Different definitions of trigonometric functions

In school, we learn that sin is "opposite over hypotenuse" and cos is "adjacent over hypotenuse".

Later on, we learn the power series definitions of sin and cos.

How can one prove that these two definitions are equivalent?

Solution 1:

If you allow yourself a tiny bit of calculus ( "$\sin x / x \to 1$" as "$x \to 0$" ) and apply some combinatorics, there's a really nice geometric interpretation of the terms of the power series for the functions. Consider this diagram and the polygonal "spiral" that starts at $P_0$ and closes in on the point $P$ (where $|P_0 P| = 1$).

The horizontal segments $P_{2n} P_{2n+1}$ alternately overshoot and undershoot the length of the cosine segment; the vertical segments $P_{2n+1} P_{2n+2}$ do the same for the sine segment. So,

$\cos \theta = \sum_{n=0}^{\infty}(-1)^n | P_{2n} P_{2n+1} |$

$\sin \theta = \sum_{n=0}^{\infty} (-1)^n | P_{2n+1} P_{2n+2} |$

Now, the lengths $|P_{k} P_{k+1}|$ are equal to the lengths of the curves $|I_k|$, which constitute a series of successive involutes (with $I_0$ defined to be a segment, and $I_1$ defined to be an arc of the unit circle). Combinatorics and the calculus result I mentioned show that the involute lengths satisfy ...

$|I_k| = \theta^k / k!$

... so that the above are, in fact, power series.

Interestingly, the same thing can be done with secant and tangent, using an involute zig-zag:

where

$\sec \theta = \sum_{n=0}^{\infty} | P_{2n} P_{2n+1} | = \sum_{n=0}^{\infty} | I_{2n} |$

$\tan \theta = \sum_{n=0}^{\infty} | P_{2n+1} P_{2n+2} | = \sum_{n=0}^{\infty} | I_{2n+1} |$

and the lengths $|I_k|$ turn out to be the appropriate multiples of powers of $\theta^k$.

Reference to the argument for sine and cosine (attributed to Y. S. Chaikovsky, as reported by Leo Gurin), a complete discussion of the trickier argument for secant and tangent, and then a refinement of the argument for sine and cosine, are in my note "Zig-Zag Involutes, Up-Down Permutations, and Secant and Tangent" (PDF).

BTW: I have not (yet) cracked the case for cosecant and cotangent.

Solution 2:

Most of the proofs in elementary calculus textbooks use the definition of $\sin x$ via geometry to prove that the derivative of $\sin x$ is $\cos x$ (namely, the fact that $\lim_{x \to 0} \frac{ \sin x}{x} = 1$). Consequently, it follows that $\sin x$ and $\cos x$ are the two linearly independent solutions of $y'' = -y$. The power series equations are also two linearly independent solutions of this differential equation. Moreover, $\sin x$ and its derivative coincide with the derivative of the power series for $ \sin x$ at zero (no surprise, it's a Taylor series). Same for $\cos x$. By uniqueness of solutions to ordinary differential equations, this proves that $\sin x$ and $\cos x$ as defined in school are equal to their power series. (This is an expansion of Qiaochu's comment.)

Solution 3:

Robison, "A new approach to circular functions, π, and lim sin(x)/x", Math. Mag. 41.2 (March 1968), 66–70 [jstor].

In this paper it is shown that the addition law for cosine (and a couple other simple assumptions) uniquely determines cosine and sine. So then it's enough to prove geometrically that the high school functions satisfy that law (this is essentially Ptolemy's theorem), and prove that the power series functions satisfy it (using the binomial theorem and such manipulations).

Solution 4:

There is another proof that the derivative of sine is cosine that doesn't use the sandwich theorem mentioned by Qiaochu and Akhil above. Instead, one can use the definition of arcsine and the standard calculus formula for arc length in terms of an integral to show that arcsine = the integral of (1 - x^2)^(-.5). It follows that the derivative of arcsine is (1 - x^2)^(-.5), and (by the chain rule) one can use this fact to prove that the derivative of sine is cosine. In fact, I'm not sure why this proof is presented less frequently then the one via the sandwich theorem. The unit circle definition of sine is based on arc length, and in calculus we learn a formula for arc length based on integration. Why not connect these two concepts for a natural proof that the derivative of sine is cosine?

Solution 5:

As a rough outline, the circular definitions of sine and cosine (the y- and x-coordinates of the image of (1,0) under a rotation about the origin) lead to being able to differentiate sine and cosine, and once you know how to differentiate them (infinitely), Taylor's Theorem justifies that the power series is equal to the function.