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Simple factorisation

  1. Nov 15, 2008 #1
    I want [tex]\frac{dI}{d \theta} = 0[/tex] when [tex]I = 0.8 \cdot 10^{-5}[\frac{273}{16} + \frac{17}{2}cos(kd sin \theta) + 2cos(2kd sin \theta)][/tex]

    If I've calculated correctly, this means that [tex]-8.5sin(8.1sin \theta)8.1cos \theta - 32.4sin(16.2sin \theta)cos \theta = 0[/tex]. Can somebody help me factorise this?
  2. jcsd
  3. Nov 15, 2008 #2
    If only the stuff in brackets is the argument of the first sine of each term, you could get away with dividing through by [tex] cos \theta [/tex] to leave you with

    -8.5sin(8.1sin \theta)8.1 - 32.4sin(16.2sin \theta) = 0

    which is equivalent to
    -68.85sin(8.1sin \theta) - 32.4sin(16.2sin \theta) = 0
  4. Nov 15, 2008 #3
    Then recognising 16.2sinθ = 2(8.1sinθ) we can apply the sin double angle formula (sin2α = 2cosαsinα) on the second term
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