# A simple question about sine and cosine

I have been thinking about all of the different ways that I have encountered sine and cosine in my studies. There are no courses on trigonometry at my school, so perhaps that's why I feel like something is missing, something that ties all of these ideas together.

• The Unit Circle. The unit circle is a way to organize all possible right triangles up to similarity. The sine and cosine can be defined as ratios of sides of these right triangles, though in practice the sine becomes the vertical component and the cosine the horizontal. What I want to know is how do I relate this beautiful diagram to the other constructions of sine and cosine.

• The orthonormal basis of the solution space to the differential equation: $y'' = -y$. In some sense this is as good of a definition as the unit circle.

• Taylor Series. These grow clearly out of the differential equation above. How could I connect Taylor series to the concept of the unit circle? I think of partial Taylor series as "better and better" approximations of these function centered at zero (or maybe some place else, I guess it would not matter) -- so how is that concept linked to what $y'' = -y$ says about these equations?

Lastly, are there any other major representations I should also consider?

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What do you mean by "Taylor Series. These grow clearly out of the differential equation above."? – t.b. Jul 18 '11 at 23:07
$sin(z)= \frac{e^{iz}-e^{-iz}}{2i}$ for complex $z$ is the first that comes to mind. There are others, see en.wikipedia.org/wiki/Sine for a continued fraction expression for example. Also, I remember Apostol defining sine and cosine in his Calculus by some fundamental properties... – Bruno Stonek Jul 18 '11 at 23:10
@Theo, if you solve the differential equation in series with the right initial conditions you get sine and cosine as the only solutions, with no initial conditions you still get series that are constant multiples of sine and cosine as solutions. – Noteventhetutorknows Jul 18 '11 at 23:13
Ah, you mean that you can compute them by plugging in the power series ansatz. Yes, right. Thanks for clarifying! – t.b. Jul 18 '11 at 23:18
@noteventhetutorknows: My note ( daylateanddollarshort.com/math/pdfs/sectan.pdf ) briefly discusses the sine-cosine spiral construction, which is due to Y.S. Chaikovsky, as revealed by Leo Gurin; here's the bibliographical item from my note: Leo S. Gurin, A problem, American Mathematical Monthly, 103, 1996, 683-686. (The secant-tangent zig-zag is my own original result.) – Blue Jul 19 '11 at 13:30

It all comes down to representation theory. The assignment $\theta \mapsto \left[ \begin{array}{cc} \cos \theta & - \sin \theta \\\ \sin \theta & \cos \theta \end{array} \right]$ sending an angle $\theta$ to a matrix describing rotation by $\theta$ in the Euclidean plane $\mathbb{R}^2$ is a group homomorphism, so abstractly one cay say that sine and cosine naturally appear as matrix coefficients of this particular representation of the group $\mathbb{R}$ or perhaps instead of the circle group $S^1$.

The connection to Fourier series is given abstractly by Pontrjagin duality, which describes the representation theory of a large class of abelian groups, and the connection to differential equations is given by the fact that the space of solutions to $y'' = -y$

• naturally inherits the structure of a representation of $\mathbb{R}$ given by translation $y(x) \mapsto y(x+t)$, and
• has the property that $y^2 + (y')^2 = \text{const}$ for any solution $y$ by inspection, which is strongly suggestive of (but does not immediately prove) periodicity. Note that we get a parameterization of the unit circle this way.

The second property can be understood as conservation of energy for a harmonic oscillator and comes from the first via Noether's theorem.

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You have missed a y in the spelling of pontryagin – user9413 Jul 19 '11 at 3:38
Pontrjagin is an accepted alternate spelling. – Qiaochu Yuan Jul 19 '11 at 3:41
Oh i didn't know that. Sorry for the trouble :) – user9413 Jul 19 '11 at 3:43
"Pontrjagin is an accepted alternate spelling." - i.e., all depends on how you Anglicize your Cyrillic... – J. M. Jul 20 '11 at 12:05

My intuitive interpretation is usually in the terms of projections, \begin{align} x \cdot y &= |x||y|\cos(\theta) \end{align} If necessary, sine can be viewer similarly, as projected towards $y_\perp$.

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Can you flush this out a little more? I don't know what x and y are in this context... or the meaning of "y perpendicular" (it's probably obvious, but I'd appreciate the help, thanks.) – Noteventhetutorknows Jul 18 '11 at 23:48
I will have to draw a figure to. It'll take a moment. – Mikael Öhman Jul 18 '11 at 23:50
I think I see what you are saying now. Though, I don't see how this realtes to the differential equation, are you suggesting this as a definition of cosine that I missed? en.wikipedia.org/wiki/Dot_product – Noteventhetutorknows Jul 18 '11 at 23:51
I didn't see anything specifying interpretations only related to differential equations in your question. However, this interpretation of cosine can be usefull even when doing projections of in function spaces, as the case of fourier series or finite element. – Mikael Öhman Jul 19 '11 at 0:25
there are three things I'm looking for: a connection between taylor series and the unit circle, a connection between the ODE y''=-y and the unit circle, and as an aside any other important representations, I guess I ought to have made that more clear, thanks! – Noteventhetutorknows Jul 19 '11 at 0:57

Start with $x'' = -x$. Introduce $y = x'$, so that \begin{align}x' &= y, \\ y' &= x'' = -x.\end{align}

Now let $\mathbf x = [x, y]^T$ be the coordinates of a point in the 2D plane. Then we have $$\mathbf x'' = \begin{bmatrix}0 & 1 \\ -1 & 0\end{bmatrix} \mathbf x.$$ Assuming $\lVert \mathbf x(0) \rVert = 1$, this is the equation of a particle on the unit circle (exercise: show this) moving with unit speed (exercise: show this). (Hint: observe that the matrix in question produces a vector perpendicular to $\mathbf x$.)

Therefore,the angle $\theta$ that the particle makes with the horizontal axis increases at one radian per unit time, so $\theta = t$. From the geometrical picture, this means that the coordinates of the particle are $x(t) = \cos(t)$ and $y(t) = \sin(t)$.

In general, an $n$th-order ODE in one dimension can be thought of as a first-order ODE — a "flow" — in $n$-dimensional space. Here, the flow corresponding to $x'' = -x$ is uniform rotation around the origin in a 2D plane.

(In my view, the connection between Taylor series and the geometrical picture is only via the differential equation, and that has little do to with circles specifically. So if you don't get an answer addressing that here, I think your penultimate sentence would make a good question worth asking separately.)

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