几何定义
直角三角形定义
有一直角三角形如下
则三角函数定义如下
正弦 | 余弦 | 正切 | 余切 | 正割 | 余割 |
---|---|---|---|---|---|
sin θ = a h \sin\theta=\frac{a}{h} sinθ=ha | cos θ = b h \cos\theta=\frac{b}{h} cosθ=hb | tan θ = a b \tan\theta=\frac{a}{b} tanθ=ba | cot θ = b a \cot\theta=\frac{b}{a} cotθ=ab | sec θ = h b \sec\theta=\frac{h}{b} secθ=bh | csc θ = h a \csc\theta=\frac{h}{a} cscθ=ah |
由于是直角三角形,因此此定义只能定义 θ ∈ ( 0 , π 2 ) \theta\in(0,\frac{\pi}{2}) θ∈(0,2π) 范围内
单位圆定义
有一单位圆 x 2 + y 2 = 1 x^2+y^2=1 x2+y2=1 如下
P ( x , y ) P(x,y) P(x,y) 为单位圆上一点
则三角函数定义如下
正弦 | 余弦 | 正切 | 余切 | 正割 | 余割 |
---|---|---|---|---|---|
sin θ = y 1 = y \sin\theta=\frac{y}{1}=y sinθ=1y=y | cos θ = x 1 = x \cos\theta=\frac{x}{1}=x cosθ=1x=x | tan θ = y x \tan\theta=\frac{y}{x} tanθ=xy | cot θ = x y \cot\theta=\frac{x}{y} cotθ=yx | sec θ = 1 x \sec\theta=\frac{1}{x} secθ=x1 | csc θ = 1 y \csc\theta=\frac{1}{y} cscθ=y1 |
此时 θ ∈ ( − ∞ , + ∞ ) \theta\in(-\infty,+\infty) θ∈(−∞,+∞)
- θ > 0 \theta>0 θ>0 时按逆时针旋转
- θ < 0 \theta<0 θ<0 时按顺时针旋转
函数图像
正弦函数
y = sin x y=\sin x y=sinx
由图知正弦函数
- 最小正周期为 2 π 2\pi 2π
- 对称轴为 x = π 2 + k π , k ∈ Z x=\frac{\pi}{2}+k\pi,\ k\in\mathbb{Z} x=2π+kπ, k∈Z
- 对称中心为 ( k π , 0 ) , k ∈ Z (k\pi,0),\ k\in\mathbb{Z} (kπ,0), k∈Z
- 在一二象限取值为正,三四象限取值为负
余弦函数
y = cos x y=\cos x y=cosx
由图知余弦函数
- 最小正周期为 2 π 2\pi 2π
- 对称轴为 x = k π , k ∈ Z x=k\pi,\ k\in\mathbb{Z} x=kπ, k∈Z
- 对称中心为 ( π 2 + k π , 0 ) , k ∈ Z (\frac{\pi}{2}+k\pi,0),\ k\in\mathbb{Z} (2π+kπ,0), k∈Z
- 在一四象限取值为正,二三象限取值为负
正切函数
y = tan x y=\tan x y=tanx
由图知正切函数
- 最小正周期为 π \pi π
- 对称中心为 ( k π , 0 ) , k ∈ Z (k\pi,0),\ k\in\mathbb{Z} (kπ,0), k∈Z
- 在一三象限取值为正、二四象限取值为负
余切函数
y = cot x y=\cot x y=cotx
由图知余切函数
- 最小正周期为 π \pi π
- 对称中心为 ( π 2 + k π , 0 ) , k ∈ Z (\frac{\pi}{2}+k\pi,0),\ k\in\mathbb{Z} (2π+kπ,0), k∈Z
- 在一三象限取值为正、二四象限取值为负
正割函数
y = sec x y=\sec x y=secx
由图知正割函数
- 最小正周期为 2 π 2\pi 2π
- 对称轴为 x = k π , k ∈ Z x=k\pi,\ k\in\mathbb{Z} x=kπ, k∈Z
- 对称中心为 ( π 2 + k π , 0 ) , k ∈ Z (\frac{\pi}{2}+k\pi,0),\ k\in\mathbb{Z} (2π+kπ,0), k∈Z
- 在一四象限取值为正、二三象限取值为负
余割函数
y = csc x y=\csc x y=cscx
由图知余割函数
- 最小正周期为 2 π 2\pi 2π
- 对称轴为 x = π 2 + k π , k ∈ Z x=\frac{\pi}{2}+k\pi,\ k\in\mathbb{Z} x=2π+kπ, k∈Z
- 对称中心为 ( k π , 0 ) , k ∈ Z (k\pi,0),\ k\in\mathbb{Z} (kπ,0), k∈Z
- 在一二象限取值为正、三四象限取值为负
恒等式
互相表示
由三角函数的定义可知
