Does the product rule for differentiation has anything to do with $\sin( \alpha + \beta)$? When learning about differentation, I came along the product rule: $$D(f \cdot g) = f \cdot Dg + g \cdot Df$$
I immediately thought of this rule from trigonometry: $$ \sin( \alpha + \beta ) = \sin(\alpha) \cdot \cos(\beta) + \cos(\alpha) \cdot \sin(\beta)$$
Is there any relation between these two rules? How can this similarity be explained? Has it something to do with geometry?
For those who don't see the relation that I see: if $\alpha$ is the 'normal' function and $\beta$ is the derivative, then you 'get': $$\sin \cdot D\cos + \cos \cdot D\sin$$ And thus if we name sine $f$ and cosine $g$, we 'get': $$f \cdot Dg + g \cdot Df$$
Of course, I write 'get' but I know you can't do this like that, or can you? That's exactly my question: is there any relation between those two rules/equations stated the top? Or is this some weird thought in my brain? Thanks in advance!
 A: The closest connection I could find is (inspired by Dave's comment) the use of product rule in proving Euler's formula, and Euler's formula's connection to this trig identity.
Euler's formula, $e^{ix} = \cos x + i\sin x$, can be proved with the following:
Given a function $f(x) = e^{-ix}(\cos x + i\sin x)$, it can be shown that $f(x)$ is a constant for every real $x$ by taking its derivative using product rule...
$$f'(x) = e^{-ix}(i\cos x - \sin x)-ie^{-ix}(\cos x + i \sin x)$$
which equals $0$ for every real $x$. Knowing $f(x)$ is a constant and that $f(0) = 1$, then $f(x) = 1$ for every other real value of $x$. Algebraically this allows the following...
$$1 =  e^{-ix}(\cos x + i\sin x)$$
$$e^{ix} = \cos x + i\sin x$$
which is Euler's formula.
From here, one can prove the angle sum formula for both $\sin$ and $\cos$.
$$e^{ia} = \cos a + i\sin a$$
$$e^{ib} = \cos b + i\sin b$$
$$e^{ia}e^{ib}=(\cos a + i\sin a)(\cos b + i\sin b) = (\cos a \cos b - \sin a \sin b) + i(\cos a \sin b + \sin a \cos b)$$
$$e^{ia}e^{ib} = e^{i(a+b)}$$
However, if we apply $(a+b)$ as an argument directly into Euler's formula we yield this...
$$e^{i(a+b)} = \cos (a+b) + i\sin(a+b)$$
Thefore...
$$\cos (a+b) + i\sin(a+b) = (\cos a \cos b - \sin a \sin b) + i(\cos a \sin b + \sin a \cos b)$$
...and due to the fact that these are both equivalent complex numbers in the form $a+bi$, we know that the $a$ and $b$ components of these 2 numbers must be equivalent, which gives us our angle sum identities.
$$\cos (a+b) = \cos a \cos b - \sin a \sin b$$
$$\sin(a+b) = \cos a \sin b + \sin a \cos b$$
Now in all honesty, this seems like a far fetched connection, but perhaps there is something there. What discourages me from thinking there is a valuable or true connection is the fact that this also yields the $\cos (a+b)$ identity, which resembles product rule a lot less.
Regardless, this is what came to mind when reading your question. Hopefully it provides some fun insight or perhaps inspires someone else to make a fuller connection!
A: One connection is if you set $\alpha=\beta=x$, you get that $\sin(2x)$ is the derivative of $\sin^2(x)$, which is easy to verify.
