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三角関数

三角関数とは

三角関数 (英:trigonometric function) とは、三角法において角の大きさと線分の長さの関係を表す関数のこと。

三角関数の性質

周期性

+ 90° + 180° + 270° + 360°
sin(θ+12π)=+cosθ\displaystyle \sin\left(\theta+\frac{1}{2}\pi\right) = +\cos\theta sin(θ+π)=sinθ\sin(\theta+\pi) = -\sin\theta sin(θ+32π)=cosθ\displaystyle \sin\left(\theta+\frac{3}{2}\pi\right) = -\cos\theta sin(θ+2π)=+sinθ\sin(\theta+2\pi) = +\sin\theta
cos(θ+12π)=sinθ\displaystyle \cos\left(\theta+\frac{1}{2}\pi\right) = -\sin\theta cos(θ+π)=cosθ\cos(\theta+\pi) = -\cos\theta cos(θ+32π)=+sinθ\displaystyle \cos\left(\theta+\frac{3}{2}\pi\right) = +\sin\theta cos(θ+2π)=+cosθ\cos(\theta+2\pi) = +\cos\theta
tan(θ+12π)=cosθsinθ\displaystyle \tan\left(\theta+\frac{1}{2}\pi\right) = -\frac{\cos\theta}{\sin\theta} tan(θ+π)=tanθ\tan(\theta+\pi) = \tan\theta tan(θ+32π)=cosθsinθ\displaystyle \tan\left(\theta+\frac{3}{2}\pi\right) = -\frac{\cos\theta}{\sin\theta} tan(θ+2π)=tanθ\tan(\theta+2\pi) = \tan\theta

periodicity

還元公式

負角:

sin(θ)=sinθcos(θ)=cosθtan(θ)=tanθ \begin{aligned} \sin(-\theta) &= -\sin\theta \cr \cos(-\theta) &= \cos\theta \cr \tan(-\theta) &= -\tan\theta \end{aligned}


余角:

sin(π2θ)=cosθcos(π2θ)=sinθtan(π2θ)=1tanθ \begin{aligned} \sin\left(\frac{\pi}{2}-\theta\right) &= \cos\theta \cr \cos\left(\frac{\pi}{2}-\theta\right) &= \sin\theta \cr \tan\left(\frac{\pi}{2}-\theta\right) &= \frac{1}{\tan\theta} \end{aligned}


補角:

sin(πθ)=sinθcos(πθ)=cosθtan(πθ)=tanθ \begin{aligned} \sin(\pi-\theta) &= \sin\theta \cr \cos(\pi-\theta) &= -\cos\theta \cr \tan(\pi-\theta) &= -\tan\theta \end{aligned}

変換公式

sinθ\sin\theta cosθ\cos\theta tanθ\tan\theta
sinθ=\sin\theta= - ±1cos2θ\pm\sqrt{1-\cos^2\theta} ±tanθ1+tan2θ\displaystyle \pm\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}
cosθ=\cos\theta= ±1sin2θ\pm\sqrt{1-\sin^2\theta} - ±11+tan2θ\displaystyle \pm\frac{1}{\sqrt{1+\tan^2\theta}}
tanθ=\tan\theta= ±sinθ1sin2θ\displaystyle \pm\frac{\sin\theta}{\sqrt{1-\sin^2\theta}} ±1cos2θcosθ\displaystyle \pm\frac{\sqrt{1-\cos^2\theta}}{\cos\theta} -


各変換公式の導出:

sinθ=±1cos2θ\sin\theta=\pm\sqrt{1-\cos^2\theta}

cos2θ+sin2θ=1sin2θ=1cos2θsinθ=±1cos2θ \begin{aligned} \cos^2\theta + \sin^2\theta &= 1 \cr \sin^2\theta &= 1 - \cos^2\theta \cr \cr \therefore \sin\theta &= \pm\sqrt{1-\cos^2\theta} \end{aligned}

sinθ=±tanθ1+tan2θ\displaystyle \sin\theta = \pm\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}

cos2θ+sin2θ=1sin2θtan2θ+sin2θ=1sin2θ=tan2θ1+tan2θsinθ=±tanθ1+tan2θ \begin{aligned} \cos^2\theta + \sin^2\theta &= 1 \cr \frac{\sin^2\theta}{\tan^2\theta} + \sin^2\theta &= 1 \cr \sin^2\theta &= \frac{\tan^2\theta}{1+\tan^2\theta} \cr \cr \therefore \sin\theta &= \pm\frac{\tan\theta}{\sqrt{1+\tan^2\theta}} \end{aligned}

cosθ=±1sin2θ\cos\theta = \pm\sqrt{1-\sin^2\theta}

cos2θ+sin2θ=1cos2θ=1sin2θcosθ=±1sin2θ \begin{aligned} \cos^2\theta + \sin^2\theta &= 1 \cr \cos^2\theta &= 1-\sin^2\theta \cr \cr \therefore \cos\theta &= \pm\sqrt{1-\sin^2\theta} \end{aligned}

