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Math Help - differential trig problem

  1. #1
    Junior Member
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    Jan 2009
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    differential trig problem

    HI, can someone please shed some light on this for me. I spent last night on it without luck. Thank you

    show that
    d/dx(ln(sqrt(2)-cos x)/(sqrt(2)+cos x) = (2sqrt(2) sin x)/(1+sin^2 x)
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  2. #2
    Senior Member
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    Carried out step by step, we derive

    f'(x) = \frac{d}{dx}\left(\ln\,\frac{\sqrt{2}-\cos\,x}{\sqrt{2}+\cos\,x}\right)
    = \frac{d}{dx}(\ln\,(\sqrt{2}-\cos\,x)-\ln\,(\sqrt{2}+\cos\,x))
    = \frac{d}{dx}(\ln\,(\sqrt{2}-\cos\,x))-\frac{d}{dx}(\ln\,(\sqrt{2}+\cos\,x))
    = \frac{1}{\sqrt{2}-\cos\,x}\cdot \sin\,x-\frac{1}{\sqrt{2}+\cos\,x}\cdot(-\sin\,x)
    = \frac{\sin\,x}{\sqrt{2}-\cos\,x}+\frac{\sin\,x}{\sqrt{2}+\cos\,x}
    = \frac{\sin\,x(\sqrt{2}+\cos\,x)+\sin\,x(\sqrt{2}-\cos\,x)}{2-\cos^2 x}
    = \frac{\sqrt{2}\sin\,x+\sin\,x\,\cos\,x+\sqrt{2}\si  n\,x-\sin\,x\,\cos\,x}{1+1-\cos^2 x}
    =\frac{2\sqrt{2}\sin\,x}{1+\sin^2 x}.
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  3. #3
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    thank heaps for the effort you put into that. I really appreciate it.
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