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Differentiation of $\sin(x^\circ)$

Can we say its derivative is $\cos x^\circ$? I read somewhere that in differentiation and integration, we assume $x$ should be in radians. But why so?

user3290550
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  • could you tell me what is $x^o$? – Someone Dec 23 '19 at 12:09
  • obviously its $x$ degree – user3290550 Dec 23 '19 at 12:10
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    Are you looking for the derivative of the degree sine function $\sin_{deg}(x):=\sin(\tfrac{x \pi}{180})$? – maxmilgram Dec 23 '19 at 12:25
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    I know its derivative, but I want to know why radian is given special treatment – user3290550 Dec 23 '19 at 12:25
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    The comment by @PedroSantos is still justified in my opinion because this question seems to be rather about a clarification of $$\text{real numbers}\leftrightsquigarrow\text{degrees}$$ rather than the sine functon in particular – Maximilian Janisch Dec 23 '19 at 12:35
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    Of course the derivative of $\sin (x^\circ)$ is $A \cos(x^\circ)$ for some constant $A$. But the constant $A$ is not $1$. The accolade of constant $1$ belongs to the sine in radians. Same answer as: why do we use natural logarithm? The derivative of $\log_a x$ is $A/x$ for some constant $A$, but the only case where $A=1$ is the natural log. – GEdgar Dec 23 '19 at 13:03

2 Answers2

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I know its derivative, but I want to know why radian is given special treatment

The radian is often considered the "correct" choice of units for working with angles, above degrees. So much so, that it often isn't even considered a unit, but a "dimensionless quantity": for that reason you will rarely see expressions like $\sin(\pi\text{ radians})$, but if you write $\sin(180)=0$ it is not acceptable unless you specify the unit $\sin(180\deg)$. The reason the radian is considered more fundamental is addressed in the excellent anwsers to these two questions: Why do we require radians in calculus? and Why are radians dimensionless?

In this particular case, let's start with the limit definition of the derivative: $$\begin{split}\frac{d}{dx}\sin(x)&=\lim_{h\to0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}\\&=\lim_{h\to0}\left(\sin(x)\frac{\cos(h)-1}{h}+\cos(x)\frac{\sin(h)}{h}\right).\end{split}$$ The manipulation usually proceeds from here by noting that $\lim_{h\to0}(\cos(h)-1)/h=0$, and that $\lim_{h\to0}\sin(h)/h=1$. Hence we get $$\frac{d}{dx}\sin(x)=\lim_{h\to0}\left(\sin(x)\cdot0+\cos(x)\cdot1\right)=\lim_{h\to0}\cos(x)=\cos(x).$$ The point is that if we were to choose to use degrees, we will encounter the expression $\lim_{h\to0}\sin(h^\circ)/h$, and this quantity simply isn't equal to $1$. Yes, it is correct that $\lim_{h\to0}\sin(h^\circ)/(h^\circ)$ is $1$, but that's not what we're dealing with here. Crucially, that $h$ in the denominator corresponds geometrically to a length, not an angular value. And the choice of radians is the unique choice of unit for the angle which will guarantee $\lim_{h\to 0}\sin(h)/h=1$. A proof of this fact can be found at this question: How to prove that $\lim_{x\to0}\frac{\sin(x)}{x}=1$?. Note that all the geometric arguments provided there use somehow the fact that the length of the arc subtended by an angle of $\theta$ in the unit circle is exactly $\theta$, a fact which is only true when $\theta$ is expressed in radians.

YiFan Tey
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  • See my simple query is this,

    What is the value of $\lim_{h\to 1}h\cos h$, is it equal to $\cos 1$ or $(1^c)(\cos 1^c)$?