cosθ=±11+tan2θ\displaystyle \cos\theta = \pm\frac{1}{\sqrt{1+\tan^2\theta}}

cos2θ+sin2θ=1cos2θ+cos2θtan2θ=1cos2θ=11+tan2θcosθ=±11+tan2θ \begin{aligned} \cos^2\theta + \sin^2\theta &= 1 \cr \cos^2\theta + \cos^2\theta\tan^2\theta &= 1 \cr \cos^2\theta &= \frac{1}{1+\tan^2\theta} \cr \cr \therefore \cos\theta &= \pm\frac{1}{\sqrt{1+\tan^2\theta}} \end{aligned}

tanθ=±sinθ1sin2θ\displaystyle \tan\theta = \pm\frac{\sin\theta}{\sqrt{1-\sin^2\theta}}

cos2θ+sin2θ=1sin2θtan2θ+sin2θ=1tan2θ=sin2θ1sin2θtanθ=±sinθ1sin2θ \begin{aligned} \cos^2\theta + \sin^2\theta &= 1 \cr \frac{\sin^2\theta}{\tan^2\theta} + \sin^2\theta &= 1 \cr \tan^2\theta &= \frac{\sin^2\theta}{1-\sin^2\theta} \cr \cr \therefore \tan\theta &= \pm\frac{\sin\theta}{\sqrt{1-\sin^2\theta}} \end{aligned}

tanθ=±1cos2θcosθ\displaystyle \tan\theta = \pm\frac{\sqrt{1-\cos^2\theta}}{\cos\theta}

cos2θ+sin2θ=1cos2θ+cos2θtan2θ=1tan2θ=1cos2θcos2θtanθ=±1cos2θcosθ \begin{aligned} \cos^2\theta + \sin^2\theta &= 1 \cr \cos^2\theta + \cos^2\theta\tan^2\theta &= 1 \cr \tan^2\theta &= \frac{1-\cos^2\theta}{\cos^2\theta} \cr \cr \therefore \tan\theta &= \pm\frac{\sqrt{1-\cos^2\theta}}{\cos\theta} \end{aligned}

三角関数の加法定理

余弦関数の加法定理:cos(α±β)=cosαcosβsinαsinβ正弦関数の加法定理:sin(α±β)=sinαcosβ±cosαsinβ正接関数の加法定理:tan(α±β)=tanα±tanβ1tanαtanβ \begin{aligned} \text{余弦関数の加法定理} &: \cos(\alpha\pm\beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta \vphantom{\int} \\ \text{正弦関数の加法定理} &: \sin(\alpha\pm\beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta \vphantom{\int} \\ \text{正接関数の加法定理} &: \tan(\alpha\pm\beta) = \frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta} \\ \end{aligned}

addition_formulas


余弦の加法定理の証明:

余弦定理により、

AB2=12+12211cos(αβ)=22cos(αβ) \begin{aligned} {AB}^2 &= 1^2 + 1^2 - 2\cdot 1\cdot 1\cdot\cos(\alpha-\beta) \cr &= 2 - 2\cos(\alpha-\beta) \end{aligned}

ピタゴラスの定理により、

AB2=(cosαcosβ)2+(sinαsinβ)2=cos2α+cos2β2cosαcosβ+sin2α+sin2β2sinαsinβ=(cos2α+sin2α)+(cos2β+sin2β)2cosαcosβ2sinαsinβ=22cosαcosβ2sinαsinβ \begin{aligned} AB^2 &= (\cos\alpha-\cos\beta)^2 + (\sin\alpha-\sin\beta)^2 \cr &= \cos^2\alpha+\cos^2\beta - 2\cos\alpha\cos\beta + \sin^2\alpha + \sin^2\beta - 2\sin\alpha\sin\beta \cr &= (\cos^2\alpha+\sin^2\alpha) + (\cos^2\beta+\sin^2\beta) - 2\cos\alpha\cos\beta - 2\sin\alpha\sin\beta \cr &= 2 - 2\cos\alpha\cos\beta - 2\sin\alpha\sin\beta \end{aligned}

AB2AB^2 を共に求めていることから、

22cos(αβ)=22cosαcosβ2sinαsinβ 2 - 2\cos(\alpha-\beta) = 2 - 2\cos\alpha\cos\beta - 2\sin\alpha\sin\beta

cos(αβ)=cosαcosβ+sinαsinβ(1) \cos(\alpha-\beta) = \cos\alpha\cos\beta + \sin\alpha\sin\beta \tag{1}

(1)(1)β\betaβ-\beta^\prime とすると、

cos(α(β))=cosαcos(β)+sinαsin(β) \begin{aligned} \cos(\alpha-(-\beta^\prime)) &= \cos\alpha\cos(-\beta^\prime)+ \sin\alpha\sin(-\beta^\prime) \end{aligned}

cos(α+β)=cosαcosβsinαsinβ(2) \cos(\alpha+\beta^\prime) = \cos\alpha\cos\beta^\prime - \sin\alpha\sin\beta^\prime \tag{2}

