If we define $\sin x$ as series, how can we obtain the geometric meaning of $\sin x$?

In Terry Tao's textbook Analysis, he defines $\sin x$ as below:

1. Define rational numbers
2. Define Cauchy sequences of rational numbers, and equivalence of Cauchy sequences
3. Define reals as the space of Cauchy sequences of rationals modulo equivalence
4. Define limits (and other basic operations) in the reals
5. Cover a lot of foundational material including: complex numbers, power series, differentiation, and the complex exponential
6. Eventually (Chapter 15!) define the trigonometric functions via the complex exponential. Then show the equivalence to other definitions.

My question is how can we obtain the geometry interpretation of $\sin x$, that is, the ratio of opposite side and hypotenuse.

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Maybe you could first notice that complex multiplication with numbers of modulus one is a rotation. – André Caldas Mar 28 '12 at 16:58
Examine the curve parameterized by $x=\cos t$, $y=\sin t$. – David Mitra Mar 28 '12 at 16:59
Glad to see you rewrote the question! Looks pretty popular ;P – Tyler Mar 29 '12 at 15:26

Knowing that $(\cos x)'=-\sin x$ and that $(\sin x)'=\cos x$ (which I assume Tao proves) allows one to show that for $f(x)=\sin^2 x+\cos^2 x$, we have $$f'(x)=2\sin x \cos x-2\cos x\sin x =0.$$ Thus, $f$ is a constant function. Since $f(0)=1$, $f$ is identically 1.

So, the Pythagorean identity is valid: $$\sin^2 x+\cos^2x=1.$$ Using this, it follows that the curve $C$ parameterized by $x=\cos t$, $y=\sin t$ is a circle of radius 1 centered at the origin. Further analysis of this curve will reveal that the values of $\sin$ and $\cos$ can be read from the side lengths of right triangles.

Note, in particular that the length of arc from $t=0$ to $t=t_0$ is $\int_0^{t_0} \sqrt{\bigl[{dx\over dt} \bigr]^2 +\bigl[{dy\over dt}\bigr]^2 }\,dt=t_0$; so the angle to which the "triangle method" refers is interpreted in the correct manner.

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For some reason, I am seeing this for the first time now - 11 years after Calculus I! Could you verify the following? That the sine (cosine) is the ratio of opposite (adjacent) to hypotenuse follows from the parametrization of C. – The Chaz 2.0 Mar 28 '12 at 18:20
Note I assumed that $\cos t$ and $\sin t$ are decreasing and increasing where they should be... Does Tao provide a proof of this? – David Mitra Mar 28 '12 at 19:08
Actually, consider the movement of a moving object generating the curve $C$. If the object came to rest at some point that is on the circle but not on a coordinate axis, then both $(\cos t)'$ and $(\sin t)'$ would be $0$ at that point. This can't happen by the Pythagorean Identity. For $t \in\{0,\pi/2,\pi,3\pi/2\}$ one can explicitly calculate the values of $\cos$ and $\sin$, and thus determine that the point always travels counterclockwise. Or, see @robjohn 's answer. – David Mitra Mar 28 '12 at 19:37
@DavidMitra if $t$ is point A,$t_0$ is point B.Could you prove that every point of arcAB is corresponding to $x= \cos t,y=\sin t$,$t \in [0,t_0]$ – noname1014 Mar 28 '12 at 21:21
I accept this answer,but I think $\int_{\cos(\theta)}^1\sqrt{1-x^2}\,\mathrm{d}x$ is more intuitive. – noname1014 Mar 29 '12 at 11:57

In this hint I suggest showing from the power series that if $$\sin(x)=\sum_{k=0}(-1)^k\frac{x^{2k+1}}{(2k+1)!}\tag{1}$$ and $$\cos(x)=\frac{\mathrm{d}}{\mathrm{d}x}\sin(x)=\sum_{k=0}(-1)^k\frac{x^{2k}}{(2k)!}\tag{2}$$ that $\frac{\mathrm{d}}{\mathrm{d}x}\cos(x)=-\sin(x)$ and from there that $$\sin^2(x)+\cos^2(x)=1\tag{3}$$ Therefore, $(\cos(x),\sin(x))$ lies on the unit circle.

