TITLE: Relationship between $\sin(a+b)$ and derivative product rule?
QUESTION [5 upvotes]: I noticed this interesting correlation between the sine angle addition formula and the derivative product rule.
The sine addition formula is 
$$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$$
The derivative product rule is 
$$(f(x)g(x))'=f'(x)g(x)+g'(x)f(x)$$
As many of you probably know the derivative of $\sin$ is $\cos$ so the sine addition formula could be rewritten as
$$\sin(a+b)=\sin(a)\sin'(b)+\sin(b)\sin'(a)$$
I was wondering if there was any reason for this correlation between the two. I understand that the sine angle addition formula is about taking the sine of two different angles and that the derivative product rule is about multiplying two different functions and finding the derivative so there is very little relation between the two. Perhaps this is just a coincidence?

REPLY [2 votes]: Let's treat that as a functional equation
and see what happens.
Suppose
$f(a+b)
=f(a)f'(b)+f(b)f'(a)
$.
My solution,
not quite complete,
is this:
If $f(0) \ne 0$,
$f(x)
=f(0)e^{x/(2f(0))}
$.
If $f(0) = 0$,
$f(x)
=\dfrac{e^{x\sqrt{r}}-e^{-x\sqrt{r}}}{2\sqrt{r}}
$
where
$r = f'''(0)$.
Note:
I show below that,
in this case,
$f'(0) = 1$
and
$f''(0) = 0$,
but haven't been able to show
that
$f'''(0) = -1$.
Once this is shown,
the solution is complete.
Setting $b=0$,
we get
$f(a)
=f(a)f'(0)+f(0)f'(a)
$,
so
$f(0)f'(a)
=f(a)(1-f'(0))
$.
Setting $a=0$ also,
$f(0) = 2f(0)f'(0)$.
If $f(0) \ne 0$,
$f'(0) = \frac12$.
If $f(0) = 0$,
then,
assuming that
$f$ is not constant zero,
$f'(0) = 1$.
If $f(0) \ne 0$,
then
$f'(a)
=cf(a)
$,
where
$c 
= \frac{1-f'(0)}{f(0)}
= \frac{1}{2f(0)}
$.
From this,
$(\ln(f(x))' = c$
so
$\ln(f(x))
=cx+d
$,
or
$f(x)
=De^{cx}
$.
For this to satisfy the original equation,
since $f'(x) = cDe^{cx}$,
we want
$De^{c(a+b)}
=De^{ca}(cDe^{cb})+De^{cb}(cDe^{ca})
=2cD^2e^{c(a+b)}
$
so we must have
$2cD = 1$
or
$D 
= \frac1{2c}
= \frac1{2(\frac{1}{2f(0)})}
=f(0)
$
(Duh! Obvious from
$f(x) = De^{cx}$).
Therefore,
if $f(0) \ne 0$,
$f(x)
=f(0)e^{x/(2f(0))}
$.
If $f(0) = 0$,
since $f'(0) = 1$,
$f(a+b)
=f(a)f'(b)+f(b)f'(a)
$.
If $b$ is small
and $f'$ and $f''$
are well-behaved,
$f(a+b)
\approx f(a) +bf'(a) + b^2f''(a)/2+b^3f'''(a)/6
$
and
$f(a)f'(b)+f(b)f'(a)
\approx f(a)(f'(0)+bf''(0))+f'(a)(f(0)+bf'(0))
=f(a)+bf''(0)+bf'(a)
$
so $f''(0) = 0$.
Taking an additional term,
$f(a)f'(b)+f(b)f'(a)
\approx f(a)(f'(0)+bf''(0)+b^2f'''(0)/2)+f'(a)(f(0)+bf'(0)+b^2f''(0)/2)
=f(a)(1+b^2f'''(0)/2)+bf'(a)+O(b^3)
$
so
$b^2f''(a)/2
\approx f(a)b^2f'''(0)/2
$
or,
letting $b \to 0$,
$f''(a) = f(a)f'''(0)
$.
Let
$r = f'''(0)$,
so
$f''(a) = rf(a)$
with
$f(0) = 0$
and
$f'(0) = 1$.
The solution to this is
$f(x)
=\dfrac{e^{x\sqrt{r}}-e^{-x\sqrt{r}}}{2\sqrt{r}}
$.
Aha! Looks like $\sin$
if we can show that
$r < 0$
.
However,
I am running out of energy,
so I will stop here.
If someone can finish this,
I will gladly
upvote them.