     – user3290550 Dec 23 '19 at 12:58
  • @user3290550 Of course it's $\cos1$. (To type angles use the command ^\circ, for example $1^\circ$ gives $1^\circ$.) – YiFan Tey Dec 23 '19 at 12:59
  • ok if its $\cos1$, what is $1$ here? is it $1$ or $1^c$ – user3290550 Dec 23 '19 at 12:59
  • @user3290550 Of course it's just $1$. Whenever we do not specify the unit, $\cos (x)$ is always taken to mean $\cos(x\text{ radians})$. – YiFan Tey Dec 23 '19 at 13:00
  • you mean to say for $\cos $ function , $1=1^c$? – user3290550 Dec 23 '19 at 13:01
  • @user3290550 No, where do you get that idea? (I'm assuming you mean $1$ degree, not $1$ raised to the power of $c$, which would be a totally different story.) – YiFan Tey Dec 23 '19 at 13:02
  • By $1^c$, I mean $1$ radian and its the standard stuff – user3290550 Dec 23 '19 at 13:02
  • Oh okay, I see. Yes, when we see a real number as the argument of a trigonometric function without the degree symbol it is automatically assumed to be in radians. That's because radians are more "fundamental" than the other unit choices; see the second link in my answer. – YiFan Tey Dec 23 '19 at 13:04
  • ok, so you mean to say for trigonometric functions $1=1^c$ right? – user3290550 Dec 23 '19 at 13:04
  • @user3290550 Yes, although that is an extremely imprecise way to say what you are trying to say. Radians are dimensionless, so $1\text{ radian}=1$ for our purposes. – YiFan Tey Dec 23 '19 at 13:05
  • I don't understand "$1\text{ radian }=1$ for our purposes", $1^c=\left(\dfrac{180}{\pi}\right)^{\circ}$ and how this ever be equal to $1$ – user3290550 Dec 23 '19 at 13:07
  • so either we should have some other dimensionless quantity for defining angles, why we take use of radians as $1^c\ne1$ – user3290550 Dec 23 '19 at 13:09
  • @user3290550 Think of it this way. Draw the angle subtended on the unit circle by an angle $\theta$. It will have a certain arclength $s$. We define $\theta=s$ by choosing units appropriately for the angle---this unit that we chose is the radian measure. You might object: but one of them is an angle, another is a length, how can we say that they're the same thing? The point is that units don't matter in math (not in this context at least), so this is not a concern. Angles and lengths are both nothing but real numbers. – YiFan Tey Dec 23 '19 at 13:09
  • definition of $1$ radian is "angle subtended by the arc at the center when length of arc is equal to radius of circle", how this quantity ever be equal to $1$? – user3290550 Dec 23 '19 at 13:11
  • Can you explain your reasoning why not? – YiFan Tey Dec 23 '19 at 13:12
  • how $1$ radian can be equal to $(\dfrac{180}{\pi})^{\circ}$?. If you say its possible, then you mean to say $1=57.5^{\circ}$ approximately – user3290550 Dec 23 '19 at 13:14
  • Yes, I do, because the $^\circ$ symbol is exactly defined that way. Think of $\circ$ as a constant factor of $\pi/180$, if you like. – YiFan Tey Dec 23 '19 at 13:16
  • by the way, how we can compare angles with real numbers?, angles have units and real numbers are dimensionless, so I think when we say $\cos 1$, then $1$ is not $1^c$ – user3290550 Dec 23 '19 at 13:16
  • This is exactly what I covered a few comments ago, did you read it? "You might object: but one of them is an angle, another is a length, how can we say that they're the same thing? The point is that units don't matter in math (not in this context at least), so this is not a concern. Angles and lengths are both nothing but real numbers." – YiFan Tey Dec 23 '19 at 13:17
  • Note to others (not the OP): I'm aware that not everyone is happy with treating $^\circ$ as a multiplicative factor of $\pi/180$, but I think it suits the discussion in this context. I'm also aware of problems with the claim "units don't matter in math"---the field of dimensional analysis for instance has rigorous mathematical footing. So it's an oversimplification but one that suits our purposes here, I think. – YiFan Tey Dec 23 '19 at 13:19
  • so at last I think, we should not be using $1^c$ because $1^c\ne1$, but as you said it suits our purposes but I don't see any good logic behind that. – user3290550 Dec 23 '19 at 13:21
  • Please do not be so sure of your claim that $1^c\neq1$. It is, do not be so stubborn about changing your mind. Are you trying to say convention which the whole world abides by is wrong and you're better than everyone who thinks it's right? – YiFan Tey Dec 23 '19 at 13:22
  • I am not able to digest because how one can compare angles with real values? If someone comes and tells me he needs $1$ cloth, what I will ask, I would ask what do you mean by $1$,right? He can say I mean $1$ in number or he can say $1$ meter or he can say $1$ centimeter right? – user3290550 Dec 23 '19 at 13:25
  • I think the problem is that we have mixed algebra with quantities which take angle as input. Either we should not have done that or we should have defined some dimensionless quantity for angles, and clearly radian cannot be the choice as it has proper dimension as degree has. – user3290550 Dec 23 '19 at 13:28
  • @user3290550 Okay, I understand the urge to think of this as a problem. Indeed that's very much a physicist's way of thinking. In math however, a function is defined as a mapping from a set $A$ to a set $B$, which for a single "input" $a\in A$ gives one "output" $b\in B$. Here $\sin$ and $\cos$ are functions from the set of real numbers to the set of real numbers between $-1$ and $1$ inclusive, it is a purely mathematical construct and the geometric description regarding angles and such is purely a geometric description---a good analogy, if you will. It provides intuition, no more. – YiFan Tey Dec 23 '19 at 13:32
  • In case my POV is hard to grasp... there are also different ways to explain the same concept which may help you. See this question: https://math.stackexchange.com/questions/3447626/are-trigonometry-functions-ratios-or-distance/3448390#3448390 especially the answer by David K. – YiFan Tey Dec 23 '19 at 13:34
  • I read it and there the question is about, what is the output of trigonometric functions, does it have any dimension or dimensionless, as its a ratio, so obviously its dimensionless, but my question is about the input to trigonometric ratio, right? – user3290550 Dec 23 '19 at 13:43
  • Yes; they're not the same question per se, but fundamentally the issue about equating quantities with different units is the same. – YiFan Tey Dec 23 '19 at 13:45
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Use the definition of the derivative. Let $f(x)=\sin(x^{\circ})$. Then $$f'(x)=\lim_{h\to0}\frac{\sin(x+h)^{\circ}-\sin x^{\circ}}{h}=\lim_{h\to0}\frac{\sin x^{\circ}\cos h^{\circ}+\cos x^{\circ}\sin h^{\circ}-\sin x^{\circ}}{h}$$ But then the problem is that $\displaystyle\lim_{h\to 0}\frac{\sin h^{\circ}}{h}$ is not equal to $1$ - it is equal to $\pi/180$.

Only in radians do we have $\displaystyle\lim_{h\to 0}\frac{\sin h}{h}=1$.

A. Goodier
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