Find x in Degrees, not Radians

Date: 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/