The Unapologetic Mathematician

Mathematics for the interested outsider

Properties of the Sine and Cosine

Blaise got most of the classic properties of the sine and cosine in the comments to the last post, so I’ll crib generously from his work. As a note: I know many people write powers of the sine and cosine functions as \sin^2(x) (for example) instead of \sin(x)^2. As I tell my calculus students every year I refuse to do that myself because that should mean \sin(\sin(x)), and I guarantee people will get confused between \sin^{-1}(x)=\arcsin(x) or \sin^{-1}(x)=\frac{1}{\sin(x)}

First, let’s consider the function g(x)=\sin(x)^2+\cos(x)^2. We can take its derivative using the rules for derivatives of trigonometric functions from last time:

\displaystyle g'(x)=2\sin(x)\cos(x)-2\cos(x)\sin(x)=0

So this function is a constant. We easily check that g(0)=1, and so \sin(x)^2+\cos(x)^2=1.

What does this mean? It tells us that if \sin(x) and \cos(x) are the lengths of the legs of a right triangle, the hypotenuse will have length {1}. Alternately, the point with coordinates (\cos(x),\sin(x)) in the standard coordinate plane will lie on the unit circle. We haven’t talked yet about using integration to calculate the length of a path in the plane, but when we do we’ll see that the length of the arc on the circle from (1,0) to (\cos(x),\sin(x)) is exactly x.

This gives us another definition for the sine and cosine functions — one closer to the usual one people see in a trigonometry class. Given an input value x, walk that far around the unit circle, starting from the point (1,0). The coordinates of the point you end up at are the sine and cosine of x. And this gives us our “original” definitions: given a right triangle, it is similar to a right triangle whose hypotenuse has length {1}, and the sine and cosine are the lengths of the two legs.

Now, since \sin(x)^2 and \cos(x)^2 are both nonnegative, they must each be bounded above by g(x)=1. Thus -1\leq\sin(x)\leq1 and -1\leq\cos(x)\leq1. More specifically, any time that \sin(x_0)=0 we must have \cos(x_0)=\pm1.

We know that \sin(0)=0 and \cos(0)=1, so if we ever have another point t where \sin(t)=0 and \cos(t)=1 we have a period. This is because the differential equation will determine the future behavior of \sin(t+x) the same way it determined the behavior of \sin(0). In fact, if \sin(p)=0 and \cos(p)=-1, then the future behavior of \sin(p+x) will be exactly the negative of the behavior of \sin(x), and so eventually \sin(2p)=0 and \cos(2p)=1 again.

Admittedly, I’m sort of waving my hands here without an existence/uniqueness proof for solving differential equations. But the geometric intuition should suffice for the idea that since the function’s value and first derivative at {0} are enough to determine the function, then the specific point we know them at shouldn’t matter.

So, does the sine function have a positive zero? That is, is there some p>0 so that \sin(p)=0? If so, the lowest such one would have to have \cos(p)=-1 (because positive numbers near {0} have positive sines). The next one would then be \sin(2p)=0 with \cos(2p)=1, and the whole thing repeats with period 2p.

The function \sin(x) starts out increasing, and so \cos(x) decreases (since \cos(x)^2=1-\sin(x)^2. If \sin(x) has a maximum, then \cos(x) (its derivative) must cross zero. Then \sin(x) is decreasing, and it cannot increase again unless \cos(x) crosses zero again. But if \cos(x) crosses zero again it must have passed through a local extremum (Rolle) and so \sin(x) cannot increase again before it crosses zero itself.

So if we are to avoid \sin(x) having a positive zero, it must either increase to some asymptote below {1}, or it must increase to a maximum and then decrease to some asymptote below {1}. But for a function to have an asymptote it must approach a horizontal line, and its derivative must approach {0}. That is, we can only have \sin(x) approaching an asymptote at y=1, while \cos(x) approaches an asymptote at y=0.

But if \cos(x) approaches an asymptote, its derivative must also asymptotically approach {0}. But this derivative is -\sin(x), which we are assuming approaches -1! And so none of these asymptotes are possible!

So the sine function must have a positive zero: \sin(p)=0. And thus the sine and cosine (and all other solutions to this differential equation) will have period 2p.

Finally, what the heck is this value p? In point of fact, we have no way of telling. But it might come in handy, so we’ll define this number and give it a new name: \pi. Whenever we say \pi we’ll mean “the first positive zero of the sine function”.

Here I want to point out that I’ve fulfilled my boast of a few months ago on some other weblog. In my tireless rant against the \pi-fetishism that infests the geek community, I told someone that \pi can be derived, ultimately, from solely the properties of the real number system. Studying this field — itself uniquely specified on algebraic and topological grounds — leads us to both differential calculus and to power series, and from there to series solutions to differential equations. One of the most natural differential equations in the world thus gives rise to the trigonometric functions, and the definition \pi follows from their properties. There is no possible way it could be anything other than what it is when you see it from this side, while the geometric definition hinges on some very deep assumptions on the geometry of spacetime.

October 14, 2008 Posted by | Analysis, Calculus | 12 Comments