To see that $(\cos(x),\sin(x))$ moves around the unit circle at unit speed, note that $(3)$ implies $$\left|\frac{\mathrm{d}}{\mathrm{d}x}(\cos(x),\sin(x))\right|=\left|(-\sin(x),\cos(x))\right|=1\tag{4}$$ Thus, $(3)$ and $(4)$ say that $(\cos(x),\sin(x))$ moves around the unit circle at unit speed. Note also that $(-\sin(x),\cos(x))$ is at a right angle counter-clockwise from $(\cos(x),\sin(x))$. Therefore, $(\cos(x),\sin(x))$ moves counter-clockwise around the unit circle at unit speed, starting at $(1,0)$. This should be sufficient to show that $\sin(x)$ and $\cos(x)$ are the standard trigonometric functions.

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From the series, it is easy to see Euler's forumla,

$$e^{ix} = \cos(x) + i\sin(x)$$

With more series manipulation, we can obtain the Pythagorean theorem,

$$|e^{ix}| = e^{ix}e^{-ix} = (\cos(x) + i\sin(x))(\cos(x) - i\sin(x)) = \cos^{2}(x) + \sin^{2}(x) = 1$$

Knowing that $\sin(x)$ and $\cos(x)$ have range $[-1,1]$, and are odd and even functions respectively, we see that $e^{ix}$ traces out the unit circle in $\mathbb{C}$. From this, we can extract the geometric interpretation of sine and cosine.

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But how do you "see" that $e^{ix}$ traces out a circle from the power series definition? And how do you show that $e^{2\pi i}=1$? Once you have those two statements, and that $e^{z+w}=e^ze^w$ for all complex $z,w$, then you can "see" that $\sin$ defined by power series is the same as $\sin$ defined "geometrically." – Thomas Andrews Mar 28 '12 at 17:16
That's a good point, and I've tried to clarify the answer appropriately. Is it any better now? – Isaac Solomon Mar 28 '12 at 17:42
but what does x mean,how do you establish relation of x(some real number) and $\theta$(radian)? – noname1014 Mar 28 '12 at 17:52

To get to the geometry, take the "non-geometric" versions of cosine and sine, say $C(t)$ and $S(t)$. We can use these to parametrize the unit circle, simply because of the fact that $C^2(t)+S^2(t)=1$.

Now comes the crucial bit, calculating the arclength. We find that the arclength from $0$ to $t$ is $t$. This connects $C(t)$ and $S(t)$ with "angle," which in the formal theory is just arclength.

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The best approach I've seen uses the fact that $\exp z=\lim_{n\to\infty} (1+\frac{z}{n})^n$ for all complex $z$. Then, for real $x$, we show that, for large $n$, $1+\frac{ix}{n}$, is "close enough" to $\cos \frac{x}{n} + i\sin\frac{x}{n}$, so that $(1+\frac{ix}{n})^n$ is "close enough" to $\cos x + i\sin x$. (Here, $\cos$ and $\sin$ are the geometric definitions of the trig functions.)

So, first define $\exp z$ via the normal power series, and then show that $\exp z = \lim_{n\to\infty} (1+\frac{z}{n})^n$.

Then, define $\operatorname{cis} x =\cos x + i\sin x$, and prove, using (geometric) properties of the trig functions, that $\operatorname{cis} x\operatorname{cis} y =\operatorname{cis} (x+y)$. Then note that by induction, $(\operatorname{cis} x)^n = \operatorname{cis} nx$.

Next, we are going to compute $\exp(ix)$ by writing $1+\frac{ix}n$ in polar coordiates, $r_n \operatorname{cis}(\theta_n)$.

As mentioned above, the goal is to show that, for large enough $n$, $1+\frac{ix}{n}$ is "close enough" to $\operatorname{cis} \frac{x}n$, so that the limit of $(1+\frac{ix}n)^n$ is the same as the limit of $(\operatorname{cis}\frac x n)^n = \operatorname{cis} x$, and therefore $\exp(ix)=\operatorname{cis} x$.

First, note that $r_n=\sqrt{1+\frac{x^2}{n^2}}$, and show that $r_n^n\to 1$ as $n\to \infty$. (This is relatively easy, noting that $(r_n)^{2n^2}\to \exp (x^2)$.)

