Find x in Degrees, not RadiansDate: 01/14/98 at 11:26:59 From: Ronny Subject: Trigonometry/Derivatives Hi Dr. Math, I Just have one quick question. We have learned in Math class that the derivative of sin(x) is cos(x). My teacher said this implies that the x is in radians, not degrees. My teacher gave us a homework assignment and one of the problems is to find the derivative of sin(x) when x is in degrees, not radians. I have no idea how to prove this. Could you please help? Thank you very very much in advance. Mathematically Yours, Ronny Date: 02/10/98 at 15:45:01 From: Doctor Sonya Subject: Re: Trigonometry/Derivatives The key to this problem is the lim (sin x)/x x->0 When you work in radians the answer is 1. Look back in your book to where they do the proof of that result. The important point is that if x is in radians, then for values of x close to 0, sin x and x are approximately the same. You also need to verify what happens to (1 - cos x)/x as x-> 0 when x is in degrees. This one will still be 0, just as in radians. Let's suppose, instead, that x is in degrees. Then sin x is approximately the same as x * Pi/180. This is because sin x is approximately equal to x. If we want to convert x from radians to degrees, we have to multiply x by Pi/180, so sin(x) = x(Pi/180). Therefore we see that for x in degrees, sin x Pi lim ------- = ----- x->0 x 180 Now, let's go to derivatives. d(sin x) sin(x+h) - sin x sin x cos h + sin h cos x - sin x ------- = lim --------------- = lim --------------------------------- dx h->0 h h->0 h sin h 1 - cos h Pi = lim cos x ----- - lim sin x --------- = ---- cos x - sin x (0) h->0 h h->0 h 180 Pi = ---- cos x 180 I hope this helps. -Doctors Sonya and Fred, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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