三角函数公式简述

Sat, 05 Sep 2026 00:00:00 GMT

三角勾股三角勾股三角勾股 sin⁡2α+cos⁡2α=1\sin^2{\alpha} + \cos^2{\alpha} = 1sin2α+cos2α=1 和差角公式和差角公式和差角公式 sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡βsin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡βcos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡βcos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β\begin{aligned} \sin{(\alpha + \beta)} &= \sin{\alpha} \cos{\beta} + \cos{\alpha} \sin{\beta} \\ \sin{(\alpha - \beta)} &= \sin{\alpha} \cos{\beta} - \cos{\alpha} \sin{\beta} \\ \cos{(\alpha + \beta)} &= \cos{\alpha} \cos{\beta} - \sin{\alpha} \sin{\beta} \\ \cos{(\alpha - \beta)} &= \cos{\alpha} \cos{\beta} + \sin{\alpha} \sin{\beta} \end{aligned}sin(α+β)sin(α−β)cos(α+β)cos(α−β)=sinαcosβ+cosαsinβ=sinαcosβ−cosαsinβ=cosαcosβ−sinαsinβ=cosαcosβ+sinαsinβ 倍角公式倍角公式倍角公式 sin⁡2α=sin(α+α)=sin⁡αcos⁡α+cos⁡αsin⁡α=2sin⁡αcos⁡αcos⁡2α=cos⁡(α+α)=cos⁡αcos⁡α−sin⁡αsin⁡α=cos⁡2α−(1−cos⁡2α)=2cos⁡2α−1=1−2sin⁡2α\begin{aligned} \sin{2\alpha} &= sin{(\alpha+\alpha)} \\ &= \sin{\alpha} \cos {\alpha} + \cos{\alpha} \sin{\alpha} \\ &= 2 \sin{\alpha} \cos{\alpha} \\ \cos{2\alpha} &= \cos{(\alpha + \alpha)} \\ &= \cos{\alpha} \cos{\alpha} - \sin{\alpha} \sin{\alpha} \\ &= \cos^{2} {\alpha} - (1 - \cos^{2} {\alpha}) \\ &= 2 \cos^{2} {\alpha} - 1 \\ &= 1 - 2\sin^2{\alpha} \end{aligned}sin2αcos2α=sin(α+α)=sinαcosα+cosαsinα=2sinαcosα=cos(α+α)=cosαcosα−sinαsinα=cos2α−(1−cos2α)=2cos2α−1=1−2sin2α 半角公式半角公式半角公式 sin⁡α2=1−cos⁡α2cos⁡α2=1+cos⁡α2\begin{aligned} \sin{\frac{\alpha}{2}} &= \sqrt{\frac{1 - \cos{\alpha}}{2}} \\ \cos{\frac{\alpha}{2}} &= \sqrt{\frac{1 + \cos{\alpha}}{2}} \end{aligned}sin2αcos2α=21−cosα=21+cosα 半角公式证明半角公式证明半角公式证明 ∵设 θ=α2 , 由倍角公式可得sin⁡2α2=1−cos⁡α2 , cos⁡2α2=1+cos⁡α2∴sin⁡α2=1−cos⁡α2 , cos⁡α2=1+cos⁡α2\begin{aligned} &\because {\text{设} \ \theta = \frac{\alpha}{2} \ \text{, 由倍角公式可得} \sin^2{\frac{\alpha}{2}} = \frac{1 - \cos{\alpha}}{2} \ , \ \cos^2{\frac{\alpha}{2}} = \frac{1 + \cos{\alpha}}{2}} \\ &\therefore {\sin{\frac{\alpha}{2}} = \sqrt{\frac{1 - \cos{\alpha}}{2}} \ , \ \cos{\frac{\alpha}{2}} = \sqrt{\frac{1 + \cos{\alpha}}{2}}} \end{aligned}∵设 θ=2α , 由倍角公式可得sin22α=21−cosα , cos22α=21+cosα∴sin2α=21−cosα , cos2α=21+cosα 积化和差公式积化和差公式积化和差公式 cos⁡αcos⁡β=12(cos⁡(α+β)+cos⁡(α−β))sin⁡αsin⁡β=12(cos⁡(α−β)−cos⁡(α+β))sin⁡αcos⁡β=12(sin⁡(α+β)+sin⁡(α−β))\begin{aligned} \cos{\alpha} \cos{\beta} &= \frac{1}{2}(\cos{(\alpha + \beta)} + \cos{(\alpha - \beta)}) \\ \sin{\alpha} \sin{\beta} &= \frac{1}{2}(\cos{(\alpha - \beta) - \cos{(\alpha + \beta)}}) \\ \sin{\alpha} \cos{\beta} &= \frac{1}{2}(\sin{(\alpha + \beta) + \sin{(\alpha - \beta)}}) \end{aligned}cosαcosβsinαsinβsinαcosβ=21(cos(α+β)+cos(α−β))=21(cos(α−β)−cos(α+β))=21(sin(α+β)+sin(α−β)) 