(1),(2)(1),(2) により、

cos(α±β)=cosαcosβsinαsinβ \therefore \cos(\alpha\pm\beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta


正弦の加法定理の証明:

余弦の加法定理の β\betaβ+12π\displaystyle \beta^\prime+\frac{1}{2}\pi とすると、

cos[α(β+12π)]=cosαcos(β+12π)+sinαsin(β+12π)cos[(αβ)12π]=cosαcos(β+12π)+sinαsin(β+12π)sin(αβ)=cosα(sinβ)+sinαcosβ \begin{aligned} \cos\left[\alpha-\left(\beta^\prime+\frac{1}{2}\pi\right)\right] &= \cos\alpha\cos\left(\beta^\prime+\frac{1}{2}\pi\right) + \sin\alpha\sin\left(\beta^\prime+\frac{1}{2}\pi\right) \cr \cos\left[(\alpha-\beta^\prime)-\frac{1}{2}\pi\right] &= \cos\alpha\cos\left(\beta^\prime+\frac{1}{2}\pi\right) + \sin\alpha\sin\left(\beta^\prime+\frac{1}{2}\pi\right) \cr \sin(\alpha-\beta^\prime) &= \cos\alpha(-\sin\beta^\prime) + \sin\alpha\cos\beta^\prime \end{aligned}

sin(αβ)=sinαcosβcosαsinβ(1) \sin(\alpha-\beta^\prime) = \sin\alpha\cos\beta^\prime - \cos\alpha\sin\beta^\prime \tag{1}

(1)(1)β\beta^\primeβ-\beta^{\prime\prime} とすると、 sin(α(β))=sinαcos(β)cosαsin(β) \sin(\alpha-(-\beta^{\prime\prime})) = \sin\alpha\cos(-\beta^{\prime\prime}) - \cos\alpha\sin(-\beta^{\prime\prime})

sin(α+β)=sinαcos(β)+cosαsinβ(2) \sin(\alpha+\beta^{\prime\prime}) = \sin\alpha\cos(\beta^{\prime\prime}) + \cos\alpha\sin\beta^{\prime\prime} \tag{2}

(1),(2)(1),(2) により、

sin(α±β)=sinαcosβ±cosαsinβ \therefore \sin(\alpha\pm\beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta


正接の加法定理の証明:

余弦の加法定理と正弦の加法定理により、

tan(α+β)=sin(α+β)cos(α+β)=sinαcosβ+cosαsinβcosαcosβsinαsinβ=sinα/cosα+sinβ/cosβ1sinαsinβ/cosαcosβ \begin{aligned} \tan(\alpha+\beta) &= \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} \cr &= \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta} \cr &= \frac{\sin\alpha/\cos\alpha+\sin\beta/\cos\beta}{1-\sin\alpha\sin\beta/\cos\alpha\cos\beta} \end{aligned}

tan(α+β)=tanα+tanβ1tanαtanβ(1) \tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \tag{1}

(1)(1)β\betaβ-\beta^\prime とすると、

tan[α+(β)]=tanα+tan(β)1tanαtan(β) \tan[\alpha+(-\beta^\prime)] = \frac{\tan\alpha+\tan(-\beta^\prime)}{1-\tan\alpha\tan(-\beta^\prime)}

tan(αβ)=tanαtanβ1+tanαtanβ(2) \tan(\alpha-\beta^\prime) = \frac{\tan\alpha-\tan\beta^\prime}{1+\tan\alpha\tan\beta^\prime} \tag{2}

(1),(2)(1),(2) により、

tan(α±β)=tanα±tanβ1tanαtanβ \therefore \tan(\alpha\pm\beta) = \frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}

三角関数の積和の公式

余弦関数と正弦関数の積:cosαsinβ=12(sin(α+β)sin(αβ)):sinαcosβ=12(sin(α+β)+sin(αβ))余弦関数同士の積:cosαcosβ=12(cos(α+β)+cos(αβ))正弦関数同士の積:sinαsinβ=12(cos(α+β)cos(αβ)) \begin{aligned} \text{余弦関数と正弦関数の積} &: \cos\alpha\sin\beta = \frac{1}{2}\left(\sin(\alpha+\beta)-\sin(\alpha-\beta)\right) \\ &: \sin\alpha\cos\beta = \frac{1}{2}\left(\sin(\alpha+\beta)+\sin(\alpha-\beta)\right) \\ \text{余弦関数同士の積} &: \cos\alpha\cos\beta = \frac{1}{2}\left(\cos(\alpha+\beta)+\cos(\alpha-\beta)\right) \\ \text{正弦関数同士の積} &: \sin\alpha\sin\beta = -\frac{1}{2}\left(\cos(\alpha+\beta)-\cos(\alpha-\beta)\right) \\ \end{aligned}


余弦関数と正弦関数の積:

