nomadreid said:
Ah, one detail. In doing the MacLauren series for sin and cos, in the expansion one only needs that sin(0) = 0 and cos(0) = 1. This is true whether or not your units are in degrees or in Radians. Therefore the answer from D H seems to be begging the question. There's something else to this... What?
You may be thinking that, because (sin x)'= cos x, (sin x)''= -sin x, (sin x)'''= -cos x, etc. and similarly for the derivatives of cos x, in order to find the MacLauren series, we only need to evaluate those at x= 0.
But the point is that those derivative formula are only true if x is in
radians.
In particular, the derivative of sin x is given by
\lim_{h\to 0}\frac{sin(x+h)- sin(x)}{h}
Using the trig identity sin(x+h)= sin(x)cos(h)+ cos(x)sin(h) that becomes
\lim_{h\to 0}\frac{sin(x)cos(h)- cos(x)sin(h)- sin(x)}{h}
= \lim_{h\to 0} sin(x)\frac{cos(h)- 1}{h}+ cos(x)\frac{sin(h)}{h}
so to have the derivative of sin(x) be cos(x) we must have
\lim_{h\to 0}\frac{cos(h)- 1}{h}= 0
and
\lim_{h\to 0}\frac{sin(h)}{h}= 1
and, as Dickfore noted, those are only true if h is in
radian.
In fact, if you are going to use the trig functions as general
functions and not just as ways to "solve triangles", you should stop thinking of the arguments as being angles at all- and just think of them as number, without any "degrees"
or "radians". One way of doing that is to use the "circle" definition: Given a unit circle (circle with center at the origin, radius 1, in some coordinate system), to find sin(t) or cos(t), start from the point with coordinates (1, 0) and measure counter-clockwise (clockwise if t is negative) a distance t around the circumference of the circle. What ever (x,y) point you end at gives you cos(t)= x, sin(t)= y. No angles at all there! (Yes, you have measured a distance but the "units" are whatever "units" you used to construct the coordinate system.)
