Step-by-step explanation:
[tex]\text{By First Principle,}[/tex]
[tex]\begin{align}\dfrac{d}{dx}\sin x&=\lim_{h\to0}\frac{\sin(x+h)-\sin x}{h}\\&=\lim_{h\to0}\frac{\sin x\cos h+\sin h\cos x-\sin x}{h}\\&=\lim_{h\to0}\left(\frac{\cos h-1}{h}\cdot\sin x\right)+\lim_{h\to0}\left(\frac{\sin h}{h}\cdot\cos x\right)\\&=\sin x\cdot\left(\lim_{h\to0}\frac{\cos h-1}{h}\right)+\cos x\cdot\lim_{h\to0}\left(\frac{\sin h}{h}\right)\\&=\sin x\cdot 0+\cos x\cdot1\\&=\cos x\end{align}[/tex]
[tex]\color{blue}\text{Note: This proof is in fact circular without the geometric proof of the value of the two limits, but for the sake of this question it is done in this way.}[/tex]
