Are there theoretical applications of trigonometry?

I am a high school student currently taking pre-calculus. We have just finished a unit on analytic trigonometry.

Are any purely theoretical uses for trigonometry? More specifically, can trigonometric concepts (or even functions) be used to prove/disprove general mathematical conjectures?

I have been told it is used a lot in calculus, but by my (extremely) limited knowledge it mainly consists of applying calculus concepts to trigonometric functions. Is this correct?

Solution 1:

I tell my students in differential equations (a one-semester-past-calculus class) that people have been lying to them about why trig is important! Solving triangles, who cares. Trig matters for all sorts of reasons in more advanced math.

One big deal is that the sine and the cosine both satisfy $$f(x+2\pi)=f(x);$$that is, they are functions with period $2\pi$ (in radians). The amazing and hugely important thing about that is that you can use sine and cosine to "generate" any other function with period $2\pi$. That is, leaving out a lot of technical details, if $f$ is any function with period $2\pi$ then there are constants $a_n$ and $b_n$ so that $$f(x)=a_0+(a_1\cos(x)+b_1\sin(x))+(a_2\cos(2x)+b_2\sin(2x))+\dots.$$ That's the "Fourier series" for $f$; Fourier series are awesomely useful and important in many areas of math, theoretical and applied both.

Confession: It appears people are reading this. I can't stand it; I told a lie above. For the record, one of the "technical details" I omitted is that it's not actually true that every $2\pi$-periodic function can be expanded in a Fourier series.

Fourier told the exact same lie. In some sense though it was true in his day, at least truer than it is now. Because the modern notion of "function" is very different from what people thought of as a "function" a few centuries ago. And it was one of the most fruitful lies in the history of mathematics.

In the same spirit, I think that although it's not actually true, in a pre-calculus context it's a more appropriate statement than any of the actually true assertions it approximates.

Solution 2:

All the answers so far are concerned with applications in advanced mathematics, which is hardly comprehensible for high school students. Therefore, this is an answer that concentrates on applications on the high school level. It is going to be a long but hopefully exciting and thought-provoking journey.

Introduction

Although trigonometric functions seem to be concerned with only triangles, they do go far beyond that, even in elementary mathematics. Trigonometric functions are ubiquitous in mathematics. Due to lack of knowledge, however, trigonometric functions are usually applied in tricks or techniques when tackling problems that are otherwise difficult to solve. These methods exploit the properties of trigonometric functions and extend them to other concepts in mathematics. For instance, if you regard the sides of a triangle as merely positive real numbers, and by using some basic theorems, you can deduce statements of real numbers from their corresponding facts in trigonometry:

The law of cosines tells us that, in a triangle $ABC$, for any permutations of the sides $a, b, c$, $$c^2 = a^2 + b^2 - 2ab \cos C,$$ or equivalently, $$\cos C = \frac {a^2 + b^2 - c^2}{2ab}.$$

By using a well-known result from trigonometry, i.e. if $A, B, C$ are angles of a triangle, then $$\cos A + \cos B + \cos C \le \frac 32,$$ we have $$\frac {a^2 + b^2 - c^2}{2ab} + \frac {b^2 + c^2 - a^2}{2bc} + \frac {c^2 + a^2 - b^2}{2ca} \le \frac 32,$$ for $a, b, c$ sides of a triangle.

Remarks:

The above inequality has various proofs (try to find one yourself!). You may find some of them here. However, the simplest of all these proofs is to use Jensen's inequality, which becomes easy once you know basic differential calculus and the concept of convexity.
If the inequality seems trivial to you, we may go one step further, using something that is occasionally called Ravi's transformation. This trick is especially useful because it transforms sides of a triangle into pure numbers, discarding its geometrical significance. It states that a sufficient and necessary condition for $a, b, c$ to be the sides of a triangle is that there exist positive real numbers $p, q, r$ such that $$a = p + q,\, b = q + r,\, c = r + p.$$ (The proof is omitted here. Incidentally, the numbers $p, q, r$ here have geometrical interpretations. You can try to prove the theorem yourself, using geometry.) Now apply this to our inequality, and we get a statement about positive real numbers (instead of sides of a triangle). Moreover, after some algebraic manipulations the final inequality can seem as nontrivial as you like!

