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Dervative of trig functions.

  1. Nov 28, 2009 #1
    1. The problem statement, all variables and given/known data

    y' = csc2([tex]\vartheta[/tex] / 2 )
    Find y"



    3. The attempt at a solution

    So far, i have 2 csc ([tex]\vartheta[/tex]/2) *csc([tex]\vartheta[/tex]/2)cot([tex]\vartheta[/tex]/2) * 1/2

    but i'm wondering, how do you take the derivative of a half angle identity?
    or does it just simplify down to csc2([tex]\vartheta[/tex] / 2) * cot([tex]\vartheta[/tex]/2)
     
  2. jcsd
  3. Nov 28, 2009 #2
    You don't need to worry about derivatives of half-angle identities here. You're right about how it simplifies, and you used the chain rule properly, except you're missing a minus sign (the derivative of csc(x) is -csc(x)cot(x)).
     
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