两角和与差公式:
sin(α±β)=sinαcosβ±cosαsinβ\sin (\alpha \pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \betasin(α±β)=sinαcosβ±cosαsinβ
cos(α±β)=cosαcosβ∓sinαsinβ\cos (\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin \alpha \sin \betacos(α±β)=cosαcosβ∓sinαsinβ
tan(α±β)=tanα±tanβ1∓tanαtanβ\tan (\alpha \pm \beta)=\frac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}tan(α±β)=1∓tanαtanβtanα±tanβ
证:sin(α+β)=sinαcosβ+cosαsinβ\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \betasin(α+β)=sinαcosβ+cosαsinβ
在笛卡尔坐标系中以原点O为圆心作单位圆,在单位圆中作以下线段
如图中所示,容易看出:sin(α+β)=CF;sinα=AB;cosα=OB;sinβ=CD;cosβ=OD\sin (\alpha+\beta)=C F ; \quad \sin \alpha=A B ; \quad \cos \alpha=OB ; \quad \sin \beta=C D ; \cos \beta=ODsin(α+β)=CF;sinα=AB;cosα=OB;sinβ=CD;cosβ=OD
则:SΔOCA=12×1×CF=12×1×sin(α+β)S_{\Delta O C A}=\frac{1}{2} \times 1 \times C F=\frac{1}{2} \times 1 \times \sin (\alpha+\beta)SΔOCA=21×1×CF=21×1×sin(α+β)
SΔOCE=SΔOCB−SΔCEBS_{\Delta O C E}=S_{\Delta O C B}-S_{\Delta C E B}SΔOCE=SΔOCB−SΔCEB
=12×OB×CD−SΔCEB=\frac{1}{2} \times O B \times C D-S_{\Delta C E B}=21×OB×CD−SΔCEB
=12×cosα×sinβ−SΔCEB\quad=\frac{1}{2} \times \cos \alpha \times \sin \beta-S_{\Delta C E B}=21×cosα×sinβ−SΔCEB
SΔOAE=SΔOAD+S△AEDS_{\Delta O A E}=S_{\Delta O A D}+S_{\triangle A E D}SΔOAE=SΔOAD+S△AED
=12×OD×AB+S△AED=\frac{1}{2} \times O D \times A B+S_{\triangle A E D}=21×OD×AB+S△AED
=12×cosβ×sinα+S△AED=\frac{1}{2} \times \cos \beta \times \sin \alpha+S_{\triangle A E D}=21×cosβ×sinα+S△AED
SΔOCA=12×1×sin(α+β)=SΔOCE+SΔOAE=(12×cosα×sinβ−SΔCEB)+(12×cosβ×sinα+S△AED)S_{\Delta O C A}=\frac{1}{2} \times 1 \times \sin (\alpha+\beta)= S_{\Delta O C E}+S_{\Delta O A E}=\left(\frac{1}{2} \times \cos \alpha \times \sin \beta-S_{\Delta C E B})+\left(\frac{1}{2} \times \cos \beta \times \sin \alpha+S_{\triangle A E D}\right)\right.SΔOCA=21×1×sin(α+β)=SΔOCE+SΔOAE=(21×cosα×sinβ−SΔCEB)+(21×cosβ×sinα+S△AED)
容易看出, S△ABD=SΔABC=12×AB×BDS_{\triangle A B D}=S_{\Delta A B C}=\frac{1}{2} \times A B \times B DS△ABD=SΔABC=21×AB×BD ,而 S△ABES_{\triangle A B E}S△ABE 是公共三角形, ∴S△AED=S△CEB,\quad \therefore S_{\triangle{AED}}=S_{\triangle{CEB}},∴S△AED=S△CEB,
可得到:sin(α+β)=sinαcosβ+cosαsinβ\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \betasin(α+β)=sinαcosβ+cosαsinβ
对于sin(α−β)=sinαcosβ−cosαsinβ\sin (\alpha-\beta)=\sin \alpha \cos \beta-\cos \alpha \sin \betasin(α−β)=sinαcosβ−cosαsinβ,只要将 sin(α+β)\sin (\alpha+\beta)sin(α+β) 中的 β\betaβ 唤成 (−β)(-\beta)(−β) 即可。
**二倍角公式:**由和公式,当α=β\alpha = \betaα=β,得到:
sin2α=2sinαcosα\sin 2 \alpha=2 \sin \alpha \cos \alphasin2α=2sinαcosα
cos2α=cos2α−sin2α\cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alphacos2α=cos2α−sin2α
tan2α=2tanα1−tan2α\tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}tan2α=1−tan2α2tanα
由于sin2α+cos2α=1\sin^{2} \alpha+\cos ^{2} \alpha=1sin2α+cos2α=1,故cos2α=cos2α−sin2α=1−2sin2α=2cos2α−1\cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha=1-2 \sin ^{2} \alpha=2 \cos ^{2} \alpha-1cos2α=cos2α−sin2α=1−2sin2α=2cos2α−1
降幂公式:由二倍角公式得到
sin2α=1−cos2α2\sin ^{2} \alpha=\frac{1-\cos 2 \alpha}{2}sin2α=21−cos2α
cos2α=1+cos2α2\cos ^{2} \alpha=\frac{1+\cos 2 \alpha}{2}cos2α=21+cos2α
半角公式:由降幂公式令α\alphaα为α2\frac{\alpha}{2}2α
sin2α2=1−cosα2,cos2α2=1+cosα2\sin ^{2} \frac{\alpha}{2}=\frac{1-\cos \alpha}{2} , \cos ^{2} \frac{\alpha}{2}=\frac{1+\cos \alpha}{2}sin22α=21−cosα,cos22α=21+cosα tan2α2=sin2α2cos2α2=1−cosα1+cosα\tan ^{2} \frac{\alpha}{2}=\frac{\sin ^{2} \frac{\alpha}{2}}{ \cos ^{2} \frac{\alpha}{2}}=\frac{1-\cos \alpha}{1+\cos \alpha}tan22α=cos22αsin22α=1+cosα1−cosα