- tan θ = sin θ cos θ \tan\theta=\frac{\sin\theta}{\cos\theta} tanθ=cosθsinθ
- cot θ = 1 tan θ = cos θ sin θ \cot\theta=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta} cotθ=tanθ1=sinθcosθ
- sec θ = 1 cos θ \sec\theta=\frac{1}{\cos\theta} secθ=cosθ1
- csc θ = 1 sin θ \csc\theta=\frac{1}{\sin\theta} cscθ=sinθ1
毕达哥拉斯恒等式
由直角三角形定义和勾股定理或单位圆定义可知
- sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1 sin2θ+cos2θ=1
由此可推出
- tan 2 θ + 1 = sec 2 θ \tan^2\theta+1=\sec^2\theta tan2θ+1=sec2θ
- cot 2 θ + 1 = csc 2 θ \cot^2\theta+1=\csc^2\theta cot2θ+1=csc2θ
诱导公式
形如 sin / cos / tan / cot / sec / csc ( θ + k π 2 ) , k ∈ Z \sin/\cos/\tan/\cot/\sec/\csc(\theta+\frac{k\pi}{2}),\ k\in\mathbb{Z} sin/cos/tan/cot/sec/csc(θ+2kπ), k∈Z 其中“ / / /”表示或者之意
变换口诀:奇变偶不变,符号看象限
- 奇偶的意思是 k k k 的奇偶性。若为奇数,则 sin \sin sin 和 cos \cos cos 互换、 tan \tan tan 和 cot \cot cot 互换、 sec \sec sec 和 csc \csc csc 互换;偶数则不变
- 将 θ \theta θ 看作锐角,判断 θ + k π 2 \theta+\frac{k\pi}{2} θ+2kπ 所在象限。若在该象限下三角函数取负值,则在前面添加负号
如:
- sin ( θ + π 2 ) = cos θ \sin(\theta+\frac{\pi}{2})=\cos\theta sin(θ+2π)=cosθ
k = 1 k=1 k=1 为奇数, sin \sin sin 变 cos \cos cos
θ + π 2 \theta+\frac{\pi}{2} θ+2π 在第二象限, sin \sin sin 值为正,符号为正 - sin ( θ + π ) = − sin θ \sin(\theta+\pi)=-\sin\theta sin(θ+π)=−sinθ
k = 2 k=2 k=2 为偶数,不需要变
θ + π \theta+\pi θ+π 在第三象限, sin \sin sin 值为负,符号为负 - cos ( θ + 3 π 2 ) = sin θ \cos(\theta+\frac{3\pi}{2})=\sin\theta cos(θ+23π)=sinθ
k = 3 k=3 k=3 为奇数, cos \cos cos 变 sin \sin sin
θ + 3 π 2 \theta+\frac{3\pi}{2} θ+23π 在第四象限, cos \cos cos 值为正,符号为正 - tan ( θ − π 2 ) = − cot θ \tan(\theta-\frac{\pi}{2})=-\cot\theta tan(θ−2π)=−cotθ
k = − 1 k=-1 k=−1 为奇数, tan \tan tan 变 cot \cot cot
θ − π 2 \theta-\frac{\pi}{2} θ−2π 在第四象限, tan \tan tan 值为负,符号为负 - sin ( − θ ) = − sin θ \sin(-\theta)=-\sin\theta sin(−θ)=−sinθ
k = 0 k=0 k=0 为偶数,不需要变
− θ -\theta −θ 在第四象限, sin \sin sin 值为负,符号为负
以上结论由函数图形易得出
和差公式
正弦
- sin ( θ 1 + θ 2 ) = sin θ 1 cos θ 2 + cos θ 1 sin θ 2 \sin(\theta_1+\theta_2)=\sin \theta_1\cos \theta_2+\cos \theta_1\sin \theta_2 sin(θ1+θ2)=sinθ1cosθ2+cosθ1sinθ2
- sin ( θ 1 − θ 2 ) = sin θ 1 cos θ 2 − cos θ 1 sin θ 2 \sin(\theta_1-\theta_2)=\sin \theta_1\cos \theta_2-\cos \theta_1\sin \theta_2 sin(θ1−θ2)=sinθ1cosθ2−cosθ1sinθ2
设有一单位圆 θ 1 2 + θ 2 2 = 1 \theta_1^2+\theta_2^2=1 θ12+θ22=1