三角関数の加法定理より、

sin(α±β)=sinαcosβ±cosαsinβsin(α+β)+sin(αβ)=(sinαcosβ+cosαsinβ)+(sinαcosβcosαsinβ)=2sinαcosβsin(α+β)sin(αβ)=(sinαcosβ+cosαsinβ)(sinαcosβcosαsinβ)=2cosαsinβsinαcosβ=12(sin(α+β)+sin(αβ))cosαsinβ=12(sin(α+β)sin(αβ)) \begin{aligned} \sin(\alpha\pm\beta) &= \sin\alpha\cos\beta \pm \cos\alpha\sin\beta \cr \sin(\alpha+\beta) + \sin(\alpha-\beta) &= (\sin\alpha\cos\beta + \cos\alpha\sin\beta) + (\sin\alpha\cos\beta - \cos\alpha\sin\beta) \cr &= 2\sin\alpha\cos\beta \cr \sin(\alpha+\beta) - \sin(\alpha-\beta) &= (\sin\alpha\cos\beta + \cos\alpha\sin\beta) - (\sin\alpha\cos\beta - \cos\alpha\sin\beta) \cr &= 2\cos\alpha\sin\beta \cr \cr \therefore \frac{}{}\sin\alpha\cos\beta &= \frac{1}{2}\left(\sin(\alpha+\beta)+\sin(\alpha-\beta)\right) \cr \cos\alpha\sin\beta &= \frac{1}{2}\left(\sin(\alpha+\beta)-\sin(\alpha-\beta)\right) \end{aligned}


余弦関数同士の積:

三角関数の加法定理より、

cos(α±β)=cosαcosβsinαsinβcos(α+β)+cos(αβ)=(cosαcosβsinαsinβ)+(cosαcosβ+sinαsinβ)=2cosαcosβcosαcosβ=12(cos(α+β)+cos(αβ)) \begin{aligned} \cos(\alpha\pm\beta) &= \cos\alpha\cos\beta \mp \sin\alpha\sin\beta \cr \cos(\alpha+\beta) + \cos(\alpha-\beta) &= (\cos\alpha\cos\beta - \sin\alpha\sin\beta ) + (\cos\alpha\cos\beta + \sin\alpha\sin\beta ) \cr &= 2\cos\alpha\cos\beta \cr \cr \therefore \cos\alpha\cos\beta &= \frac{1}{2}\left(\cos(\alpha+\beta)+\cos(\alpha-\beta)\right) \end{aligned}


正弦関数同士の積:

三角関数の加法定理より、

cos(α±β)=cosαcosβsinαsinβcos(α+β)cos(αβ)=(cosαcosβsinαsinβ)(cosαcosβ+sinαsinβ)=2sinαsinβsinαsinβ=12(cos(α+β)cos(αβ)) \begin{aligned} \cos(\alpha\pm\beta) &= \cos\alpha\cos\beta \mp \sin\alpha\sin\beta \cr \cos(\alpha+\beta) - \cos(\alpha-\beta) &= (\cos\alpha\cos\beta - \sin\alpha\sin\beta ) - (\cos\alpha\cos\beta + \sin\alpha\sin\beta ) \cr &= -2\sin\alpha\sin\beta \cr \cr \therefore \sin\alpha\sin\beta &= -\frac{1}{2}\left(\cos(\alpha+\beta)-\cos(\alpha-\beta)\right) \end{aligned}

三角関数の和積の公式

余弦関数同士の和:cosx+cosy=2cosx+y2cosxy2余弦関数同士の差:cosxcosy=2sinx+y2sinxy2正弦関数同士の和:sinx+siny=2sinx+y2cosxy2正弦関数同士の差:sinxsiny=2cosx+y2sinxy2 \begin{aligned} \text{余弦関数同士の和} &: \cos x +\cos y = 2\cos\frac{x+y}{2}\cos\frac{x-y}{2} \\ \text{余弦関数同士の差} &: \cos x-\cos y = -2\sin\frac{x+y}{2}\sin\frac{x-y}{2} \\ \text{正弦関数同士の和} &: \sin x + \sin y = 2\sin\frac{x+y}{2}\cos\frac{x-y}{2} \\ \text{正弦関数同士の差} &: \sin x - \sin y = 2\cos\frac{x+y}{2}\sin\frac{x-y}{2} \\ \end{aligned}


余弦関数同士の和の導出:

積和の公式から、

cosαcosβ=12(cos(α+β)+cos(αβ))(1) \cos\alpha\cos\beta = \frac{1}{2}\left(\cos(\alpha+\beta)+\cos(\alpha-\beta)\right) \tag{1}

x=α+β, y=αβx=\alpha+\beta, ~y=\alpha-\beta とすると、

x+y=(α+β)+(αβ)α=x+y2xy=(α+β)(αβ)β=xy2(2) \begin{aligned} x + y &= (\alpha+\beta) + (\alpha-\beta) \cr \alpha &= \frac{x+y}{2} \cr x - y &= (\alpha+\beta) - (\alpha-\beta) \cr \beta &= \frac{x-y}{2} \end{aligned} \tag{2}

