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The derivative of a periodic function would of course be periodic, but I am just curious to know why the derivative of a trigonometric function does not go out of the trigonometric world?

Deepak
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    Since all the trig functions have formulas in terms of the sine function, the product rule and the chain rule guarantee that if the derivative of the sine function stays in Trig World (my favorite attraction at Disneyland!) then so will all derivatives of trig functions. So all we have to do is explain why the derivative of $\sin x$ is $\sqrt{1-\sin^2 x}$. – Gerry Myerson Oct 07 '20 at 05:04
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    Every derivative formula is valid because of a proof. There are proofs of the derivative formulas for trigonometric functions in many sources including this online textbook. To me that answers the "why" question; if it doesn't for you, what sort of reason are you looking for? – Greg Martin Oct 07 '20 at 05:04
  • Pretty sure 3b1b has made a video about something related to this – Amadeus Oct 07 '20 at 05:08
  • @GerryMyerson you mean the derivative of $\sin{x}$ is $\pm\sqrt{1-\sin^2x}$. – A-Level Student Oct 07 '20 at 11:36
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    Perhaps the figure in this answer of mine will provide some insights, at least in cases of sine and cosine. – Blue Oct 07 '20 at 12:42

3 Answers3

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So to understand that, one has to understand how the trigonometric functions are defined. Perhaps the most important property is $$ \sin^2 + \cos^2 = 1 $$ which tells that we are looking at the coordinates of points on the circle. However, there are a lot of ways to go over the points of the circle. For example taking $\sin(x^2)$ and $\cos(x^2)$ would work as well. Or we could also count in an other unit than radian, which would also change the derivative. This tells us that the other condition to define $\sin$ is the condition to go over the points of the circle at speed $1$ (The value of the angle corresponds to the arclength of the portion of circle). So, the derivative of the vector $u_x = (\sin x,\cos x)$ has to be again an element of the unit circle, and so is expressed with the same trigonometric functions (it is just the orthogonal vector $(\cos,-\sin)$.

If you know complex numbers, this is even easier to see there, since the vector corresponds to $u_x = \cos x + i \sin x = e^{ix}$. And the definition of this exponential is exactly the function such that itsderivative is $u_x' = i u_x$, which is the vector orthogonal and of size $1$.

LL 3.14
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  • Using Euler's relation with $e^{ix}$ is a bit of circular reasoning in my opinion. It is based on Maclaurin series which are based on derivatives which are based on what we are trying to prove! – A-Level Student Oct 07 '20 at 12:06
  • Yes, if you read what I wrote, expressing things with Euler relation is just another way to see things. What is Olt? But my goal is to explain why we are remaining in the trigonometric world, not doing a proof of the derivative of the sine, which is not the question here ... – LL 3.14 Oct 07 '20 at 12:10
  • Oh, my mistake; were you not trying to prove the derivatives of $\sin{x}$ and $\cos{x}$ from differentiating $e^{ix}$? I think I corrected the 'Olt' error btw, you were probably typing while I corrected it :) – A-Level Student Oct 07 '20 at 12:13
  • @A-levelStudent circular reasoning. Hah! – Cameron Williams Oct 07 '20 at 12:13
  • @CameronWilliams XD I didn't even realize that! :) – A-Level Student Oct 07 '20 at 12:14
  • @LL3.14 ok. My answer is still explaining why we remain in the trig world though; the proof is the reason. – A-Level Student Oct 07 '20 at 12:15
  • For example the reason of choosing the normalization of $\sin x/x$ converging to $1$ is the same as asking to go on the circle at speed $1$. Without this justification, any other normalization would work as well. – LL 3.14 Oct 07 '20 at 12:15
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To answer this, we'd need a definition of "trigonometric", so I'd say trigonometric functions are real-valued rational functions of $e^{ix}$, whose closure rules are trivial. By contrast, logarithms of rational functions of $ix,\,\sqrt{1-x^2}$ don't achieve the same, which is why the situation is different with inverse trigonometric functions.

J.G.
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A simple way to find the derivative of $\sin{x}$ is shown below. A similar method works for $\cos{x}$ and $\tan{x}$: $$\frac{d}{dx}\sin{x}=\lim_{h\to0}\frac{\sin{(x+h)}-\sin{x}}{h}=\lim_{h\to0}\frac{\sin{x}\cos{h}+\sin{h}\cos{x}-\sin{x}}{h}$$ $$=\lim_{h\to0}\frac{\sin{h}}{h}\cos{x}+\lim_{h\to0}(\sin{x}(\frac{\cos{h}-1}{h}))$$ I'm afraid you'll have to take my word that $$\lim_{h\to0}\frac{\sin{h}}{h}=1,~~\lim_{h\to0}\frac{\cos{h}-1}{h}=0$$ which can easily be checked graphically eg on Desmos. So we have $$\frac{d}{dx}\sin{x}=\lim_{h\to0}\frac{\sin{h}}{h}\cos{x}+\lim_{h\to0}(\sin{x}(\frac{\cos{h}-1}{h}))=\lim_{h\to0}\cos{x}+\lim_{h\to0}0=\cos{x}$$ as we expected.

A-Level Student
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