则 ∣ A G ∣ = sin θ 2 , ∣ O G ∣ = cos θ 2 |AG|=\sin\theta_2,\ |OG|=\cos\theta_2 ∣AG∣=sinθ2, ∣OG∣=cosθ2
∴ ∣ M P ∣ = ∣ G R ∣ = ∣ O G ∣ sin θ 1 = sin θ 1 cos θ 2 \therefore |MP|=|GR|=|OG|\sin\theta_1=\sin\theta_1\cos\theta_2 ∴∣MP∣=∣GR∣=∣OG∣sinθ1=sinθ1cosθ2
∵ △ O N P ∼ △ G N M \because\triangle ONP\sim\triangle GNM ∵△ONP∼△GNM(直角三角形一个角相等)
∴ ∠ N G M = ∠ N O P = θ 1 \therefore\angle NGM=\angle NOP=\theta_1 ∴∠NGM=∠NOP=θ1
∵ ∠ N G P + ∠ A G M = ∠ M A G + ∠ A G M = 90 ° \because\angle NGP+\angle AGM=\angle MAG+\angle AGM=90\degree ∵∠NGP+∠AGM=∠MAG+∠AGM=90°
∴ ∠ M A G = ∠ N G M = θ 1 \therefore\angle MAG=\angle NGM=\theta_1 ∴∠MAG=∠NGM=θ1
∴ ∣ A M ∣ = ∣ A G ∣ cos θ 1 = cos θ 1 sin θ 2 \therefore |AM|=|AG|\cos\theta_1=\cos\theta_1\sin\theta_2 ∴∣AM∣=∣AG∣cosθ1=cosθ1sinθ2
∴ \therefore ∴
sin ( θ 1 + θ 2 ) = ∣ A P ∣ = ∣ A M ∣ + ∣ M P ∣ = sin θ 1 cos θ 2 + cos θ 1 sin θ 2 \begin{aligned}\sin(\theta_1+\theta_2)&=|AP|=|AM|+|MP|\\&=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2\end{aligned} sin(θ1+θ2)=∣AP∣=∣AM∣+∣MP∣=sinθ1cosθ2+cosθ1sinθ2
令 θ 2 = − θ 2 \theta_2=-\theta_2 θ2=−θ2 则
sin ( θ 1 − θ 2 ) = sin [ θ 1 + ( − θ 2 ) ] = sin θ 1 cos ( − θ 2 ) + cos θ 1 sin ( − θ 2 ) = sin θ 1 cos θ 2 − cos θ 1 sin θ 2 \begin{aligned}\sin(\theta_1-\theta_2)&=\sin[\theta_1+(-\theta_2)]\\&=\sin\theta_1\cos(-\theta_2)+\cos\theta_1\sin(-\theta_2)\\&=\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2\end{aligned} sin(θ1−θ2)=sin[θ1+(−θ2)]=sinθ1cos(−θ2)+cosθ1sin(−θ2)=sinθ1cosθ2−cosθ1sinθ2
余弦
- cos ( θ 1 + θ 2 ) = cos θ 1 cos θ 2 − sin θ 1 sin θ 2 \cos(\theta_1+\theta_2)=\cos \theta_1\cos \theta_2-\sin \theta_1\sin \theta_2 cos(θ1+θ2)=cosθ1cosθ2−sinθ1sinθ2
- cos ( θ 1 − θ 2 ) = cos θ 1 cos θ 2 + sin θ 1 sin θ 2 \cos(\theta_1-\theta_2)=\cos \theta_1\cos \theta_2+\sin \theta_1\sin \theta_2 cos(θ1−θ2)=cosθ1cosθ2+sinθ1sinθ2
令 θ 2 = θ 2 + π 2 \theta_2=\theta_2+\frac{\pi}{2} θ2=θ2+2π 则
cos ( θ 1 + θ 2 ) = sin ( θ 1 + θ 2 + π 2 ) = sin θ 1 cos ( θ 2 + π 2 ) + cos θ 1 sin ( θ 2 + π 2 ) = cos θ 1 cos θ 2 − sin θ 1 sin θ 2 \begin{aligned}\cos(\theta_1+\theta_2)&=\sin(\theta_1+\theta_2+\frac{\pi}{2})\\&=\sin\theta_1\cos(\theta_2+\frac{\pi}{2})+\cos\theta_1\sin(\theta_2+\frac{\pi}{2})\\&=\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2\end{aligned} cos(θ1+θ2)=sin(θ1+θ2+2π)=sinθ1cos(θ2+2π)+cosθ1sin(θ2+2π)=cosθ1cosθ2−sinθ1sinθ2