(2)(2)(1)(1) に代入すると、

cosx+y2cosxy2=12(cosx+cosy)cosx+cosy=2cosx+y2cosxy2 \begin{aligned} \cos\frac{x+y}{2}\cos\frac{x-y}{2} &= \frac{1}{2}(\cos x+\cos y) \cr \cr \therefore \cos x+\cos y &= 2\cos\frac{x+y}{2}\cos\frac{x-y}{2} \end{aligned}


余弦関数同士の差の導出:

積和の公式から、

sinαsinβ=12(cos(α+β)cos(αβ))(1) \sin\alpha\sin\beta = -\frac{1}{2}\left(\cos(\alpha+\beta)-\cos(\alpha-\beta)\right) \tag{1}

x=α+β, y=αβx=\alpha+\beta, ~y=\alpha-\beta とすると、

x+y=(α+β)+(αβ)α=x+y2xy=(α+β)(αβ)β=xy2(2) \begin{aligned} x + y &= (\alpha+\beta) + (\alpha-\beta) \cr \alpha &= \frac{x+y}{2} \cr x - y &= (\alpha+\beta) - (\alpha-\beta) \cr \beta &= \frac{x-y}{2} \end{aligned} \tag{2}

(2)(2)(1)(1) に代入すると、

sinx+y2sinxy2=12(cosxcosy)cosxcosy=2sinx+y2sinxy2 \begin{aligned} \sin\frac{x+y}{2}\sin\frac{x-y}{2} &= -\frac{1}{2}(\cos x-\cos y) \cr \cr \therefore \cos x-\cos y &= -2\sin\frac{x+y}{2}\sin\frac{x-y}{2} \end{aligned}


正弦関数同士の和の導出:

積和の公式から、

sinαcosβ=12(sin(α+β)+sin(αβ))(1) \sin\alpha\cos\beta = \frac{1}{2}\left(\sin(\alpha+\beta)+\sin(\alpha-\beta)\right) \tag{1}

x=α+β, y=αβx=\alpha+\beta, ~y=\alpha-\beta とすると、

x+y=(α+β)+(αβ)α=x+y2xy=(α+β)(αβ)β=xy2(2) \begin{aligned} x + y &= (\alpha+\beta) + (\alpha-\beta) \cr \alpha &= \frac{x+y}{2} \cr x - y &= (\alpha+\beta) - (\alpha-\beta) \cr \beta &= \frac{x-y}{2} \end{aligned} \tag{2}

(2)(2)(1)(1) に代入すると、

sinx+y2cosxy2=12(sinx+siny)sinx+siny=2sinx+y2cosxy2 \begin{aligned} \sin\frac{x+y}{2}\cos\frac{x-y}{2} &= \frac{1}{2}\left(\sin x +\sin y\right) \cr \cr \therefore \sin x + \sin y &= 2\sin\frac{x+y}{2}\cos\frac{x-y}{2} \end{aligned}


正弦関数同士の差の導出:

積和の公式から、

cosαsinβ=12(sin(α+β)sin(αβ))(1) \cos\alpha\sin\beta = \frac{1}{2}\left(\sin(\alpha+\beta)-\sin(\alpha-\beta)\right) \tag{1}

x=α+β, y=αβx=\alpha+\beta, ~y=\alpha-\beta とすると、

x+y=(α+β)+(αβ)α=x+y2xy=(α+β)(αβ)β=xy2(2) \begin{aligned} x + y &= (\alpha+\beta) + (\alpha-\beta) \cr \alpha &= \frac{x+y}{2} \cr x - y &= (\alpha+\beta) - (\alpha-\beta) \cr \beta &= \frac{x-y}{2} \end{aligned} \tag{2}

(2)(2)(1)(1) に代入すると、

cosx+y2sinxy2=12(sinxsiny)sinxsiny=2cosx+y2sinxy2 \begin{aligned} \cos\frac{x+y}{2}\sin\frac{x-y}{2} &= \frac{1}{2}\left(\sin x -\sin y\right) \cr \cr \therefore \sin x - \sin y &= 2\cos\frac{x+y}{2}\sin\frac{x-y}{2} \end{aligned}

三角関数の微分

余弦関数の微分:(cosx)=sinx正弦関数の微分:(sinx)=cosx正接関数の微分:(tanx)=1cos2x \begin{aligned} \text{余弦関数の微分} &: (\cos x)^\prime = -\sin x \vphantom{\int} \\ \text{正弦関数の微分} &: (\sin x)^\prime = \cos x \vphantom{\int} \\ \text{正接関数の微分} &: (\tan x)^\prime = \frac{1}{\cos^2 x} \end{aligned}


