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삼각함수의 미분법 📂함수

삼각함수의 미분법

공식1

삼각함수의 도함수는 다음과 같다.

ddxsin⁡x=cos⁡xddxcsc⁡x=−csc⁡xcot⁡xddxcos⁡x=−sin⁡xddxsec⁡x=sec⁡xtan⁡xddxtan⁡x=sec⁡2xddxcot⁡x=−csc⁡2x \begin{align*} \dfrac{d}{dx} \sin x &= \cos x \qquad & \dfrac{d}{dx} \csc x &= -\csc x \cot x \\[1em] \dfrac{d}{dx} \cos x &= - \sin x \qquad & \dfrac{d}{dx} \sec x &= \sec x \tan x \\[1em] \dfrac{d}{dx} \tan x &= \sec^{2} x \qquad & \dfrac{d}{dx} \cot x &= -\csc^{2} x \end{align*}

증명

삼각함수의 덧셈공식 sin⁡(α±β)=sin⁡αcos⁡β±cos⁡αsin⁡βcos⁡(α∓β)=cos⁡αcos⁡β±sin⁡αsin⁡β \sin\left( \alpha \pm \beta \right) =\sin\alpha \cos\beta \pm\cos\alpha \sin\beta \\ \cos\left( \alpha \mp \beta \right) =\cos\alpha \cos\beta \pm \sin\alpha \sin\beta 사인함수의 극한

lim⁡x→0sin⁡xx=1 \lim \limits_{x \to 0} \dfrac{\sin x}{x} = 1

코사인함수의 극한 lim⁡x→01−cos⁡xx=0 \lim \limits_{x \to 0} \dfrac{1 - \cos x}{x} = 0

(sin⁡x)′=cos⁡x(\sin x)^{\prime} = \cos x

도함수의 정의에 의해,

lim⁡h→0sin⁡(x+h)−sin⁡xh=lim⁡h→0sin⁡xcos⁡h+sin⁡hcos⁡x−sin⁡xh(∵삼각함수의 덧셈공식)=sin⁡xlim⁡h→0cos⁡h−1h+cos⁡xlim⁡h→0sin⁡hh=sin⁡x⋅0+cos⁡x⋅1=cos⁡x \begin{align*} & \lim\limits_{h \to 0} \dfrac{\sin (x + h) - \sin x}{h} \\ &= \lim\limits_{h \to 0} \dfrac{\sin x \cos h + \sin h \cos x - \sin x}{h} &\small (\because \text{삼각함수의 덧셈공식}) \\ &= \sin x \lim\limits_{h \to 0} \dfrac{\cos h - 1}{h} + \cos x \lim\limits_{h \to 0} \dfrac{\sin h}{h} \\ &= \sin x \cdot 0 + \cos x \cdot 1 \\ &= \cos x \end{align*}

■

(cos⁡x)′=−sin⁡x(\cos x)^{\prime} = -\sin x

정의에 의해,

lim⁡h→0cos⁡(x+h)−cos⁡xh=lim⁡h→0cos⁡xcos⁡h−sin⁡hsin⁡x−cos⁡xh(∵삼각함수의 덧셈공식)=cos⁡xlim⁡h→0cos⁡h−1h−sin⁡xlim⁡h→0sin⁡hh=cos⁡x⋅0−sin⁡x⋅1=−sin⁡x \begin{align*} & \lim\limits_{h \to 0} \dfrac{\cos (x + h) - \cos x}{h} \\ &= \lim\limits_{h \to 0} \dfrac{\cos x \cos h - \sin h \sin x - \cos x}{h} &\small (\because \text{삼각함수의 덧셈공식}) \\ &= \cos x \lim\limits_{h \to 0} \dfrac{\cos h - 1}{h} - \sin x \lim\limits_{h \to 0} \dfrac{\sin h}{h} \\ &= \cos x \cdot 0 - \sin x \cdot 1 \\ &= - \sin x \end{align*}

