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Thread: differentiation of inverse trignometric equation continuation

  1. #1
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    differentiation of inverse trignometric equation continuation

    sir,

    please find below the question


    differentiation of inverse trignometric equation continuation-question.jpg

    kindly enlighten me.
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  2. #2
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    Re: differentiation of inverse trignometric equation continuation

    You ignored the absolute value signs. Don't ignore them and you will get the two answers you show. They are each correct on different intervals. Find the intervals that make each true.
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  3. #3
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    Re: differentiation of inverse trignometric equation continuation

    In the book they have given it as x/2 and on differentiation, it is 1/2. But, my friend proceeded by rationalizing the denominator by multiplying the conjugate. I could not understand which method I have to adopt and why? it was given only that x lies between 0 and Pi/2. That means x is in the first quadrant.
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  4. #4
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    Re: differentiation of inverse trignometric equation continuation

    $\sin \dfrac{x}{2} + \cos \dfrac{x}{2}\ge 0$ when $4n\pi -\dfrac{\pi}{2} \le x \le \dfrac{3\pi}{2}, n\in \mathbb{N}$ (this covers quadrant 1 when $n=0$).

    $\sin \dfrac{x}{2}-\cos \dfrac{x}{2} < 0$ when $4n\pi - \dfrac{3\pi}{2} < x < 4n\pi + \dfrac{\pi}{2}, n \in \mathbb{N}$ (this covers quadrant 1 when $n=0$).

    So, $\left| \sin \dfrac{x}{2} + \cos \dfrac{x}{2} \right| = \sin \dfrac{x}{2} + \cos \dfrac{x}{2}$ in quadrant 1 while $\left| \sin \dfrac{x}{2} - \cos \dfrac{x}{2} \right| = \cos \dfrac{x}{2} - \sin \dfrac{x}{2}$ when x is in quadrant 1.

    Then, you have:

    $\displaystyle \dfrac{ \left| \sin \dfrac{x}{2} + \cos \dfrac{x}{2} \right| + \left| \sin \dfrac{x}{2} - \cos \dfrac{x}{2} \right| }{ \left| \sin \dfrac{x}{2} + \cos \dfrac{x}{2} \right| - \left| \sin \dfrac{x}{2} - \cos \dfrac{x}{2} \right| } = \dfrac{2\cos \dfrac{x}{2} }{2\sin \dfrac{x}{2}}$ when x is in quadrant 1.

    Thus, you have $\cot^{-1}\cot \dfrac{x}{2} = \dfrac{x}{2}$ when x is in quadrant 1. Your method works just fine. You just needed to check the intervals where each answer was true.
    Last edited by SlipEternal; Apr 25th 2018 at 11:52 AM.
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  5. #5
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    Re: differentiation of inverse trignometric equation continuation

    Excellent Sir. Your reply was very useful.
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