삼각함수의 미분

In order for $f$ to be differentiable at $x=0$, it must also be continuous at $x=0$. This forces $$\lim_{x\to 0^-}f(x) = \lim_{x\to 0^+}f(x) =f(0)\qquad\text{or}\qquad \lim_{x\to 0^-}\cos(x) = \lim_{x\to 0^+}(ax+b) =1$$ or $b=1$. In order for $f$ to be differentiable at $x=0$, we need the limit $$\lim_{h\to 0}\frac{f(0+h)-f(0)}{h}$$ to exist. This is the case if and only if the two one–sided limits

$$\begin{aligned} \lim_{h\to 0^-}\frac{f(0+h)-f(0)}{h} &=\lim_{h\to 0^-}\frac{\cos(h)-\cos(0)}{h} \\ \text{and} \\ \lim_{h\to 0^+}\frac{f(0+h)-f(0)}{h} &=\lim_{h\to 0^+}\frac{(ah+b)-\cos(0)}{h} =a\qquad\text{since $b=1$} \\ \text{exist and are equal. Because $\cos(x)$ is differentiable at $x=0$ we have} \\ \lim_{h\to 0^-}\frac{\cos(h)-\cos(0)}{h} &= \frac{\mathrm{d}}{\mathrm{d}x}\cos(x)\bigg|_{x=0} = -\sin(x)\Big|_{x=0}=0 \end{aligned}$$

So, we need $a=0$ and $b=1$.