Trigonometric Identities as Functional Equations

We take some basic trigonometric identities and essentially ask whether trigonometric functions are the only ones that satisfy these identities. In other words, what are all the functions that satisfy a given trigonometric identity? Thus, we take trigonometric identities and use them as functional equation problems.

I suspect that generally these identities are “tight” enough that they’ll constrain the solution uniquely, or at least “relatively uniquely,” so that all the solutions are in fact trigonometric or based on trigonometric functions in some way. If that’s the case, then that identity can in fact be used as a basis for uniquely defining a trigonometric function.

We can even expand this research by looking at collections of trigonometric identities as inspiring functional equation system problems, if the individual equations don’t yield relatively unique solutions (as is the case in the first proposed equation below.)

The Equation $\mathbf{f}\left( \mathbf{x} \right)^{\mathbf{2}}\mathbf{+ f}\left( \frac{\mathbf{\pi}}{\mathbf{2}}\mathbf{- x} \right)^{\mathbf{2}}\mathbf{= 1}$

We suspect that the generic answer to this is $f(x) = \sin(x - a)$ for constant $a$ . But is that true?

In fact, we show this is false. First, substitute $x = \frac{\pi}{4} - y$ , then

$\displaystyle f\left( \frac{\pi}{4} - y \right)^{2} + f\left( y - \frac{\pi}{4} \right)^{2} = 1.$

Defining $g(x) = f\left( x - \frac{\pi}{4} \right)$ , the equation is

$\displaystyle g(x)^{2} + g( - x)^{2} = 1.$

We clearly have $2g(0)^{2} = 1 \rightarrow g(0) = \pm \sqrt{\frac{1}{2}}$ . Otherwise, the solutions are relatively unconstrained. In fact, we can quickly see that the solution must be: for any $h:\mathbb{R}_{+} \rightarrow \mathbb{R}$ with $- 1 \leq h(x) \leq 1$ for all $x$ ,

$\displaystyle g(x) = \left\{ \begin{matrix} \pm h(x),\ \ x > 0 \\ \pm \sqrt{1 - h(x)^{2}},\ \ x < 0 \\ \pm \frac{\sqrt{2}}{2},\ \ x = 0. \\ \end{matrix} \right.\$

(This is clearly necessary, and we can check that this works so it is sufficient.) Thus, our conjectured solution is false. We can even make $h$ continuous or differentiable without it being trigonometric (just make it bounded and scale it to be below 1), thus those conditions don’t guarantee trigonometric solutions either.

The Equation $\mathbf{f}\left( \mathbf{x} \right)^{\mathbf{2}}\mathbf{+ f}\left( \mathbf{a - x} \right)^{\mathbf{2}}\mathbf{=}\mathbf{b}^{\mathbf{2}}$ for Constants $\mathbf{a}$ and $\mathbf{b}$

We can easily reduce this equation to the previous one by scaling: substitute $x = \frac{a}{\frac{\pi}{2}}y$ and define $g:\mathbb{R \rightarrow R}$ by $f = bg$ , then $g\left( \frac{a}{\frac{\pi}{2}}y \right)^{2} + g\left( \frac{a}{\frac{\pi}{2}}\left( \frac{\pi}{2} - y \right) \right)^{2} = 1$ . Let $h:\mathbb{R \rightarrow R}$ be defined by $h(x) = g\left( \frac{a}{\frac{\pi}{2}}x \right)$ , then $h(y)^{2} + h\left( \frac{\pi}{2} - y \right)^{2} = 1$ , which is the previously studied equation. Thus, denoting a solution to that equation as $S$ , all solutions to this are given by $f(x) = bg(x) = bS\left( \frac{\pi}{2a}x \right)$ .

(This equation is not a trigonometric identity, but it is directly inspired from the previously studied equation.)

The Equation $\mathbf{f}\left( \mathbf{a + b} \right)\mathbf{= f}\left( \mathbf{a} \right)\mathbf{f}\left( \frac{\mathbf{\pi}}{\mathbf{2}}\mathbf{- b} \right)\mathbf{+ f}\left( \frac{\mathbf{\pi}}{\mathbf{2}}\mathbf{- a} \right)\mathbf{f}\left( \mathbf{b} \right)$

We suspect that the unique solution could be $f(x) = \sin x$ . (It would be very cool if this were true.) (EDIT: we have that $f(x)=\frac{1}{2}$ is a solution, so we amend this guess to saying that the unique non-constant solution is $f(x)=\sin x$ .)

We now solve this equation. Substitute $a \rightarrow \frac{\pi}{2} - a,b \rightarrow \frac{\pi}{2} - b$ :

$\displaystyle f\left( \pi - (a + b) \right) = f\left( \left( \frac{\pi}{2} - a \right) + \left( \frac{\pi}{2} - b \right) \right) = f\left( \frac{\pi}{2} - a \right)f(b) + f(a)f\left( \frac{\pi}{2} - b \right) = f(a + b).$

Thus, $f(x) = f(\pi - x)$ .

Now, let $b = 0$ :

$\displaystyle f(a) = f(a)f\left( \frac{\pi}{2} \right) + f\left( \frac{\pi}{2} - a \right)f(0)$

Let $a = 0$ :

$\displaystyle f(0) = 2f(0)f\left( \frac{\pi}{2} \right)$

Let $a = \frac{\pi}{2}$ :

$\displaystyle f\left( \frac{\pi}{2} \right) = f\left( \frac{\pi}{2} \right)^{2} + f(0)^{2}$

If $f(0) \neq 0$ then $2f\left( \frac{\pi}{2} \right) = 1 \rightarrow f\left( \frac{\pi}{2} \right) = \frac{1}{2}$ , so $f(0)^{2} = \frac{1}{4} \rightarrow f(0) = \pm \frac{1}{2}$ . Otherwise $f(0) = 0$ and $f\left( \frac{\pi}{2} \right)^{2} = f\left( \frac{\pi}{2} \right) \rightarrow f\left( \frac{\pi}{2} \right) \in \left\{ 0,1 \right\}$ .