Proof: d/dx sinx = cosx – Sam Artigliere's Blog

$\frac d{dx}\sin x=\displaystyle\lim_{\triangle x \to 0}\frac{\sin(x+\triangle x)-\sin x}{\triangle x}$

From the basic trigonometry identity for the sine of two added angles:

$\sin(x+\triangle x)=\cos x \sin \triangle x + \sin x \cos \triangle x$ therefore,

$\begin{array}{rcl}\frac{d}{dx}\sin x&=&\displaystyle\lim_{\triangle x \to 0}\frac{\cos x \sin \triangle x + \sin x \cos \triangle x -\sin x}{\triangle x}\\ &=&\displaystyle \lim_{\triangle x \to 0}\left( \frac{\cos x \sin \triangle x}{\triangle x} + \frac{\sinx \cos \triangle x - \sin x}{\triangle x}\right)\\ &=& \displaystyle \lim_{\triangle x \to 0} \cos x \left( \frac{\sin\triangle x}{\triangle x}\right) + \displaystyle \lim_{\triangle x \to 0} \sin x \left(\frac{(\cos \triangle x - 1)}{\triangle x}\right)\\ &=& \displaystyle \lim_{\triangle x \to 0} \cos x \left( \frac{\sin\triangle x}{\triangle x}\right) - \displaystyle \lim_{\triangle x \to 0} \sin x \left(\frac{(1-\cos \triangle x )}{\triangle x}\right)\\ &=& \displaystyle \lim_{\triangle x \to 0} \cos x \left( \frac{\sin\triangle x}{\triangle x}\right) - \displaystyle \lim_{\triangle x \to 0} \sin x \left(\frac{(1-\cos \triangle x )}{\triangle x}\right)\\ &=& \cos x \displaystyle \lim_{\triangle x \to 0} \left( \frac{\sin\triangle x}{\triangle x}\right) - \sin x \displaystyle \lim_{\triangle x \to 0} \left(\frac{(1-\cos \triangle x )}{\triangle x}\right)\end{array}$

However,

$\displaystyle \lim_{\triangle x \to 0} \left( \frac{\sin\triangle x}{\triangle x}\right) = 1$ (see proof)

and

$\displaystyle \lim_{\triangle x \to 0} \left(\frac{(1-\cos \triangle x )}{\triangle x}\right)=0$ (see proof)

Therefore,

$\begin{array}{rcl}\frac{d}{dx}\sin x &=& \cos x (1) - \sin x(0)\\&=&\cos x - 0\end{array}$

That means that

$\frac{d}{dx}\sin x=\cos x$