삼각함수의 미분

(a) By definition, we know that $$f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h}\,.$$ By the sum of two angles identity for the sine function, $\sin(x+h) = \sin(x)\cos(h) + \cos(x)\sin(h)$, and thus $$f'(x) = \lim_{h \to 0} \frac{\sin(x)\cos(h) + \cos(x)\sin(h) - \sin(x)}{h}$$ .. Combining the first and last terms and removing a factor of $\sin(x)$, $$f'(x) = \lim_{h \to 0} \frac{ \sin(x)(\cos(h)-1) + \cos(x)\sin(h) - \sin(x) }{h}$$ .. (b) Continuing our work in (a), we divide each part of the numerator by $h$ and see it follows $$f'(x) = \lim_{h \to 0} \left( \frac{\sin(x) (\cos(h)-1) }{h} + \frac{\cos(x)\sin(h)}{h} \right)\,.$$ Next, we note that $\sin(x)$ does not depend on $h$, and likewise $\cos(x)$ is independent of $h$. It thus makes sense to view those constants as being outside the limit being taken, and also that the limit of a sum is the sum of the limits. Hence we write $$f'(x) = \sin(x) \cdot \lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) + \cos(x) \cdot \lim_{h \to 0} \left( \frac{\sin(h)}{h} \right)\,.$$ (c) Using small positive and negative values of $h$ to estimate the desired limits, we find that it appears $$\lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) = 0$$ and $$\lim_{h \to 0} \left( \frac{\sin(h)}{h} \right) = 1$$ It suffices in each case to consider $h$-values of $h = \pm 0.01, \pm 0.001, \pm 0.0001$. (d) Combining our results in (b) and (c), we have found that $$f'(x) = \sin(x) \cdot \lim_{h \to 0} \left( \frac{\cos(h)-1}{h} \right) + \cos(x) \cdot \lim_{h \to 0} \left( \frac{\sin(h)}{h} \right) = \sin(x) \cdot 0 + \cos(x) \cdot 1 = \cos(x)\,.$$ (e) The argument for showing $\frac{d}{dx}[\cos(x)] = -\sin(x)$ is almost identical. Do note that the sum of two angles identity for $\cos(\alpha + \beta)$ is $\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)$.