx - siny)/(cosx - cosy)
    = - 2sin((X-Y)/2) cos( (X+Y)/2)/- 2sin( (X-Y)/2) sin( (X+Y)/2)
    = cos((X+Y)/2) / sin(X+Y)/2)
    = cot((X+Y)/2)

    = LS

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