#math/calculus The derivative formulas for trigonometric functions assume angles are measured in **radians**. They begin with the two limits $\lim_{h\to0}\sin h/h=1$ and $\lim_{h\to0}(\cos h-1)/h=0$. Using the angle-addition identities in the definition of the derivative gives $ \begin{aligned} \frac{d}{d\theta}\sin\theta &=\sin\theta\lim_{h\to0}\frac{\cos h-1}{h} +\cos\theta\lim_{h\to0}\frac{\sin h}{h}=\cos\theta,\\ \frac{d}{d\theta}\cos\theta &=\cos\theta\lim_{h\to0}\frac{\cos h-1}{h} -\sin\theta\lim_{h\to0}\frac{\sin h}{h}=-\sin\theta. \end{aligned} $ Thus $ \boxed{\frac{d}{dx}\sin x=\cos x}, \qquad \boxed{\frac{d}{dx}\cos x=-\sin x}. $ With the chain rule, $\frac{d}{dx}\sin u(x)=\cos(u(x))u'(x)$ and $\frac{d}{dx}\cos u(x)=-\sin(u(x))u'(x)$.