Proof 1 ¶
식
$\displaystyle \frac{\cos(x+h)-\cos x}{h}$
$\displaystyle =\frac{\cos x\cos h-\sin x\sin h-cos x}{h}$
$\displaystyle =\cos x\left(\frac{\cos h-1}{h}\right)-\sin x\left(\frac{\sin h}{h}\right)$
$\displaystyle h\to 0$ 일 때의 극한을 구하면$\displaystyle =\frac{\cos x\cos h-\sin x\sin h-cos x}{h}$
$\displaystyle =\cos x\left(\frac{\cos h-1}{h}\right)-\sin x\left(\frac{\sin h}{h}\right)$
$\displaystyle =\cos x\cdot 0 -\sin x \cdot 1$
$\displaystyle =-\sin x$
i.e.$\displaystyle =-\sin x$
Let $\displaystyle f(x)=\cos x$
$\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
로피탈_정리,L_Hopital_s_rule를 쓰면
$\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\cos x\cdot\cos h-\sin x\cdot\sin h-\cos x}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\cos x(\cos h-1)-\sin x\cdot\sin h}{h}$
$\displaystyle =\lim_{h\to 0}\left(\cos x\frac{\cos h-1}{h}-\sin x\cdot\frac{\sin h}{h}\right)$
좌측에 $\displaystyle =\lim_{h\to 0}\frac{\cos x\cdot\cos h-\sin x\cdot\sin h-\cos x}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\cos x(\cos h-1)-\sin x\cdot\sin h}{h}$
$\displaystyle =\lim_{h\to 0}\left(\cos x\frac{\cos h-1}{h}-\sin x\cdot\frac{\sin h}{h}\right)$
로피탈_정리,L_Hopital_s_rule를 쓰면$\displaystyle =\cos x\cdot 0 - \sin x \cdot 1$
$\displaystyle =-\sin x$
$\displaystyle =-\sin x$
$\displaystyle \frac{d}{dx}(\cos x)$ 는,
$\displaystyle =\lim_{h\to0}\frac{\cos(x+h)-\cos x}{h}$
$\displaystyle =\lim_{h\to0}\frac{\cos x\cdot\cos h-\sin x\cdot\sin h-\cos x}{h}$
$\displaystyle =\lim_{h\to0}\frac{\cos x\cdot\cos h-\cos x-\sin x\cdot\sin h}{h}$
$\displaystyle =\lim_{h\to0}\frac{\cos x\cdot(\cos h-1)}{h}-\lim_{h\to0}\frac{\sin x\cdot\sin h}{h}$
$\displaystyle =\cos x\cdot\lim_{h\to0}\frac{\cos h-1}{h}-\sin x\cdot\lim_{h\to0}\frac{\sin h}{h}$
$\displaystyle =\cos x\cdot 0-\sin x\cdot 1$
$\displaystyle =-\sin x$
$\displaystyle =\lim_{h\to0}\frac{\cos x\cdot\cos h-\sin x\cdot\sin h-\cos x}{h}$
$\displaystyle =\lim_{h\to0}\frac{\cos x\cdot\cos h-\cos x-\sin x\cdot\sin h}{h}$
$\displaystyle =\lim_{h\to0}\frac{\cos x\cdot(\cos h-1)}{h}-\lim_{h\to0}\frac{\sin x\cdot\sin h}{h}$
$\displaystyle =\cos x\cdot\lim_{h\to0}\frac{\cos h-1}{h}-\sin x\cdot\lim_{h\to0}\frac{\sin h}{h}$
$\displaystyle =\cos x\cdot 0-\sin x\cdot 1$
$\displaystyle =-\sin x$
Proof 2 ¶
Let $\displaystyle f(x)=\cos x$
$\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
삼각함수,trigonometric_function#s-8.2 합·차 → 곱 공식을 쓰면,
$\displaystyle f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$
$\displaystyle =\lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h}$
삼각함수,trigonometric_function#s-8.2 합·차 → 곱 공식을 쓰면,$\displaystyle =\lim_{h\to 0}\left(-2\sin\frac{x+h+x}{2}\cdot\sin\frac{x+h-x}{2}\right)/h$
$\displaystyle =\lim_{h\to 0}\left(-2\sin\frac{2x+h}{2}\cdot\sin\frac{h}{2}\right)/h$
$\displaystyle =\lim_{h\to 0}\frac{-\sin\left(x+\frac{h}{2}\right)\cdot\sin\frac{h}{2}}{\frac{h}{2}}$
$\displaystyle =-\sin(x+0)=-\sin x=f'(x)$
$\displaystyle =\lim_{h\to 0}\left(-2\sin\frac{2x+h}{2}\cdot\sin\frac{h}{2}\right)/h$
$\displaystyle =\lim_{h\to 0}\frac{-\sin\left(x+\frac{h}{2}\right)\cdot\sin\frac{h}{2}}{\frac{h}{2}}$
$\displaystyle =-\sin(x+0)=-\sin x=f'(x)$
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