사인 함수 미분 증명

= Theorem =
 $\frac{d}{dx}\sin x=\cos x$

= Proof 1 =
Let $f(x)=\sin x$
$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
 $=\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$
 $=\lim_{h\to 0}\frac{\sin x\cdot\cos h+\cos x\cdot\sin h-\sin x}{h}$
 $=\lim_{h\to 0}\left(\sin x\cdot\frac{\cos h-1}{h}+\cos x\cdot\frac{\sin h}{h}\right)$
 $=\sin x\cdot\lim_{h\to 0}\frac{\cos h-1}{h}+\cos x\cdot\lim_{h\to 0}\frac{\sin h}{h}$
좌측에 [[VG:로피탈_정리,L_Hopital_s_rule]]를 쓰면
 $=\sin x\cdot 0+\cos x\cdot 1$
 $=\cos x$

i.e.
 $\frac{\sin(x+h)-\sin x}{h}$
 $=\frac{\sin x\cos h+\cos x\sin h-\sin x}{h}$
 $=\sin x\left(\frac{\cos h-1}{h}\right)+\cos x\left(\frac{\sin h}{h}\right)$
$h\to 0$ 일 때의 극한을 구하면
 $=\sin x \cdot 0 + \cos x \cdot 1$
 $=\cos x$

= Proof 2 =
Let $f(x)=\sin x$
 $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
 $=\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$
[[VG:삼각함수,trigonometric_function#s-8.2]] 합·차 → 곱 공식을 쓰면,
 $=\lim_{h\to 0}\left(2\cos\frac{x+h+x}{2}\cdot\sin\frac{x+h-x}{2}\right)/h$
 $=\lim_{h\to 0}\left(2\cos\frac{2x+h}{2}\cdot\sin\frac{h}{2}\right)/h$
 $=\lim_{h\to 0}\frac{2\cos\left(x+\frac{h}{2}\right)\cdot\sin\frac{h}{2}}{h}$
 $=\lim_{h\to 0}\frac{\cos\left(x+\frac{h}{2}\right)\cdot\sin\frac{h}{2}}{\frac{h}{2}}$
 $=\cos(x+0)\cdot 1$
 $=\cos x$
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from [[VG:삼각함수_미분표]]

Up: [[여러가지증명]]