积化和差公式证明积化和差公式证明积化和差公式证明 ∵cos⁡αcos⁡β+sin⁡αsin⁡β+cos⁡αcos⁡β−sin⁡αsin⁡β=cos⁡(α+β)+cos⁡(α−β)∴2cos⁡αcos⁡β=cos⁡(α+β)+cos⁡(α−β)∴cos⁡αcos⁡β=12cos⁡(α+β)+cos⁡(α−β)∵−(cos⁡αcos⁡β−sin⁡αsin⁡β)+cos⁡αcos⁡β+sin⁡αsin⁡β=cos⁡(α−β)−cos⁡(α+β)∴2sin⁡αsin⁡β=cos⁡(α−β)−cos⁡(α+β)∴sin⁡αsin⁡β=12(cos⁡(α−β)−cos⁡(α+β))∵sin⁡αcos⁡β+cos⁡αsin⁡β+sin⁡αcos⁡β−cos⁡αsin⁡β=sin⁡(α+β)+sin⁡(α−β)∴2sin⁡αcos⁡β=sin⁡(α+β)+sin⁡(α−β)∴sin⁡αcos⁡β=12(sin⁡(α+β)+sin⁡(α−β))\begin{aligned} &\because{\cos{\alpha} \cos{\beta} + \sin{\alpha} \sin{\beta} + \cos{\alpha} \cos{\beta} - \sin{\alpha} \sin{\beta} = \cos{(\alpha + \beta)} + \cos{(\alpha - \beta)}} \\ &\therefore{2 \cos{\alpha} \cos{\beta} = \cos{(\alpha + \beta)} + \cos{(\alpha - \beta)}} \\ &\therefore{\cos{\alpha} \cos{\beta} = \frac{1}{2}\cos{(\alpha + \beta)} + \cos{(\alpha - \beta)}} \\ \\ &\because{-(\cos{\alpha} \cos{\beta} - \sin{\alpha} \sin{\beta}) + \cos{\alpha} \cos{\beta} + \sin{\alpha} \sin{\beta} = \cos{(\alpha - \beta)} - \cos{(\alpha + \beta)}} \\ &\therefore{2 \sin{\alpha} \sin{\beta} = \cos{(\alpha - \beta)} - \cos{(\alpha + \beta)}} \\ &\therefore{\sin{\alpha} \sin{\beta} = \frac{1}{2}(\cos{(\alpha - \beta) - \cos{(\alpha + \beta)}})} \\ \\ &\because{\sin{\alpha} \cos{\beta} + \cos{\alpha} \sin{\beta} + \sin{\alpha} \cos{\beta} - \cos{\alpha} \sin{\beta} = \sin{(\alpha + \beta)} + \sin{(\alpha - \beta)}} \\ &\therefore{2 \sin{\alpha} \cos{\beta} = \sin{(\alpha + \beta)} + \sin{(\alpha - \beta)}} \\ &\therefore{\sin{\alpha} \cos{\beta} = \frac{1}{2}(\sin{(\alpha + \beta) + \sin{(\alpha - \beta)}})} \end{aligned}∵cosαcosβ+sinαsinβ+cosαcosβ−sinαsinβ=cos(α+β)+cos(α−β)∴2cosαcosβ=cos(α+β)+cos(α−β)∴cosαcosβ=21cos(α+β)+cos(α−β)∵−(cosαcosβ−sinαsinβ)+cosαcosβ+sinαsinβ=cos(α−β)−cos(α+β)∴2sinαsinβ=cos(α−β)−cos(α+β)∴sinαsinβ=21(cos(α−β)−cos(α+β))∵sinαcosβ+cosαsinβ+sinαcosβ−cosαsinβ=sin(α+β)+sin(α−β)∴2sinαcosβ=sin(α+β)+sin(α−β)∴sinαcosβ=21(sin(α+β)+sin(α−β)) 和差化积公式和差化积公式和差化积公式 sin⁡α±sin⁡β=2sin⁡α±β2cos⁡α∓β2cos⁡α+cos⁡β=2cos⁡α+β2cos⁡α−β2cos⁡α−cos⁡β=2sin⁡α+β2sin⁡α−β2\begin{aligned} \sin{\alpha} \pm \sin{\beta} &= 2 \sin{\frac{\alpha \pm \beta}{2}} \cos{\frac{\alpha \mp \beta}{2}} \\ \cos{\alpha} + \cos{\beta} &= 2 \cos{\frac{\alpha + \beta}{2}} \cos{\frac{\alpha - \beta}{2}} \\ \cos{\alpha} - \cos{\beta} &= 2 \sin{\frac{\alpha + \beta}{2}} \sin{\frac{\alpha - \beta}{2}} \end{aligned}sinα±sinβcosα+cosβcosα−cosβ=2sin2α±βcos2α∓β=2cos2α+βcos2α−β=2sin2α+βsin2α−β 和差化积公式证明和差化积公式证明和差化积公式证明 ∵前提:设α=α+β2, β=α−β2∴由积化和差公式化简可得和差化积公式\begin{aligned} &\because{\text{前提:设}\alpha = \frac{\alpha + \beta}{2} , \ \beta = \frac{\alpha - \beta}{2}} \\ \\ &\therefore{\text{由积化和差公式化简可得和差化积公式}} \end{aligned}∵前提:设α=2α+β, β=2α−β∴由积化和差公式化简可得和差化积公式

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三角函数公式简述