3

While graphing the sine function, it is always assumed that x is in radians.Similarly while differentiating the sinx function x is always in radians and we have to multiply by a constant if x is given to us in degrees, before differentiating. Is radian just a convention for trigonometric functions or is there some other reason?Why are we able to differentiate a function directly when x is in radians and not when it is in degrees?

user456
  • 709

2 Answers2

4

Let $f(x)=\sin x$ radians, and $g(x)=\sin x$ degrees. Then $$f'(x)=\cos x\text{ radians}$$ and $$g'(x)=\frac{\pi}{180}\cos x\text{ degrees}.$$ Which is a more convenient sine function: $f$ or $g$?

The slope of the graph of $f$ at the origin is $1$, that of $g$ is $\pi/180$.

When you get to Maclaurin series you find $$f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$ and $$g(x)=\frac{\pi x}{180}-\frac{\pi^3x^3}{180^33!}+\frac{\pi^5x^5}{180^55!}-\cdots.$$

A particle moving so that its at $(\sin t\text{ radians},\cos t\text{ radians})$ after $t$ seconds is rotating round the unit circle an one unit per second, a particle moving so that its at $(\sin t\text{ degrees},\cos t\text{ degrees})$ after $t$ seconds is rotating round the unit circle at $\pi/180$ units per second.

Are you still keen on degrees?

Angina Seng
  • 158,341
  • I don't think it would create any difference if everything is done in degrees.Like derivative of $\sin x^\circ$ will be $1^\circ \cos x^\circ$. –  Apr 23 '17 at 09:58
3

You could do everything in degrees... but then the formulas would look considerably uglier.

Suppose I give you an angle $x$ degrees. Then it corresponds to $\pi x/180$ radians. Let's define $\sin^\circ(x)$ and $\cos^\circ(x)$ to be the sine and cosine of the angle $x^\circ$, in degrees. (We still let $\sin(x),\cos(x)$ be the sine and cosine of $x$ in radians.) Then

$$ \frac{d}{dx}\sin^\circ(x) = \frac{d}{dx}\sin\left(\frac{\pi x}{180}\right) = \frac{\pi}{180}\cos\left(\frac{\pi x}{180}\right) = \frac{\pi}{180}\cos^\circ(x) $$

Bleh. That factor of $\pi/180$ is really ugly and annoying. The fact that we get no ugly constant when doing calculus with radians is precisely the reason we use radians for trigonometry.

eepperly16
  • 7,232
  • I guess, the question is the following: The rules are simple in radians: $\sin'(x)=\cos(x)$ and $\cos'(x)=-\sin(x)$. Why? The answer cannot be that the rules are simple because we "bleh" if there are ugly constants. – zoli Apr 23 '17 at 05:22
  • @zoli In general, if we measure $x$ in any units, $\sin'(x) = a \cos(x)$ where $a$ is some constant. At a simple level, you can thing of "radians" as just being the units that make $a = 1$. At a deeper mathematical level, there is a close link between trig functions expressed in radians, and natural log and exponential functions (to base $e$), and therefore between $e$ and $\pi$. None of this was known when people first decided to divide a circle into 360 degrees - which was probably chosen because there are approximately 360 days in a year, not for any mathematical reason. – alephzero Apr 23 '17 at 09:13