■

(tan⁡x)′=sec⁡2x(\tan x)^{\prime} = \sec^{2} x

몫의 미분법

(fg)′(x)=f′(x)g(x)−f(x)g′(x)g2(x) \left( \frac{f}{g} \right)^{\prime}(x) = \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{g^{2}(x)}

몫의 미분법에 의해,

tan⁡′x=(sin⁡xcos⁡x)′=sin⁡′xcos⁡x−sin⁡xcos⁡′xcos⁡2x=cos⁡xcos⁡x+sin⁡xsin⁡xcos⁡2x=1cos⁡2x=(1cos⁡x)2=sec⁡2x \begin{align*} & \tan^{\prime}x \\ &= \left( \dfrac{\sin x}{\cos x} \right)^{\prime} \\ &= \dfrac{\sin^{\prime} x \cos x - \sin x \cos^{\prime} x}{\cos^{2} x} \\ &= \dfrac{\cos x \cos x + \sin x \sin x}{\cos^{2} x} \\ &= \dfrac{1}{\cos^{2} x} \\ &= \left( \dfrac{1}{\cos x} \right)^{2} \\ &= \sec^{2}{x} \end{align*}

■

(csc⁡x)′=−csc⁡xcot⁡x(\csc x)^{\prime} = -\csc x \cot x

연쇄법칙에 의해,

ddxcsc⁡x=ddx(1sin⁡x)=dd(sin⁡x)(1sin⁡x)d(sin⁡x)dx=−1sin⁡2xcos⁡x=−1sin⁡xcos⁡xsin⁡x=−csc⁡xcot⁡x \begin{align*} \dfrac{d}{dx} \csc x &= \dfrac{d}{dx} \left( \dfrac{1}{\sin x} \right) \\ &= \dfrac{d}{d(\sin x)} \left( \dfrac{1}{\sin x} \right) \dfrac{d(\sin x)}{dx} \\ &= -\dfrac{1}{\sin^{2} x} \cos x \\ &= -\dfrac{1}{\sin x}\dfrac{\cos x}{\sin x} \\ &= -\csc x \cot x \end{align*}

■

(sec⁡x)′=sec⁡xtan⁡x(\sec x)^{\prime} = \sec x \tan x

연쇄법칙에 의해,

ddxsec⁡x=ddx(1cos⁡x)=dd(cos⁡x)(1cos⁡x)d(cos⁡x)dx=−1cos⁡2x(−sin⁡x)=1cos⁡xsin⁡xcos⁡x=sec⁡xtan⁡x \begin{align*} \dfrac{d}{dx} \sec x &= \dfrac{d}{dx} \left( \dfrac{1}{\cos x} \right) \\ &= \dfrac{d}{d(\cos x)} \left( \dfrac{1}{\cos x} \right) \dfrac{d(\cos x)}{dx} \\ &= -\dfrac{1}{\cos^{2} x} (- \sin x) \\ &= \dfrac{1}{\cos x}\dfrac{\sin x}{\cos x} \\ &= \sec x \tan x \end{align*}

■

(cot⁡x)′=−csc⁡2x(\cot x)^{\prime} = -\csc^{2} x

(cot⁡x)′=(cos⁡xsin⁡x)′=(cos⁡x)′sin⁡x−cos⁡x(sin⁡x)′sin⁡2x=−sin⁡2x−cos⁡2xsin⁡2x=−1sin⁡2x=−csc⁡2x \begin{align*} (\cot x)^{\prime} &= \left( \dfrac{\cos x}{\sin x} \right)^{\prime} \\ &= \dfrac{(\cos x)^{\prime} \sin x - \cos x (\sin x)^{\prime}}{\sin^{2} x} \\ &= \dfrac{-\sin^{2} x - \cos^{2} x}{\sin^{2} x} \\ &= -\dfrac{1}{\sin^{2} x} \\ &= -\csc^{2} x \end{align*}

  1. James Stewart, Daniel Clegg, and Saleem Watson, Calculus (early transcendentals, 9E), p191-197 ↩︎