cos and sin

$\cos$ and $\sin$ are what you use when you want to draw a circle, or animate something moving in a circle.

In the post on unit vectors, we defined the unit $n$-sphere and found a parametrization of it. In the previous post, we encountered the arclength parametrization, which for most purposes is more convenient than other parametrizations. In this post, we discuss the arclength parametrization of the unit circle $\cl{S}{S^1}$, which is given by the $\cos$ and $\sin$ functions. This will be useful for drawing things along a circle or animating something that's rotating.

Radians

We need to start by naming the length of the unit circle.

Definition

We write $\Twopi$ (pronounced either "circle", "var-pi", or "two-pi") for the arclength, or circumference, of the unit circle

$$\cl{S}{S^1} \Defeq \{(x, y) \in \VecR2 \mid x^2 + y^2 = 1 \}.$$

In the exercises of the previous unit, you calculated that $\Twopi\approx6.28\dotsc$

By scaling, it follows that the circumference of a circle of radius $r$ is given by $\Twopi r$. For example, the circumference of a circle of radius $4$ is $4\Twopi\approx25.13\dots$. Equivalently, we could define $\Twopi$ as the ratio of any circle's circumference to its radius.

Remark

You may be more used to the constant $\pi=\Twopi/2\approx3.14\dots$, which is the ratio of a circle's circumference to its diameter (twice the radius). It turns out the ancient Greeks made a mistake in defining $\pi$: while diameter is relevant in real-life situations, mathematically it's much more convenient to index circles by their radius rather than their diameter.

This is not something to get too worked up about, however: although $\Twopi$ is an important number, numbers are not that important in mathematics.

Many sources use the letter $\tau$ instead of $\varpi$. This is patently absurd: on top of having completely the wrong vibes to be a circle constant, $\tau$ is a B-tier Greek letter while $\Twopi$ is an S-tier constant.

cos and sin

Definition

We define the functions $\cos$ and $\sin$ such that

$$\ivec{\cos\theta,\, \sin\theta}, \qquad \theta\in \ClsdIntvl0{\twopi}$$

is the arclength parametrization of the unit circle $\cl{S}{S^1}$, starting at the point $\ivec{1,0}$ and traveling counterclockwise. Equivalently, $\ivec{\cos\theta,\sin\theta}$ is the $\ivec{x,y}$-coordinates of the point at angle $\theta$ from the origin.

Here's a graph to experiment with that.

This completely defines $\cos$ and $\sin$, but doesn't tell us how to compute them. However, in just a few posts we'll derive an algorithm to compute these.

You may be used to a definition of $\cos$ and $\sin$ in terms of ratios of right-angle triangles; I'll review that interpretation below. That interpretation is often useful, but it doesn't work for angles larger than $90^\circ$.

Animation

Let's say more about animation. So far, $\cos$ and $\sin$ tell us how to parametrize the unit circle, i.e. the circle of radius $1$ centered at the origin $\O=(0,0)$. What about other circles?

Let's do this in steps. First, let

$$\cl S{S^1}(r) = \{v\in\VecR2 \mathrel: \|v\| = r \}$$

be the circle of radius $r$ centered at the origin. We have

$$\begin{align*} \cl S{S^1}(r) &\phantom:= r \cl S{S^1}\\ &\Defeq \{rv \mid v\in \cl S{S^1}\} \end{align*}$$

so can parametrize this by

$$r\ivec{\cos t,\sin t} = \ivec{r\cos t,r\sin t}, \qquad t\in\CIntvl0\twopi.$$

Next, let's consider circles centered at other points. For a point $P\in\AffR2$ and a radius $r>0$, let

$$\cl S{S^1}(P,r) \Defeq \{Q \in \AffR2 \colon \|P-Q\| = r \}$$

be the circle of radius $r$ centered at $P$. Note that $\cl S{S^1}(r)$ lives in the vector space $\VecR2$, while $\cl S{S^1}(P,r)$ lives in the affine space $\AffR2$; recall that "the origin" is a feature of vector space. We can get $\cl S{S^1}(P, r)$ by translating $\cl S{S^1}(r)$ to the point $P$: in other words,

$$\begin{align*} \cl S{S^1}(P, r) &\phantom:= P + \cl S{S^1}(r)\\ &\Defeq \{P + v \mid v \in \cl S{S^1}(r) \} \end{align*}$$

Therefore, a parametrization of $\cl S{S^1}(P, r)$ is given by

$$P+r\ivec{\cos t,\sin t} = \ipt{\cl{p}{p_1}+r\cos t,\cl{p}{p_2}+r\sin t}, \qquad t\in\CIntvl0\twopi,$$

where $P=\ipt{\cl{p}{p_1},\cl{p}{p_2}}$.

That does everything we need if we want to draw a circle (or a part of a circle). Often though, we want to animate something moving in a circle. This means we need to further take into account its speed (or angular velocity) and its starting position.

By default, $(\cos t,\sin t)$ starts at $(1,0)$ and goes around the circle one every $\Twopi$ units of time, going counterclockwise. To have it start at a different angle $\cl{θ}{\theta_0}$, we just add that on to $t$:

$$\ivec{\cos(\cl{θ}{\theta_0} + t), \sin(\cl{θ}{\theta_0} + t)}$$

To change the speed, we multiply $t$ by a constant. For example, if $t$ is measured in seconds and we want to go around the circle once every two seconds, we'd use

$$\ivec{\cos\left(\cl{θ}{\theta_0}+\frac{\Twopi t}2\right), \sin\left(\cl{θ}{\theta_0}+\frac{\Twopi t}2\right)}$$

since going around the circle once every two seconds is the same as going halfway around the circle every second. We can also scale by a negative number to go clockwise instead of counter-clockwise.