余弦関数の微分の導出: (cosx)=limh0cos(x+h)cosxh=limh0cosxcoshsinxsinhcosxh=limh0[cosx(cosh1)hsinxsinhh]=sinx(cosx)=sinx \begin{aligned} (\cos x)^\prime &= \lim_{h\to 0}\frac{\cos(x+h)-\cos x}{h} \\ &= \lim_{h\to 0}\frac{\cos x\cos h - \sin x\sin h - \cos x}{h} \\ &= \lim_{h\to 0}\left[\frac{\cos x(\cos h - 1)}{h} - \sin x\frac{\sin h}{h}\right] \\ &= -\sin x \vphantom{\int} \\ \\ \therefore (\cos x)^\prime &= -\sin x \end{aligned}


正弦関数の微分の導出: (sinx)=limh0sin(x+h)sinxh=limh0sinxcosh+cosxsinhsinxh=limh0[sinx(cosh1)h+cosxsinhh]=cosx(sinx)=cosx \begin{aligned} (\sin x)^\prime &= \lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h} \\ &= \lim_{h\to 0}\frac{\sin x\cos h + \cos x\sin h - \sin x}{h} \\ &= \lim_{h\to 0}\left[\frac{\sin x (\cos h-1)}{h} + \cos x\frac{\sin h}{h}\right] \\ &= \cos x \vphantom{\int} \\ \cr \therefore (\sin x)^\prime &= \cos x \end{aligned}


正接関数の微分の導出: (tanx)=(sinxcosx)=(sinx)cosxsinx(cosx)cos2x商の微分法則=cos2x+sin2xcos2x(sinx)=cosx, (cosx)=sinx=1cos2x(tanx)=1cos2x \begin{aligned} (\tan x)^\prime &= \left(\frac{\sin x}{\cos x}\right)^\prime \cr &= \frac{(\sin x)^\prime\cos x - \sin x(\cos x)^\prime}{\cos^2 x} \quad \because \text{商の微分法則} \cr &= \frac{\cos^2x + \sin^2 x}{\cos^2 x} \because (\sin x)^\prime = \cos x, ~(\cos x)^\prime = -\sin x \cr &= \frac{1}{\cos^2 x} \cr \cr \therefore (\tan x)^\prime &= \frac{1}{\cos^2 x} \end{aligned}

三角関数の積分

余弦関数の不定積分:cosx dx=sinx+C正弦関数の不定積分:sinx dx=cosx+C正接関数の不定積分:tanx dx=lncosx+C \begin{aligned} \text{余弦関数の不定積分} &: \int{\cos x~dx} = \sin x + C \\ \text{正弦関数の不定積分} &: \int{\sin x}~dx = -\cos x + C \\ \text{正接関数の不定積分} &: \int{\tan x}~dx = -\ln|\cos x| + C \\ \end{aligned}


余弦関数の不定積分の導出:

cosx dx=(sinx) dx=sinx+Ccosx dx=sinx+C \begin{aligned} \int{\cos x}~dx &= \int(\sin x)^\prime~dx \\ &= \sin x + C \\ \\ \therefore \int{\cos x}~dx &= \sin x + C \end{aligned}


正弦関数の不定積分の導出:

sinx dx=(cosx) dx=cosx+Csinx dx=cosx+C \begin{aligned} \int{\sin x}~dx &= \int(-\cos x)^\prime~dx\\ &= -\cos x + C \\ \\ \therefore \int{\sin x}~dx &= -\cos x + C \end{aligned}


正接関数の不定積分の導出:

tanx dx=sinxcosx dx=(cosx)cosx dx=lncosx+Ctanx dx=lncosx+C \begin{aligned} \int{\tan x}~dx &= \int{\frac{\sin x}{\cos x}~dx} \cr &= \int{-\frac{(\cos x)^\prime}{\cos x}}~dx \cr &= -\ln|\cos x| + C \cr \cr \therefore \int{\tan x}~dx &= -\ln|\cos x| + C \end{aligned}

三角関数の直交性

余弦関数同士の積の積分:02πcos(mt)cos(nt) dt=πδm,n(m,nN0)正弦関数同士の積の積分:02πsin(mt)sin(nt) dt=πδm,n(m,nN0)余弦関数と正弦関数の積の積分:02πcos(mt)sin(nt) dt=0(m,nN0) \begin{aligned}\text{余弦関数同士の積の積分}&: \int_0^{2\pi}\cos(mt)\cos(nt)~dt = \pi\delta_{m,n} \quad (m,n\in\N_{\ge 0}) \\\text{正弦関数同士の積の積分}&: \int_0^{2\pi}\sin(mt)\sin(nt)~dt = \pi\delta_{m,n} \quad (m,n\in\N_{\ge 0}) \\\text{余弦関数と正弦関数の積の積分}&: \int_0^{2\pi}\cos(mt)\sin(nt)~dt = 0 \quad (m,n\in\N_{\ge 0})\end{aligned}


余弦関数同士の積の積分:

