Question about "Approaching Zero and Limits" in the Intuitive Proof of the Derivative of Sine

Question

I'm a high school student hoping to self-study some introductory calculus over the summer. While studying, I came across this intuitive proof of the derivative of the sine function, using trig and the unit circle...

As in the picture, as dθ approaches zero, angles A and B will approach 90° -- allowing triangle ABC to be "approaching" similar to triangle BDE, but this would mean you would never get the exact angles for triangle ABC; thus, never the exact ratio of sine and cosine in order to complete the proof.

Is this small (even negligible) inaccuracy inherent to calculus, or is there a flaw in my understanding?

P.S. Please forgive me if this is a stupid/far-too-basic question.

Answer 1

The inaccuracy is of lower order than the main quantities of interest, and becomes fully accurate in the limit. The key to this question is understanding precisely what the limit means, and for this I direct you towards the rigorous meaning of a limit. Note that an initial course in calculus will include many such "hand-wavy" arguments that seem to be slightly inaccurate, and it is typically in a later course (often called "analysis" or "real analysis") following calculus where this rigorous definition is presented and results in calculus are put on a more solid footing.

For a more concrete instance of this, consider an expression like $x+x^2$ in the limit as $x$ tends to $0$. Now of course, $x+x^2$ does not equal $x$, but now if you imagine substituting a super tiny value for $x$, you will see that $x$ becomes a great approximation for $x(1+x)=x+x^2$. Note that this is stronger than saying that $x$ and $x+x^2$ have the same limit as $x$ goes to $0$, since $x^2$ also has this property - yet $x^2$ is a terrible approximation for $x+x^2$ in the limit when $x$ goes to $0$, since $x^2$ is much, much, much smaller than $x$. (I know this example is somewhat contrived, but it actually includes the main ideas in a particularly simple setting.)

Answer 2

Proving a calculus theorem using pictures is not a very good idea.

While you may get a vague outline from the graphs, you also get confused when concepts like limits are involved.

If you study the definition of limits and read the proof of this theorem based on the exact definition, you will get a clear understanding as to why derivative of $\sin x$ is $\cos x$ and why the picture does not do a good job.

Answer 3

Here's another way to picture the derivative of sine. This argument is for sine specifically, and doesn't appeal to more general machinery such as limits or even the definition of the derivative.

Imagine the $(x, y)$ plane $\mathbb{R}^2$. Next imagine the unit circle. This is the set of points with distance $1$ from the origin.

The unit circle is defined below and named $U$ .

$$ U \stackrel{\mathrm{def}}{=} \left\{ (x, y) \;|\; \sqrt{x^2 + y^2} = 1 \right\} $$

This notation is called set-builder notation if you haven't seen it before.

Imagine starting at the point $(0, 1)$, which is due east if the top of the graph is north, and travelling counter-clockwise around the circle at a constant velocity.

The circumference of the unit circle is $2\pi$, so let's normalize our speed so it takes exactly $2\pi$ seconds to travel around the circle once.

Our position at any given point in time is given by $s(t)$ ...

$$ s(t) \stackrel{\mathrm{def}}{=} (\cos(t), \sin(t)) $$

$t$ is the amount of time in seconds that we have been travelling around the circle. It is also the angle, in radians, between our current position, $(0, 0)$ (the origin), and $(0, 1)$ (our starting point).

Our velocity is always perpendicular to our position if we're tracing out a circle as opposed to some other spiral-like shape. This gives us two options for what the formula for our velocity $v$ could be, let's call them $v_1$ and $v_2$ .

Note that a slope $m'$ is perpendicular to $m$ if and only if $m' = -\frac{1}{m}$ . There are other possibilities for the velocity formula such as $v_3(t) = (-\sin(t)\cos(t)\,,\, 1) $, but all those other possibilities do not satisfy $\mathrm{length}(v_3(t)) = 1$ .

Here are the candidates.

$$ v_1(t) \stackrel{\mathrm{def}}{=} (-\sin(t), \cos(t)) $$

$$ v_2(t) \stackrel{\mathrm{def}}{=} (\sin(t), -\cos(t)) $$

$v_2$ is ruled out because it gives us a velocity of $(0, -1)$ at $t=0$, which means we are travelling clockwise around the circle instead of travelling counterclockwise. So:

$$ \frac{\mathrm{d}s(t)}{\mathrm{d}t} = v(t) = (-\sin(t), \cos(t)) $$

So, if we look at our $y$ coordinate alone, we get $\sin(t)$ ... our velocity at any given point time in the $y$ direction is $\cos(t)$ as desired.

Answer 4

If you look at the computer generated image below your drawing, the angle $\phi$ is equal to angle $\theta$. Which means that $\Delta\sin(\theta) = h \cos(\theta)$. Now if you let $\Delta\theta$ approach 0, $\Delta\theta$ approaches $h$. So when you divide through, you get: $$\Delta\sin(\theta)/\Delta\theta = \cos(\theta)$$ Hope this helps!