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Math Help - Derivative of arccos (+ feedback :D)

  1. #1
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    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 y=arctan(x) and finding \frac{dy}{dx}. In this exam, the similar thing came up this time 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 \frac{-1}{sin(y)}, remember that 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
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    So you got \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 sin^2(t)+ cos^2(t)= 1 so that sin(t)= \pm\sqrt{1- cos^2(t). Replacing t by arccos(x), sin(arccos(x))= \pm\sqrt{1- cos(arccos(x))^2} and, of course, cos(arccos(x))= x. That is, \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 \sqrt{1- x^2} so that sin(y)= \sqrt{1- x^2} and them \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; June 10th 2011 at 05:00 AM. Reason: Changing itex to tex.
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