So this shows that $\exp(ix)$ ends up on the unit circle for real $x$, and is equal to $\lim_{n\to\infty} \operatorname{cis}(n\theta_n)$.

What we know about $\theta_n$ is that $\sin \theta_n = \frac{x}{nr_n}$.

Next you need to show is that $\lim_{n\to\infty} n\theta_n = x$. Then you've shown that $\exp(ix)=\operatorname{cis} x$.

The key to proving this last limit is to show that, for small $\theta$, $|\theta-\sin(\theta)|\leq C|\sin \theta|^2$ for some constant $C$. Note, though, that proving this is not easy since we are assuming we do not know the power series expansion for $\sin$.

If you can show that, then you can show, for large $n$, $|\theta_n - \frac{x}{nr_n}|\leq C|\frac{x}{r_nn}|^2<C|\frac{x}{n}|^2$. So $|n\theta_n - \frac{x}{r_n}|<C\frac{|x^2|}{n}$.

So, since $\frac{x}{r_n}\to x$, $n\theta_n\to x$, and therefore, $\lim \operatorname{cis} n\theta_n = \operatorname{cis} x$

So all that's left to prove is that for small enough $\theta$, $|\theta - \sin \theta| \leq C|\sin \theta|^2$. (You could actually replace $2$ with $1+\epsilon$ for fixed $\epsilon>0$.)

The geometric reason for this last one is as follows. Let $P=(1,0)$ and $Q=(\cos \theta,\sin \theta)$. Take the lines tangent to the circle at $P$ and $Q$ and let their intersection be $R$. Claims (for small $\theta>0$:)

$$|PR|=|QR|=\tan\frac{\theta}2$$ $$\sin\theta < \theta < |PR|+|QR|=2\tan\frac{\theta}2$$

The first half of the inequality is because $\sin \theta$ is the shortest distant from $Q$ to the $x$-axis, and $\theta$ is the length of the path from $Q$ to the $x$-axis along the circle. The second inequality is a little less intuitive - basically, this is due to the rule that the shortest path from $P$ to $Q$ that does not go inside the circle is along the circle.

(For small $\theta<0$, we have to reverse all the signs, but the results are the same.)

Thus we see:

$$|\theta-\sin \theta| < |2\tan\frac\theta 2 - \sin\theta|$$

But $2\tan\frac{\theta}2 = \frac{2\sin \frac{\theta}2}{\cos\frac\theta 2}$. Mutliply numerator and denominator by $\cos \frac{\theta}2$ and we see that:

$$2\tan\frac{\theta}2 = \frac{\sin \theta}{\cos^2\frac{\theta}2}$$

So:

$$|2\tan\frac\theta 2 - \sin\theta| = |\sin \theta \frac{\sin^2\frac \theta 2}{\cos^2\frac\theta 2}|$$

Multiplying the numerator and denominator by $4\cos^2\frac \theta 2$, we get:

$$|2\tan\frac\theta 2 - \sin\theta| = |\sin\theta \frac{\sin^2\theta}{4\cos^4\frac{\theta}2}|$$

So with $\theta$ small enough that $\cos\frac\theta 2>\frac{1}{2}$, we have

$$|\theta-\sin \theta| < |2\tan\frac\theta 2 - \sin\theta| < 4|\sin^3\theta|<4|\theta|^3$$

So we now know that $\exp ix = \operatorname{cis} x$. That means, in turn, that $$\sin x = \frac{\exp(ix)-\exp(-ix)}{2i}$$ and $$\cos x = \frac{\exp(ix)+\exp(-ix)}{2}$$

From there we can derive the power series for $\sin$ and $\cos$ using the power series for $\exp$.

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Euler's formula (the complex exponential) and the identification of the complex plane $\mathbb{C}$ with $\mathbb{R}^2$ does that for us. Then the equations $x=\cos\theta,~y=\sin\theta$ define the ordinate and abscissa of points on the unit circle; drawing in the relevant triangles gets you the rest of the way.

From the series definition of $\sin$ and $\cos$, the easiest link with geometry is via the complex exponential series to derive Euler's formula, $\cos\theta+i\sin\theta=e^{i\theta}$. Once we have that, the rest follows (from the background that Tao provides).

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