02πcos(mt)cos(nt) dt=1202πcos((m+n)t)+cos((mn)t) dt三角関数の積和公式={1202πcos((m+n)t)+cos((mn)t) dtif mn1202πcos(2mt)+1 dtif m=n={12[sin((m+n)t)m+n+sin((mn)t)mn]02πif mn12[sin(2mt)2m+t]02πif m=n={0if mnπif m=n02πcos(mt)cos(nt) dt=πδm,n(m,nN0) \begin{aligned}\int_0^{2\pi}\cos(mt)\cos(nt)~dt&= \frac{1}{2}\int_0^{2\pi}\cos\big((m+n)t\big)+\cos\big((m-n)t\big)~dt \quad \because \text{三角関数の積和公式} \\&= \begin{cases}\displaystyle \frac{1}{2}\int_0^{2\pi}\cos\big((m+n)t\big)+\cos\big((m-n)t\big)~dt & \text{if }m\ne n \\\displaystyle \frac{1}{2}\int_0^{2\pi}\cos(2mt)+1~dt & \text{if }m=n \\\end{cases} \\&= \begin{cases}\displaystyle \frac{1}{2}\left[\frac{\sin\big((m+n)t\big)}{m+n}+\frac{\sin\big((m-n)t\big)}{m-n}\right]_0^{2\pi} & \text{if }m\ne n \\\displaystyle \frac{1}{2}\left[\frac{\sin\big(2mt\big)}{2m}+t\right]_0^{2\pi} & \text{if }m=n \\\end{cases} \\&= \begin{cases}0 & \text{if }m\ne n \vphantom{\displaystyle\int} \\\pi & \text{if }m=n \vphantom{\displaystyle\int} \\\end{cases} \\\\\therefore \int_0^{2\pi}\cos(mt)\cos(nt)~dt&= \pi\delta_{m,n} \quad (m,n\in\N_{\ge 0})\end{aligned}


正弦関数同士の積の積分:

02πsin(mt)sin(nt) dt=1202πcos((m+n)t)cos((mn)t) dt三角関数の積和公式={1202πcos((m+n)t)cos((mn)t) dtif mn1202πcos(2mt)1 dtif m=n={12[sin((m+n)t)m+nsin((mn)t)mn]02πif mn12[sin(2mt)2mt]02πif m=n={0if mnπif m=n02πsin(mt)sin(nt) dt=πδm,n(m,nN0) \begin{aligned}\int_0^{2\pi}\sin(mt)\sin(nt)~dt&= -\frac{1}{2}\int_0^{2\pi}\cos\big((m+n)t\big)-\cos\big((m-n)t\big)~dt \quad \because \text{三角関数の積和公式} \\&= \begin{cases}\displaystyle -\frac{1}{2}\int_0^{2\pi}\cos\big((m+n)t\big)-\cos\big((m-n)t\big)~dt & \text{if }m\ne n \\\displaystyle -\frac{1}{2}\int_0^{2\pi}\cos(2mt)-1~dt & \text{if }m=n \\\end{cases} \\&= \begin{cases}\displaystyle -\frac{1}{2}\left[\frac{\sin\big((m+n)t\big)}{m+n}-\frac{\sin\big((m-n)t\big)}{m-n}\right]_0^{2\pi} & \text{if }m\ne n \\\displaystyle -\frac{1}{2}\left[\frac{\sin\big(2mt\big)}{2m}-t\right]_0^{2\pi} & \text{if }m=n \\\end{cases} \\&= \begin{cases}0 & \text{if }m\ne n \vphantom{\displaystyle\int} \\\pi & \text{if }m=n \vphantom{\displaystyle\int} \\\end{cases} \\\\\therefore \int_0^{2\pi}\sin(mt)\sin(nt)~dt&= \pi\delta_{m,n} \quad (m,n\in\N_{\ge 0})\end{aligned}


余弦関数と正弦関数の積の積分:

02πcos(mt)sin(nt) dt=1202πsin((m+n)t)sin((mn)t) dt三角関数の積和公式={1202πsin((m+n)t)sin((mn)t) dtif mn1202πsin(2mt) dtif m=n={12[cos((m+n)t)m+n+cos((mn)t)mn]02πif mn12[cos(2mt)2m]02πif m=n={0if mn0if m=n02πcos(mt)sin(nt) dt=0(m,nN0) \begin{aligned}\int_0^{2\pi}\cos(mt)\sin(nt)~dt&= \frac{1}{2}\int_0^{2\pi}\sin\big((m+n)t\big)-\sin\big((m-n)t\big)~dt \quad \because \text{三角関数の積和公式} \\&= \begin{cases}\displaystyle \frac{1}{2}\int_0^{2\pi}\sin\big((m+n)t\big)-\sin\big((m-n)t\big)~dt & \text{if }m\ne n \\\displaystyle \frac{1}{2}\int_0^{2\pi}\sin(2mt)~dt & \text{if }m=n \\\end{cases} \\&= \begin{cases}\displaystyle \frac{1}{2}\left[-\frac{\cos\big((m+n)t\big)}{m+n}+\frac{\cos\big((m-n)t\big)}{m-n}\right]_0^{2\pi} & \text{if }m\ne n \\\displaystyle \frac{1}{2}\left[-\frac{\cos\big(2mt\big)}{2m}\right]_0^{2\pi} & \text{if }m=n \\\end{cases} \\&= \begin{cases}0 & \text{if }m\ne n \vphantom{\displaystyle\int} \\0 & \text{if }m=n \vphantom{\displaystyle\int} \\\end{cases} \\\\\therefore \int_0^{2\pi}\cos(mt)\sin(nt)~dt&= 0 \quad (m,n\in\N_{\ge 0})\end{aligned}

