Proof: d/dx cos x = -sin x – Sam Artigliere's Blog

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

Recall the trigonometric identity for the cosine of two added angles:

$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$ . Therefore,

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

But

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

and

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

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

Which means that

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

This proof was taken from the MIT Calculus I course on MIT OpenCourseWare at the following link:

https://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/1.-differentiation/part-a-definition-and-basic-rules/session-7-derivatives-of-sine-and-cosine/MIT18_01SCF10_Ses7b.pdf