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Let $y$ be a value such that $y=\sin(x)$. Then, by first principals of differentiation: $$\frac{\delta y}{\delta x} = \lim_{h\to 0} \frac{\sin(x+h)-\sin(x)}{h}$$ using the difference of angles identity, $\displaystyle \sin(A)-\sin(B)=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$, $$\frac{\delta y}{\delta x} = \lim_{h\to 0}\frac{2\cos\left(\frac{2x+h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}$$ dividing top and bottom by $2$ yields, $$\frac{\delta y}{\delta x}=\lim_{h\to0}\frac{\cos\left(\frac{2x+h}{2}\right)\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$$ from here, given that $\displaystyle \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$, when the limit is taken, since there is no reason to take the limit "piece by piece" the entire expression is reduced to $$\frac{dy}{dx} = \cos\left(\frac{2x}{2}\right) = \cos(x)$$ short question is, is this a strong proof? Or is there another, potentially more simple proof for this scenario?

joshuaheckroodt
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  • It depends on where you got the trigonometric identity and $\lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1$ from. Was that proved without using the derivatives of sine and cosine? – Martin R Oct 20 '17 at 08:09
  • Related (duplicate?): Is the proof of the derivative of $\sin(x)$ circular? – Martin R Oct 20 '17 at 08:12
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    I'm not sure if you're familiar with it, but the proof for $\lim_{\theta \to 0} \frac{\sin\theta}{\theta}$ comes from a geometric inequality, concluded by the squeeze theorem. I.e. no, it doesn't relate at all to the derivates of sine or cosine – joshuaheckroodt Oct 20 '17 at 08:12
  • Related, yes, but not necessarily a duplicate, the question you added uses different steps, I'm interested specifically in this method @MartinR – joshuaheckroodt Oct 20 '17 at 08:15
  • That depends on where you start: With a geometrical definition or with an analytical (as an infinite series). – If you take that trigonometric identity and the limit for granted then your proof is fine. – Martin R Oct 20 '17 at 08:15
  • How does it depend on where I start? If I present a geometric proof for the limit of sin theta divided by theta as theta approaches 0, would I have to stick to a geometric proof throughout the case above as well? @MartinR – joshuaheckroodt Oct 20 '17 at 08:20
  • Related: https://math.stackexchange.com/questions/2319969/is-the-proof-of-lim-theta-to-0-frac-sin-theta-theta-1-in-some-high-s – velut luna Oct 20 '17 at 08:30
  • Thank you @velutluna , thats exactly the proof I was referring to. – joshuaheckroodt Oct 20 '17 at 08:31

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