三角関数のマクローリン展開

余弦関数のマクローリン展開:cosx=1x22!+x44!x66!+=k=0(1)k(2k)!x2k正弦関数のマクローリン展開:sinx=xx33!+x55!x77!+=k=0(1)k(2k+1)!x2k+1 \begin{aligned} \text{余弦関数のマクローリン展開} : \cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} +\cdots \\ &= \sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k} \\ \text{正弦関数のマクローリン展開} : \sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} +\cdots \\ &= \sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}x^{2k+1} \\ \end{aligned}


余弦関数のマクローリン展開の導出:

cosx\cos x をマクローリン展開すると、 cosx=k=0cos(k)0k!xk=cos(0)00!x0+cos(1)01!x1+cos(2)02!x2+cos(3)03!x3+cos(4)04!x4+=1sin0xcos02!x2+sin03!x3+cos04!x4+=1x22!+x44!x66!+=k=0(1)k(2k)!x2k(1) \begin{aligned} \cos x &= \sum_{k=0}^\infty\frac{\cos^{(k)}0}{k!}x^k \cr &= \frac{\cos^{(0)}0}{0!}x^0 + \frac{\cos^{(1)}0}{1!}x^1 + \frac{\cos^{(2)}0}{2!}x^2 + \frac{\cos^{(3)}0}{3!}x^3 + \frac{\cos^{(4)}0}{4!}x^4 +\cdots \cr &= 1 - \sin 0\cdot x - \frac{\cos 0}{2!}x^2 + \frac{\sin 0}{3!}x^3 + \frac{\cos 0}{4!}x^4 +\cdots \cr &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} +\cdots \cr &= \sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k} \end{aligned} \tag{1} ダランベールの収束判定法により、 limnan+1an=limn(1)n+1(2(n+1))!x2(n+1)(1)n(2n)!x2n=limnx2(2n+2)P2=0(2) \begin{aligned} \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}}{(2(n+1))!}x^{2(n+1)}}{\frac{(-1)^n}{(2n)!}x^{2n}}\right| \cr &= \lim_{n\to\infty}\left|-\frac{x^2}{_{(2n+2)}P_2}\right| \cr &= 0 \end{aligned} \tag{2} (1),(2)(1),(2) より、 cosx=1x22!+x44!x66!+=k=0(1)k(2k)!x2k \begin{aligned} \therefore \cos x &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} +\cdots \cr &= \sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k} \end{aligned}


正弦関数のマクローリン展開の導出:

sinx=k=0sin(k)0k!xk=sin(0)00!x0+sin(1)01!x1+sin(2)02!x2+sin(3)03!x3+sin(4)04!x4+=0+cos0xsin02!x2cos03!x3+sin04!x4+=xx33!+x55!x77!+=k=0(1)k(2k+1)!x2k+1(1) \begin{aligned} \sin x &= \sum_{k=0}^\infty\frac{\sin^{(k)}0}{k!}x^k \cr &= \frac{\sin^{(0)} 0}{0!}x^0 + \frac{\sin^{(1)} 0}{1!}x^1 + \frac{\sin^{(2)}0}{2!}x^2 + \frac{\sin^{(3)}0}{3!}x^3 + \frac{\sin^{(4)}0}{4!}x^4 +\cdots \cr &= 0 + \cos 0\cdot x - \frac{\sin 0}{2!}x^2 - \frac{\cos 0}{3!}x^3 + \frac{\sin 0}{4!}x^4 + \cdots \cr &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} +\cdots \cr &= \sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}x^{2k+1} \end{aligned} \tag{1} ダランベールの収束判定法により、 limnan+1an=limn(1)n+1(2(n+1)+1)!x2(n+1)+1(1)n(2n+1)!x2n+1=limnx2(2n+3)P2=0(2) \begin{aligned} \lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}}{(2(n+1)+1)!}x^{2(n+1)+1}}{\frac{(-1)^n}{(2n+1)!}x^{2n+1}}\right| \cr &= \lim_{n\to\infty}\left|-\frac{x^2}{_{(2n+3)}P_2}\right| \cr &= 0 \end{aligned} \tag{2} (1),(2)(1),(2) より、 sinx=xx33!+x55!x77!+=k=0(1)k(2k+1)!x2k+1 \begin{aligned} \therefore \sin x &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} +\cdots \cr &= \sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}x^{2k+1} \end{aligned}

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