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Recently I showed my students how to prove that the derivative of $\sin(x) = \cos(x)$, using the limit definition of the derivative, trigonometric identities, and the fact that $\lim_{h \to 0} \frac{sin(h)}{h} = 1$.

I'm trying to think of another way for them to think about the derivative of $\sin(x)$. Can I say something about the periodicity - the derivative is a function that is still $2\pi$ - periodic? That doesn't sound enlightening. Can I say that the derivative changes the evenness / oddness of the function, e.g. the derivative of $\sin(x)$, an odd function, is $\cos(x)$, an even function? That sounds a little more interesting but still not something that would grab their attention.

Is there a deeper way of thinking about the derivative of $\sin(x)$?

(This is for a first term course in Calculus, so no power series, please.)

Ak3.14
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D.Hutchinson
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    This image can be helpful: physics.unsw.edu. Also check out answers to this question: math.stackexchange.com. – jonsno Oct 02 '17 at 09:17
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    I don't think it gets much "deeper" than deriving it from the definition. But, having the students make a rough sketch of the derivative of $\sin$, by approximating slopes of tangent lines and plotting should be edifying. – David Mitra Oct 02 '17 at 09:18
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    What about complex numbers ? Consider the function $f$ defined on $\mathbb{R}$ by : $\forall t \in \mathbb{R}, ; f(t) = \exp(it)$. One could say that $f$ describes the position of a point moving on the unit circle $\lbrace z \in \mathbb{C}, ; \vert z \vert = 1 \rbrace$. The derivative of $f$ is given by : $\forall t \in \mathbb{R}, ; f'(t) = i\exp(it) = i\cos(t) - \sin(t)$. By identification, one gets that $\sin' = \cos$. – pitchounet Oct 02 '17 at 10:14

3 Answers3

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Apropos "deeper way":

1) $f(x) = f(-x),$ even fct.

Examples: $y=x^2$, $y=cos(x)$

$ f'(x) = -f'(-x),$ chain rule, odd fct.

2) $f(x)=-f(-x)$, odd fct.

Examples: $y=x^3,$ $ y=sin(x)$.

$f'(x) = f'(-x)$, chain rule, even fct.

3) Example, periodic fct:

$f(x) = f(x +2πk)$, $k \in \mathbb{N}$.

$f'(x)=f'(x+2πk)$.

4) Draw $\sin$ and $\cos$ curve, $0 \le x\le π/2$,

(in one diagram, superimpose).

Choose any $x_0$ in this interval.

Find the derivative of the $\sin$ fct at $x_0$ by inspection. ($\cos(x_0)$ on $\cos$ curve).

5) By inspection:

Find the derivative of $\sin(x)$ at the point of intersection of the 2 curves.

Given that at the point of intersection, $x_0=π/4$, $\sin(x_0) =(1/2)√2$, find the derivative of $\sin$ at this point.

Helps a little?

Peter Szilas
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Using only elementary trig identities and the chain rule will show the derivative of sine must be some multiple of cosine. Then one only needs to be convinced that the derivative at the origin is one (i.e., basically the identity $\lim_{h\to 0} \frac{\sin h}{h}=1$) to see that this multiple is precisely one.

Recall the Pythagorean identity,

$$1=\sin^2x+\cos^2x.\quad \quad (1)$$

Define $f(x):=\mathrm{D}_x \sin x.$ Since $\cos x =\sin (\pi/2-x),$ the chain rule tells us $\mathrm{D}_x \cos x$ = $-f(\pi/2-x),$ so differentiating both sides of $(1)$ wrt $x$ and simplifying gives

$$0=f(x)\sin x -f(\pi/2-x) \cos x. \quad \quad (2)$$

By inspection, $f(x)=A\cos x$ (for constant $A$) gives one set of solutions to this functional equation. Can $f(x)$ take any other form? No, because recall that

$$\sin(y-x)=\sin y \cos x -\cos y \sin x.\quad \quad (3)$$

Differentiating both sides of $(3)$ wrt $y$ (taking $x$ as a constant) gives

$$f(y-x)=f(y) \cos x +f(\pi/2-y) \sin x.$$

Letting $y=x$ gives

$$f(0)=f(x) \cos x + f(\pi/2-x) \sin x.\quad \quad (4)$$

Multiplying both sides of $(2)$ by $\sin x$ and both sides of $(4)$ by $\cos x$, adding the two equations, and using the Pythagorean identity gives

$$f(x)=f(0)\cos x.$$

Golden_Ratio
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  • Besides chain rule, this assumes familiarity with other elementary derivative rules (i.e., sum, difference, power, constant multiple, constant). – Golden_Ratio Feb 06 '25 at 15:20
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I like to start with them being the solutions to $f''(x)=-f(x)$ with the initial conditions $f(0)=0, f'(0)=1$ and $g(0)=1, g'(0)=0$.

From this many of the properties can be derived such as $f^2(x)+g^2(x)=1$.

marty cohen
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