Derivative of sin x by First Principles

Method 1

Required Trigonometric Formula

$\displaystyle \sin A-\sin B = 2 \cos \left( \frac{A+B}{2} \right) \sin \left( \frac{A-B}{2} \right)$

\require{ASMsymbols} \displaystyle \begin{align} \frac{d}{dx} \sin x &= \lim_{h \to 0} \frac{\sin (x+h)-\sin x}{h} \\ &= \lim_{h \to 0} \frac{2 \cos \left(x+\frac{h}{2} \right) \sin \frac{h}{2}}{h} &\color{green}{\text{by the given formula above}} \\ &= \lim_{u \to 0} \frac{2 \cos (x+u) \sin u}{2u} &\color{green}{\text{by letting } u = \frac{h}{2}} \\ &= \lim_{u \to 0} \cos (x+u) \times \lim_{u \to 0} \frac{\sin u}{u} \\ &= \cos x \times 1 & \color{green}{\lim_{u \to 0} \frac{\sin u}{u} = 1} \\ &= \cos x \\ \therefore \frac{d}{dx} \sin x &= \cos x \end{align}

Method 2

Required Trigonometric Formula

$\displaystyle \sin (A+B) = \sin A \cos B + \cos A \sin B$

\require{ASMsymbols} \displaystyle \begin{align} \frac{d}{dx} \sin x &= \lim_{h \to 0} \frac{\sin (x+h)-\sin x}{h} \\ &= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h-\sin x}{h} &\color{green}{\text{by the formula given above}} \\ &= \lim_{h \to 0} \frac{\sin x (\cos h-1) + \cos x \sin h}{h} \\ &= \lim_{h \to 0} \frac{\sin x (\cos h-1)}{h} + \lim_{h \to 0} \frac{\cos x \sin h}{h} \\ &= \sin x \lim_{h \to 0} \frac{\cos h-1}{h} + \cos x \lim_{h \to 0} \frac{ \sin h}{h} \\ &= \sin x \times 0 + \cos x \times 1 &\color{green}{\lim_{h \to 0} \frac{\cos h-1}{h} = 0 \text{ and } \lim_{h \to 0} \frac{\sin h}{h} = 1} \\ &= \cos x \\ \therefore \frac{d}{dx} \sin x &= \cos x \end{align}