Trigonometric Substitution

The inverse of what we have done above, i.e. substituting algebraic expressions with trigonometric quantities, is a method called trigonometric substitution. In most cases, the problem does not even seem to be related to trigonometry at all; you will find it surprising how trigonometric methods lead to such beautiful proofs! Here is an example to illustrate this idea:

For $0 \lt a, b, c \lt 1$, if $ab + bc + ca = 1$, Prove that $$\frac a{1 - a^2} + \frac b{1 - b^2} + \frac c{1 - c^2} \ge \frac {3 \sqrt 3}2.$$

Solution:

Observe that if $A, B, C$ are angles of a triangle, then $$\tan \frac A2 \tan \frac B2 + \tan \frac B2 \tan \frac C2 + \tan \frac C2 \tan \frac A2 = 1,$$

Let $a = \tan \frac A2, b = \tan \frac B2, c = \tan \frac C2$, where $A, B, C$ are angles of an acute triangle, taking into account the condition that $0 \lt a, b, c \lt 1$. Thus it suffices to prove that $$\frac {\tan \frac A2}{1 - \tan^2 \frac A2} + \frac {\tan \frac B2}{1 - \tan^2 \frac B2} + \frac {\tan \frac C2}{1 - \tan^2 \frac C2} \ge \frac {3 \sqrt 3}2,$$ or $$\tan A + \tan B + \tan C \ge 3\sqrt{3},$$ which is trivial. Q.E.D.

In this solution many results are used without proofs, but all of them are perfectly provable from trigonometric methods. Thus, we solved an algebraic statement using only trigonometric methods. Furthermore, if one were to prove this inequality from purely algebraic methods, then the solution will become tedious and ugly. If you are interested in this method, there is a good book for further reading. It contains a comprehensive chapter on these methods, as well as proofs of the identities and inequalities used here, of course.

Although we have restricted our attention to inequalities presently, the method of trigonometric substitution has much more applications. For example, it is an essential tool in the evaluation of integrals. You will encounter this technique in various situations when studying mathematics.

Parametrization

In the examples above we have exploited some properties of trigonometric functions, most of which are identities or inequalities involving the three angles of a triangle. Nonetheless, we need not restrict our focus to triangles only; we will look at the extended concept of trigonometric functions, i.e. for arbitrary real numbers.

An important property is that these functions have certain bounds. For example, $-1 \le \sin \theta \le 1$, for all $\theta\in\mathbb{R}$, and if $\theta \in [0, \frac {\pi}2)$ then $\tan \theta \gt 0$. We will use this property in the next example:

Consider two real numbers $x, y$ such that $$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1,$$ where $a, b$ are two given positive real numbers. Now, how do we find the maximum and minimum of, say, $x + y$?

Observe that the equation $x^2/a^2 + y^2/b^2 = 1$ just looks like $\sin^2 \theta + \cos^2 \theta = 1$! It is natural to substitute for $x, y$. Let $x^2/a^2 = \sin^2 \theta$ and $y^2/b^2 = \cos^2 \theta$ and we get the equations

\begin{cases} x = a \sin \theta, \\ y = b \cos \theta. \\ \end{cases}

(Note that $\theta$ can be any real number, though in this case you can restrict it to the interval $[0, 2 \pi]$.) So we need to find the extrema of $x + y$, that is, $a \sin \theta + b \cos \theta$, which is easy. The maximum is $\sqrt {a^2 + b^2}$ and the minimum $-\sqrt {a^2 + b^2}$.

Remarks:

The attentive readers will find that this is essentially the same as trigonometric substitution. However, here this method is not considered as merely substitution. It is called parametrization, where $\theta$ plays the role of the parameter. The equations we obtained are called parametric equations.
Surprisingly, not only curves like circles and ellipses but also lines have parametrizations involving trigonometric functions. A standard parametrization for a line that passes through the point $(a, b)$ with slope $\tan \theta$ will look like this: \begin{cases} x = a + t\cos \theta, \\ y = b + t\sin \theta. \\ \end{cases} The advantage of this parametrization is that $|t|$ equals the distance between the point $(x,y)$ and the given point $(a,b)$.

To conclude, we make some final remarks:

Trigonometry is a bridge between geometry and (elementary) algebra: On the one hand, many theorems in geometry can be readily applied to algebraic problems; on the other hand, algebraic theorems can be used to prove geometrical statements.
Trigonometric functions have many other applications, but most of them are too advanced for high school students. For example, we can use them to represent other periodic functions (a subject called Fourier series). Also, once you learn about Taylor series (when learning calculus) you will find that it even has a close connection with complex numbers, where you will see the beautiful formula (called Euler's formula) $$e^{ix}=\cos x+i\sin x.$$ This list may extend forever, as these functions are such fundamental in the edifice of mathematics.

Hope you had fun reading this!

Solution 3:

One momentous theoretical use of trigonometric functions is to solve the Basel problem (i.e. proving that the sum of the reciprocals of the squares equals $\pi^2/6$). See for example: http://math.cmu.edu/~bwsulliv/basel-problem.pdf.

Or better: https://en.wikipedia.org/wiki/Basel_problem#A_rigorous_elementary_proof (as suggested by Mark S. in comments).