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Thread: Derivative of arccos (+ feedback :D)

  1. #1
    May 2011

    Derivative of arccos (+ feedback :D)

    So I posted a lot of questions last night and this morning, I want to say thanks to all those who helped me. I am just back from the exam and yes most of the advice I got from you all was relevant to what came up in it... I believe I will have passed well ONLY because of the advice I got on here. One night of topic questions on here and I learn WAY more than one whole year of university tuition... well that's another rant for another day But anyway the point I want to make and the question I want to ask is:

    Last night I asked for help on $\displaystyle y=arctan(x)$ and finding $\displaystyle \frac{dy}{dx}$. In this exam, the similar thing came up this time $\displaystyle y=arccos(x)$. It is the only question I did not know and I should have been prepared for this... but, I could not find a substitution like the one used for the arctan. Now please somebody tell me what I can do once I get to $\displaystyle \frac{-1}{sin(y)}$, remember that $\displaystyle x=cos(y)$ this time and I had to find something, I think, to link sin and cos, but I couldn't do it? I know you can't just square them otherwise its easy...
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  2. #2
    MHF Contributor

    Apr 2005
    So you got $\displaystyle \frac{-1}{sin(arccos(x))}$.

    There are two ways to think about this- the more formal is to use trig identities. You know, of course, that $\displaystyle sin^2(t)+ cos^2(t)= 1$ so that $\displaystyle sin(t)= \pm\sqrt{1- cos^2(t)$. Replacing t by arccos(x), $\displaystyle sin(arccos(x))= \pm\sqrt{1- cos(arccos(x))^2}$ and, of course, $\displaystyle cos(arccos(x))= x$. That is, $\displaystyle \frac{-1}{sin(arccos(x))}= \fra{\pm 1}{\sqrt{1- x^2}}$. Which sign is appropriate depends upon which quadrant you are in.

    Less formal but perhaps simpler to see- imagine a right triangle with one angle y, "near side" of length x, and hypotenuse of length 1 (so that cos(y)= x/1= x). By the Pythagorean theorem, the "opposite side" has length $\displaystyle \sqrt{1- x^2}$ so that $\displaystyle sin(y)= \sqrt{1- x^2}$ and them $\displaystyle \frac{-1}{sin(y)}= \frac{-1}{\sqrt{1- x^2}}$ which doesn't have the sign ambiguity because in setting up that right triangle we are assuming y is in the first quadrant.
    Last edited by Ackbeet; Jun 10th 2011 at 05:00 AM. Reason: Changing itex to tex.
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