Sine Angle Addition Functional Equation

In this post, we investigate further the sine angle addition functional equation that I introduced in a previous post, Trigonometric Identities as Functional Equations. (See the work done there before reading this post.) Specifically, the question is to find all functions $f$ such that

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

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If $f$ is of the form $f(x) = c\sin x$ , what must $c$ be?

$\displaystyle c\sin(a + b) = c^{2}\sin(a)\sin\left( \frac{\pi}{2} - b \right) + c^{2}\sin\left( \frac{\pi}{2} - a \right)\sin(b)$

$\displaystyle c = c^{2}$

Thus, $c = 0$ or $c = 1$ .

Actually, $f(x) = 0$ works. So let’s assume that $f(x)$ is nonzero, then see all the possible solutions.

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If $f$ is of the form $f(x) = r\sin(sx + t) + u$ , what must $r,s,t,u$ be? (Assuming $f$ is nonzero.)

$\displaystyle r\sin\left( s(a + b) + t \right) + u = \left( r\sin(sa + t) + u \right)\left( r\sin\left( s\left( \frac{\pi}{2} - b \right) + t \right) + u \right) + \left( r\sin(sb + t) + u \right)\left( r\sin\left( s\left( \frac{\pi}{2} - a \right) + t \right) + u \right),$

for any $a,b$ .

First, say $s = 0$ . Then, $f(x) = c$ (constant.)

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Are there any nonzero constant solutions?

$\displaystyle c = c^{2} + c^{2} \rightarrow c = 2c^{2} \rightarrow c = \frac{1}{2}$

Also, we have that $f(x) = \frac{1}{2}$ works, so this is a solution.

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Are there any nonconstant linear solutions, $f(x) = cx + d$ with $c \neq 0$ ?

$\displaystyle c(a + b) + d = (ca + d)\left( c\left( \frac{\pi}{2} - b \right) + d \right) + \left( c\left( \frac{\pi}{2} - a \right) + d \right)(cb + d)$

Let $a = b = 0$ :

$\displaystyle d = 2d\left( \frac{\pi}{2}c + d \right)$

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

$\displaystyle \pi c + d = 2d\left( \frac{\pi}{2}c + d \right)$

Thus, $d = \pi c + d \rightarrow c = 0$ . Therefore, there are no nonconstant linear solutions.

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Let’s say $f(x)$ is nonzero, what can we say about $f$ ?

We have

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

$\displaystyle \rightarrow f(a)\left( f\left( \frac{\pi}{2} \right) - 1 \right) + f\left( \frac{\pi}{2} - a \right)f(0) = 0$

Letting $c = f\left( \frac{\pi}{2} \right) - 1$ and $d = f(0)$ , substituting $a \rightarrow \frac{\pi}{2} - a$ ,

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

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

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

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

$\displaystyle \left( d^{2} - c^{2} \right)f(a) = 0$

Thus, for $f(x)$ to be nonzero, we must have $d^{2} = c^{2}$ . Hence,

$\displaystyle f(0)^{2} = \left( f\left( \frac{\pi}{2} \right) - 1 \right)^{2}.$

But from before we also had that either

$\displaystyle f(0) = \pm \frac{1}{2}\ and\ f\left( \frac{\pi}{2} \right) = \frac{1}{2},$

$\displaystyle f(0) = 0\ and\ f\left( \frac{\pi}{2} \right) \in \left\{ 0,1 \right\}.$

From our new relation, we can eliminate the possibility that $f(0) = 0,f\left( \frac{\pi}{2} \right) = 0$ . However, all the others still work, so we can have

$\displaystyle f(0) = 0\ and\ f\left( \frac{\pi}{2} \right) = 1,$

$\displaystyle f(0) = \pm \frac{1}{2}\ and\ f\left( \frac{\pi}{2} \right) = \frac{1}{2}.$

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From the above, we have that if $f(0) = f\left( \frac{\pi}{2} \right) = 0$ then $f$ must be zero. We still have the possibility $f(0) = f\left( \frac{\pi}{2} \right) = \frac{1}{2}$ . We do know that $f(x) = \frac{1}{2}$ is a solution; is it true that this is the only solution for $f(0) = f\left( \frac{\pi}{2} \right) = \frac{1}{2}$ ? Equivalently, can we show that if $f(0) = f\left( \frac{\pi}{2} \right)$ then $f$ must be constant? Then, to find nonconstant solutions we would only need to consider either ( $f(0) = 0$ and $f\left( \frac{\pi}{2} \right) = 1$ ) or ( $f(0) = - \frac{1}{2}$ and $f\left( \frac{\pi}{2} \right) = \frac{1}{2}$ .)

Thus, our conjecture is: if $f(0) = f\left( \frac{\pi}{2} \right)$ then $f$ is constant. Let’s see how we can show this. We equivalently can have $f(0) = f\left( \frac{\pi}{2} \right) = \frac{1}{2}$ and show that $f(x) = \frac{1}{2}$ .

