Relationship between $\sin(a+b)$ and derivative product rule?

Question

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?

Answer 1

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.

Answer 2

I am not too sure about how much mathematical background you have, so if any of this becomes excessive do leave a comment and I can try my best to adjust it to your needs.

They actually do have a lot to do with each other. There exists a famous theorem called euler's theorem (arguably one of the most beautiful results in math) that

$$ e^{i \pi} + 1 = 0$$

Which is a special case of the general formula

$$ e^{ix} = \cos(x) + i\sin(x) $$

Where $i= \sqrt{-1} $. This formula will show the correspondence you have observed. A proof that it is true is intuitive by looking at the Taylor series of $e^x, \cos(x), \sin(x)$ and substituting/multiplying. But that isn't our focus here.

We know

$$ \sin(x)^2 + \cos(x)^2 = 1$$ $$ e^{ix} = \cos(x) + i\sin(x) $$

From here we can, derive that

$$ e^{ix} = \cos(x) + i \sqrt{1 - \cos(x)^2} $$

Now lets try to solve for $\cos(x)$ . We can move terms to the left side and square to yield

$$ \left( e^{ix} - \cos(x) \right)^2 = i^2 (1 - \cos(x)^2 ) $$

$$ e^{2ix} - 2e^{ix}\cos(x) + \cos(x)^2 = -1(1 - \cos(x)^2) = \cos(x)^2 -1 $$

We cancel out the $\cos(x)^2$ from both sides which are in common to yield

$$ e^{2ix} - 2e^{ix}\cos(x) = -1 \rightarrow \cos(x) = \frac{e^{2ix}+1}{2e^{ix}}$$

Or

$$ \cos(x) = \frac{e^{ix} + e^{-ix}}{2} $$

Then it follows that

$$ \cos(a + b) = \frac{e^{i(a+b)} + e^{-i(a+b)}}{2}$$

$$ \cos(a + b) = \frac{e^{ia}e^{ib} + e^{-ia}e^{-ib}}{2}$$

WARNING: VERY LITTLE RIGOR, HIGH LEVEL OF "FEEL" AND "INTUITION" AHEAD TO SAVE WRITING

Already you see the (function)*(function) form appearing but it doesn't have anything directly to do with product rule. What does happen to be the case though is that if you take the (multiply by -1) and then derivative of this expression we arrive at

$$ \sin(a + b)$$

The produce rule when applied to each (function)*(function) piece involving exponentials, creates 4 individual products, as opposed to the 2 we started with. But it turns out sin and cos themselves have 2 terms each, so when multiplied it, its intuitive that it can be the case they yield a total of 4 terms. Thanks to some very generous algebra, terms cancel out just right, and so you get the sin addition rule from the product formula. And they very mcuh do have quite a bit to do with each other.