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!

  • 5
    $\begingroup$ I think your left-hand side is better expressed as $(fg)'$ or $\frac{d}{dx}(fg)$. $\endgroup$ – Randall Jun 12 at 17:09
  • $\begingroup$ There is no connection in the way (I think) you are trying to see. $\endgroup$ – Anurag A Jun 12 at 17:09
  • 1
    $\begingroup$ @AnuragA Perfectly possible, just want to be sure :)! $\endgroup$ – PrincepsMaximus Jun 12 at 17:11
  • 4
    $\begingroup$ Recall the basic relation between trig functions and exponential functions, namely $e^{ix} = \cos x + i\sin x,$ and the fact that $e^{\alpha + \beta} = e^{\alpha} \cdot e^{\beta}.$ You might be able to some kind of connection from this. $\endgroup$ – Dave L. Renfro Jun 12 at 17:37
  • 1
    $\begingroup$ The original reasoning also works with $\sin$ replaced by $\sinh$. Therefore, any possible connection should be based on some property that is shared by $\sin$ and $\sinh$. $\endgroup$ – Pedro Jun 12 at 19:50

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.