To summarize:

Proposition

The parametrization of a particle travelling:

• on a circle of radius $r$
• centered at the point $P=\ipt{\cl{p}{p_1},\cl{p}{p_2}}$
• starting from angle $\cl{θ}{\theta_0}$
• making $k$ (counter-clockwise) rotations per unit of time $t$

is given by

$$P + r\ivec{\cos(\cl{θ}{\theta_0}+\Twopi kt), \sin(\cl{θ}{\theta_0}+{\Twopi}kt)}$$

which is the same as

$$\ipt{\cl{p}{p_1}+r\cos(\cl{θ}{\theta_0}+\Twopi kt), \cl{p}{p_2}+r\sin(\cl{θ}{\theta_0}+\Twopi kt)}.$$

Here's some code to try that out:

Triangle interpretation

You may be more used to a definition of $\cos$ and $\sin$ in terms of ratios of angles of right-angle triangles.

$$\begin{aligned} \cos\theta &= \frac{\text{adjacent}}{\text{hypotenuse}}\\ \sin\theta &= \frac{\text{opposite}}{\text{hypotenuse}} \end{aligned}$$

This perspective is often useful, but it doesn't work for angles larger than $90^\circ$.

How does this relate to coordinates of points on the unit circle? First, note that for any point on a circle, we can form a right-angle triangle whose hypotenuse is the line segment from the center of the circle to that point (see below). In the case of the unit circle,

Special angles

Although we don't yet have a general algorithm for computing $\cos$ and $\sin$, we can read off the values of the coordinates of East $(0)$, North $(90^\circ=\frac\Twopi4)$, West $(180^\circ=\frac\Twopi2)$, and South $(270^\circ=\frac{3\Twopi}4$).

$$\begin{align*} \ivec{\cos0,\,\sin0} &= \ivec{1,0} & \ivec{\cos\frac\Twopi2,\,\sin\frac\Twopi2} &= \ivec{-1,\, 0}\\[.5em] \ivec{\cos\frac\Twopi4,\,\sin\frac\Twopi4} &= \ivec{0,\, 1} & \ivec{\cos\frac{3\Twopi}4,\,\sin\frac{3\Twopi}4} &= \ivec{0,\, -1} \end{align*}$$

With a bit more work, we can calculate the coordinates of the point at $45^\circ$, using the triangle interpretation above.

45°

x

x

1

Since the angles of a triangle add up to $180^\circ=\Twopi/2$, the remaining angle is also $45^\circ=\Twopi/8$. This implies that the two side lengths $\cos45^\circ$ and $\sin45^\circ$ are equal; let's call that $x$. By the Pythagorean theorem,

$$\begin{align*} \cos(45^\circ)^2 + \sin(45^\circ)^2 &= 1\\ 2x^2 &= 1\\ x^2 &= \frac12\\ x &= \sqrt{\frac12}\\ &= \frac1{\sqrt2}\\ &= \frac{\sqrt2}2 \end{align*}$$

Therefore, we have

$$\cos45^\circ = \sin45^\circ = \frac{\sqrt2}2.$$

Exercises

1. Find $\cos$ and $\sin$ of $30^\circ$ and $60^\circ$.

2. Consider the following diagram:

   Source

   By calculating the length of the green line segment in two different ways, establish the identity

   $$\begin{equation}\htmlId{eq-sin-pyth}{} 2\sin\left(\frac{\alpha - \beta}2\right) = \pm\sqrt{ (\cos\alpha - \cos\beta)^2 + (\sin\alpha - \sin\beta)^2 } \end{equation}$$

3. Consider a regular $n$-gon inscribed in a circle of radius 1:

   Source

   1. Show that its perimeter is given by $2n\sin\left(\dfrac\Twopi{2n}\right)$.

   2. Show that its area is given by $\displaystyle\frac n2\sin\left(\frac\Twopi n\right)$.

4. Once we've established $(1)$, we can derive all the usual trigonometric identities using algebraic manipulations, with no further geometric insight required.

   1. Use $(1)$ to derive the "half-angle formulas" $$\begin{align} \htmlId{eq-cos-half}{}\cos(\theta/2) &= \pm\sqrt{\frac{1+\cos\theta}2}\\[1em] \htmlId{eq-sin-half}{}\sin(\theta/2) &= \pm\sqrt{\frac{1-\cos\theta}2} \end{align}$$
   2. Use $(2)$ and $(3)$ to derive the "double-angle formulas" $$\begin{align*} \cos(2\theta) &= \cos^2(\theta)-\sin^2(\theta)\\ &= 2\cos^2(\theta) - 1\\ &= 1 - 2\sin^2(\theta)\\ \sin(2\theta) &= 2\cos(\theta)\sin(\theta) \end{align*}$$
   3. Derive the "angle-difference" formulas $$\begin{align*} \cos(\alpha-\beta) &= \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\\ \sin(\alpha-\beta) &= \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta) \end{align*}$$ and the "angle-sum" formulas $$\begin{align*} \cos(\alpha+\beta) &= \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\\ \sin(\alpha+\beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \end{align*}$$ These are sometimes stated in combined form $$\begin{align*} \cos(\alpha\pm\beta) &= \cos(\alpha)\cos(\beta) \mp \sin(\alpha)\sin(\beta)\\ \sin(\alpha\pm\beta) &= \sin(\alpha)\cos(\beta) \pm \cos(\alpha)\sin\beta) \end{align*}$$ This is not the best way to derive these formulas, but it works; the point is mainly that you can get them from each other using pure algebra. We'll see a more enlightening explanation of and in a future lesson, and then the rest will follow by algebra in a similar way (in the opposite order as you derived them here).