| Posted on Monday, 05 January, 2004 - 11:40 am: |
Why does calculus of trigonometric functions only work with radian measure?
It is just one of the things we have been taught without being told why. I was wondering if somebody could give me the explanation.
| Posted on Monday, 05 January, 2004 - 01:01 pm: |
Michael,
Consider what the graph of sin x looks like where x is measured in degrees .
Just from this graph (ie. don't differentiate) think of how large the gradient is at x = 0o . (If we had d/dx (sinx) = cosx for degrees aswell, then this gradient would need to be equal to cos0o = 1). From our graph we know the gradient of the tangent of x = 0o is definitely greater than 1. (If we move one unit up on the y-axis we're going to move somewhere around 60o on the x-axis! This is definitely not going to give a gradient of 1)
In fact, what we've done by using degrees is (since p = 180°) 'stretched' the x-axis by a factor of 180/p, so any gradients we measure with our x-axis in degrees will be p/ 180 less (since gradient is over the difference in x) then when we measure them with our x-axis in radians. Can you now see that (based on the fact that d/dx(sinx)=cosx for radians) we have d/dq(sinq)=p/180cosq for q in degrees? If you're familiar with the chain rule then can you see how the previous paragraph just expresses the fact that d/dx f(ax)=af¢(ax) and in this case what we need is the derivative of sin(p/180 x) where the function sinq is for q in radians and x is our value in degrees? Marcos P.S. If you're familiar with the definition of a derivative,
| f¢(a)= |
lim x® a | (f(x)-f(a)) /(x-a) |
then try to derive for yourself that d/dx(sinx) = cosx (and see where you use the fact that x is in radians).
| Posted on Monday, 05 January, 2004 - 02:57 pm: |
A "radian" is a natural measure of an angle because it is the angle subtended by wrapping the length of the radius part-way around the circumference of a circle. Nothing arbitrary (that is, nothing from the world outside the circle) is brought into this measure.
Other measures use an arbitrary division of the circle into a certain number of parts. That number, 360, in the case of degrees, is arbitrary, and can complicate the study of trigonometric functions by introducing a confounding factor from the world outside the circle.
In addition, as Marcos pointed out, the choice of radians is "natural" in the same way the choice of base e is "natural" for logarithms. This choice allows the derivative of sine to be cosine, and the fourth (and 8th, 12th, ...) derivative of sine to be sine. With this choice, the sin function fits nicely into a family of so-called circular functions that includes the exponential function and the hyperbolic functions. (The exponential function exp(x)=ex is sometimes, along with the trigonometric functions, called a "circular function" because the locus of eix , where x is real, is a circle in the complex plane.)
--Graeme
| Posted on Monday, 05 January, 2004 - 04:17 pm: |
I think the reason is that the series for the trigonometric functions only work when using radians (i think anyway). The derivatives are found by differentiating these series, giving other trigonometric functions. For instance,
sin x = x - x3 /3! + x5 /5! - x7 /7! + x9 /9!....
Differentiating gives
cos x = 1 - x2 /2! + x4 /4! - x6 /6! + x8 /8!....
Marcos, on a sin graph, the gradient is never more than 1. 1 < = Cos x < = -1, whether in radians or degrees.
James
| Posted on Monday, 05 January, 2004 - 05:11 pm: |
Oops I meant "is definitely less than 1" (basically it's pi/180 and this is the most the gradient ever becomes for degrees)
Marcos
| Posted on Tuesday, 06 January, 2004 - 01:01 am: |
Hello
Thank you for your detailed responses. Am I right in saying that the use of radians for calculus is a matter of convenience rather than anything else, as Graeme suggested.
Marcus, I looked at deriving d/dx (sin x) = cos x. Is it the case that radians are essential with the derivation of the limit:
, which is required for
working out of the differential.Thanks
P.S. Another question... Is it valid to go from this
to this 
?
This may be an easy question, but I was just wondering if
you can do that sort of thing with limits.
| Posted on Tuesday, 06 January, 2004 - 01:14 am: |
I'd say more elegance than convenience. (Convenience is for physicists.)
| Posted on Tuesday, 06 January, 2004 - 07:42 am: |
Michael,
Yes, it's true that if lim f(x) = a then lim 1/f(x) = 1/a in general. (ie. provided 1/a, 1/f(x) exist, etc.)
Marcos
P.S. I think I'm gonna use that as a quote Grame
| Posted on Friday, 09 January, 2004 - 12:06 am: |
Hi Michael, Yes, it is necessary to measure in radians for
|
lim x ® 0 | (sinx/x)=1 |
. Here's a quick proof: Let the area of the small triangle be A, the sector be B and the big triangle be C.
| A= |
1 2 | r2sinq |
| B= |
1 2 | r2q |
(this involves use of radian measure)
| C= |
1 2 | r2tanq |
It is obvious from the diagram that A < B < C so
|
1 2 | r2sinq < |
1 2 | r2q < |
1 2 | r2tanq |
Þ tanq < q > sinq Þ 1/cosq > q/sinq > 1 Þ cosq < sinq/q < 1 Taking the limit as q® 0, cosq® 1, so
|
lim x® 0 | (sinx/x)=1 |
The reason this works is because of the formula of the area for a sector,
|
1 2 | r2q |
. To see why this doesn't work for other angular measure, we can divide the circle into d sectors, and then use this as our angular measure. The formula for an area of a sector now becomes Area =qpr2/d, where q is measured in our new angular measurement. Thus to find the limit of sinq/q, we use the inequality from above: A < B < C
| Þ |
1 2 | r2sinq < r2qp/d < |
1 2 | r2q |
Þ sinq < 2qp/d < tanq Þ 1 < 2qp/dsinq < 1/cosq Þ 1 > dsinq/2qp > cosq Þ 2p/d > sinq/q > 2pcosq/d Taking the limit as q® 0, cosq® 1, so
|
lim q® 0 | (sinq/q)=2p/d |
in our angular measure. It is necessary (as I think you know) to have
|
lim x® 0 | (sinx/x)=1 |
to obtain d/dx sinx = cosx, so it makes sense to set d=2p and use this as our angular measure. Hope this helps, Tim