tanα2=sinα2cosα2=sin2α2sinα2⋅cosα2=12(1−cosα)12sinα=1−cosαsinα\tan \frac{\alpha}{2}=\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}=\frac{\sin ^{2} \frac{\alpha}{2}}{\sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2}}=\frac{\frac{1}{2}(1-\cos \alpha)}{\frac{1}{2} \sin \alpha}=\frac{1-\cos \alpha}{\sin \alpha}tan2α=cos2αsin2α=sin2α⋅cos2αsin22α=21sinα21(1−cosα)=sinα1−cosα
tanα2=sinα2cosα2=sinα2⋅cosα2cos2α2=sinα1+cosα\tan \frac{\alpha}{2}=\frac{\sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}}=\frac{\sin \frac{\alpha}{2} \cdot \cos \frac{\alpha}{2}}{\cos ^{2} \frac{\alpha}{2}}=\frac{\sin \alpha}{1+\cos \alpha}tan2α=cos2αsin2α=cos22αsin2α⋅cos2α=1+cosαsinα
万能公式:利用二倍角证明,
sin2α=2sinαcosα=2sinαcosαcos2α+sin2α=2tanα1+tan2α\sin 2 \alpha=2 \sin \alpha \cos \alpha=\frac{2 \sin \alpha \cos \alpha}{\cos ^{2} \alpha+\sin ^{2} \alpha}=\frac{2 \tan \alpha}{1+\tan ^{2} \alpha}sin2α=2sinαcosα=cos2α+sin2α2sinαcosα=1+tan2α2tanα
cos2α=cos2α−sin2αcos2α+sin2α=1−tan2α1+tan2α\cos 2 \alpha=\frac{\cos ^{2} \alpha-\sin ^{2} \alpha}{\cos ^{2} \alpha+\sin ^{2} \alpha}=\frac{1-\tan ^{2} \alpha}{1+\tan ^{2} \alpha}cos2α=cos2α+sin2αcos2α−sin2α=1+tan2α1−tan2α
tan2α=2tanα1−tan2α\tan 2 \alpha=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}tan2α=1−tan2α2tanα
积化和差公式:由两角和与差公式,sin(α±β)=sinαcosβ±cosαsinβ\sin (\alpha \pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \betasin(α±β)=sinαcosβ±cosαsinβ,将sin(α+β)\sin (\alpha + \beta)sin(α+β)与sin(α−β)\sin (\alpha - \beta)sin(α−β)相加,得到sinαcosβ=12[sin(α+β)+sin(α−β)]\sin \alpha \cos \beta=\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)]sinαcosβ=21[sin(α+β)+sin(α−β)],
将sin(α+β)\sin (\alpha + \beta)sin(α+β)与sin(α−β)\sin (\alpha - \beta)sin(α−β)相减,得到cosαsinβ=12[sin(α+β)−sin(α−β)]\cos \alpha \sin \beta=\frac{1}{2}[\sin (\alpha+\beta)-\sin (\alpha-\beta)]cosαsinβ=21[sin(α+β)−sin(α−β)],
cos(α±β)=cosαcosβ∓sinαsinβ\cos (\alpha \pm \beta)=\cos \alpha \cos \beta \mp \sin \alpha \sin \betacos(α±β)=cosαcosβ∓sinαsinβ,将cos(α+β)\cos (\alpha + \beta)cos(α+β)与cos(α−β)\cos (\alpha - \beta)cos(α−β)相加,得到cosαcosβ=12[cos(α+β)+cos(α−β)]\cos \alpha \cos \beta=\frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)]cosαcosβ=21[cos(α+β)+cos(α−β)]
将cos(α+β)\cos(\alpha + \beta)cos(α+β)与cos(α−β)\cos(\alpha - \beta)cos(α−β)相减,得到sinαsinβ=−12[cos(α+β)−cos(α−β)]\sin \alpha \sin \beta=-\frac{1}{2}[\cos (\alpha+\beta)-\cos (\alpha-\beta)]sinαsinβ=−21[cos(α+β)−cos(α−β)]
和差化积公式:由积化和差公式sinαcosβ=12[sin(α+β)+sin(α−β)]\sin \alpha \cos \beta=\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)]sinαcosβ=21[sin(α+β)+sin(α−β)],令x=α+βx=\alpha+\betax=α+β,y=α−βy=\alpha-\betay=α−β,代入替换α\alphaα与β\betaβ, 可得到:
sinα+sinβ=2sinα+β2cosα−β2\sin \alpha+\sin \beta=2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}sinα+sinβ=2sin2α+βcos2α−β
sinα−sinβ=2cosα+β2sinα−β2\sin \alpha-\sin \beta=2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}sinα−sinβ=2cos2α+βsin2α−β
cosα+cosβ=2cosα+β2cosα−β2\cos \alpha+\cos \beta=2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}cosα+cosβ=2cos2α+βcos2α−β
cosα−cosβ=−2sinα+β2sinα−β2\cos \alpha-\cos \beta=-2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}cosα−cosβ=−2sin2α+βsin2α−β
三角诱导公式 两角和与差 二倍角公式 降幂公式 半角公式 万能公式 积化和差公式 和差化积公式
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