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Can we identify any other solutions quickly? So far we have $f(x) = 0,f(x) = \frac{1}{2},f(x) = \sin x$ .

Well, we have $\sin 0 = 0,\sin\frac{\pi}{2} = 1$ . We suspect that finding the sinusoidal function so that $f(0) = - \frac{1}{2},f\left( \frac{\pi}{2} \right) = \frac{1}{2}$ may yield another solution.

So say $f(x) = r\sin(sx + t) + u$ , and

$\displaystyle f(0) = r\sin t + u = - \frac{1}{2},$

$\displaystyle f\left( \frac{\pi}{2} \right) = r\sin\left( \frac{\pi}{2}s + t \right) + u = \frac{1}{2}.$

We know that $f(x) = \sin x - \frac{1}{2}$ is a sinusoidal function that works. Is this a solution?

$\displaystyle \sin(a + b) - \frac{1}{2} = \left( \sin a - \frac{1}{2} \right)\left( \sin\left( \frac{\pi}{2} - b \right) - \frac{1}{2} \right) + \left( \sin\left( \frac{\pi}{2} - a \right) - \frac{1}{2} \right)\left( \sin b - \frac{1}{2} \right)$

$\displaystyle = \left( \sin a - \frac{1}{2} \right)\left( \cos b - \frac{1}{2} \right) + \left( \cos a - \frac{1}{2} \right)\left( \sin b - \frac{1}{2} \right)$

$\displaystyle = \sin a\cos b - \frac{1}{2}\cos b - \frac{1}{2}\sin a + \frac{1}{4} + \cos a\sin b - \frac{1}{2}\sin b - \frac{1}{2}\cos a + \frac{1}{4}$

$\displaystyle = \sin(a + b) + \frac{1}{4} - \left( \frac{1}{2}\cos b + \frac{1}{2}\sin a + \frac{1}{2}\sin b + \frac{1}{2}\sin a \right)$

$\displaystyle \rightarrow \frac{1}{4} + \frac{1}{2} = \frac{1}{2}\cos b + \frac{1}{2}\sin a + \frac{1}{2}\sin b + \frac{1}{2}\sin a$

$\displaystyle \rightarrow \frac{3}{2} = \sin a + \cos a + \sin b + \cos b$

which is clearly false. (Just pick $a = b = 0$ .) Thus, this is not a solution.

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Let’s say $f,g$ are two solutions. What can we say about their difference, $h = f - g$ ?

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

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

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

Expand the RHS:

$\displaystyle h(a)h\left( \frac{\pi}{2} - b \right) + h(a)g\left( \frac{\pi}{2} - b \right) + g(a)h\left( \frac{\pi}{2} - b \right) + g(a)g\left( \frac{\pi}{2} - b \right) + h\left( \frac{\pi}{2} - a \right)h(b) + h\left( \frac{\pi}{2} - a \right)g(b) + g\left( \frac{\pi}{2} - a \right)h(b) + g\left( \frac{\pi}{2} - a \right)g(b)$

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

$\displaystyle \rightarrow h(a + b) - \left( h(a)h\left( \frac{\pi}{2} - b \right) + h\left( \frac{\pi}{2} - a \right)h(b) \right) = h(a)g\left( \frac{\pi}{2} - b \right) + g(a)h\left( \frac{\pi}{2} - b \right) + h\left( \frac{\pi}{2} - a \right)g(b) + g\left( \frac{\pi}{2} - a \right)h(b)$

Unfortunately it doesn’t look like there’s a good or easy way to simplify the RHS to derive something about $h$ …

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Now let’s go back to finding all (nonconstant) sinusoidal solutions. This means $s \neq 0$ and also $r \neq 0$ . The hope was that information about the difference between solutions could help us say something about $u$ , but oh well … in fact, this might be easier with a more specific sinusoidal form.

We have

Let $a = b = 0$ :

$\displaystyle r\sin t + u = \left( r\sin t + u \right)\left( r\sin\left( \frac{\pi}{2}s + t \right) + u \right)$

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

$\displaystyle r\sin(\pi s + t) + u = \left( r\sin t + u \right)\left( r\sin\left( \frac{\pi}{2}s + t \right) + u \right)$

Thus,

$\displaystyle r\sin t = r\sin(\pi s + t),$

and since $r \neq 0$ ,

$\displaystyle \sin t = \sin(\pi s + t).$

Also, either $r\sin t + u = 0$ or

$\displaystyle r\sin\left( \frac{\pi}{2}s + t \right) + u = 1.$

Take the first case, $r\sin t + u = 0$ . If $\sin t = 0$ , then $u = 0$ and $\sin(\pi s + t) = 0$ . We have $t = n\pi$ for some $n\mathbb{\in Z}$ , so $\sin\left( \pi(s + n) \right) = 0$ and $s\mathbb{\in Z}$ . Hence, the form is