令 θ 2 = − θ 2 \theta_2=-\theta_2 θ2=−θ2 则
cos ( θ 1 − θ 2 ) = cos [ θ 1 + ( − θ 2 ) ] = cos θ 1 cos ( − θ 2 ) − sin θ 1 sin ( − θ 2 ) = cos θ 1 cos θ 2 + sin θ 1 sin θ 2 \begin{aligned}\cos(\theta_1-\theta_2)&=\cos[\theta_1+(-\theta_2)]\\&=\cos\theta_1\cos(-\theta_2)-\sin\theta_1\sin(-\theta_2)\\&=\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2\end{aligned} cos(θ1−θ2)=cos[θ1+(−θ2)]=cosθ1cos(−θ2)−sinθ1sin(−θ2)=cosθ1cosθ2+sinθ1sinθ2
正切
tan ( θ 1 + θ 2 ) = tan θ 1 + tan θ 2 1 − tan θ 1 tan θ 2 \tan(\theta_1+\theta_2)=\frac{\tan \theta_1+\tan \theta_2}{1-\tan \theta_1\tan \theta_2} tan(θ1+θ2)=1−tanθ1tanθ2tanθ1+tanθ2
tan ( θ 1 − θ 2 ) = tan θ 1 − tan θ 2 1 + tan θ 1 tan θ 2 \tan(\theta_1-\theta_2)=\frac{\tan \theta_1-\tan \theta_2}{1+\tan \theta_1\tan \theta_2} tan(θ1−θ2)=1+tanθ1tanθ2tanθ1−tanθ2
tan ( θ 1 + θ 2 ) = sin ( θ 1 + θ 2 ) cos ( θ 1 + θ 2 ) = sin θ 1 cos θ 2 + cos θ 1 sin θ 2 cos θ 1 cos θ 2 − sin θ 1 sin θ 2 = sin θ 1 cos θ 2 cos θ 1 cos θ 2 + cos θ 1 sin θ 2 cos θ 1 cos θ 2 cos θ 1 cos θ 2 cos θ 1 cos θ 2 − sin θ 1 sin θ 2 cos θ 1 cos θ 2 = sin θ 1 cos θ 1 + sin θ 2 cos θ 2 1 − sin θ 1 cos θ 1 sin θ 2 cos θ 2 = tan θ 1 + tan θ 2 1 − tan θ 1 tan θ 2 \begin{aligned}\tan(\theta_1+\theta_2)&=\frac{\sin(\theta_1+\theta_2)}{\cos(\theta_1+\theta_2)}\\&=\frac{\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2}\\&=\frac{\frac{\sin\theta_1\cos\theta_2}{\cos\theta_1\cos\theta_2}+\frac{\cos\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2}}{\frac{\cos\theta_1\cos\theta_2}{\cos\theta_1\cos\theta_2}-\frac{\sin\theta_1\sin\theta_2}{\cos\theta_1\cos\theta_2}}\\&=\frac{\frac{\sin\theta_1}{\cos\theta_1}+\frac{\sin\theta_2}{\cos\theta_2}}{1-\frac{\sin\theta_1}{\cos\theta_1}\frac{\sin\theta_2}{\cos\theta_2}}\\&=\frac{\tan\theta_1+\tan\theta_2}{1-\tan\theta_1\tan\theta_2}\end{aligned} tan(θ1+θ2)=cos(θ1+θ2)sin(θ1+θ2)=cosθ1cosθ2−sinθ1sinθ2sinθ1cosθ2+cosθ1sinθ2=cosθ1cosθ2cosθ1cosθ2−cosθ1cosθ2sinθ1sinθ2cosθ1cosθ2sinθ1cosθ2+cosθ1cosθ2cosθ1sinθ2=1−cosθ1sinθ1cosθ2sinθ2cosθ1sinθ1+cosθ2sinθ2=1−tanθ1tanθ2tanθ1+tanθ2
令 θ 2 = − θ 2 \theta_2=-\theta_2 θ2=−θ2 则
tan ( θ 1 − θ 2 ) = tan [ θ 1 + ( − θ 2 ) ] = tan θ 1 + tan ( − θ 2 ) 1 − tan θ 1 tan ( − θ 2 ) = tan θ 1 − tan θ 2 1 + tan θ 1 tan θ 2 \begin{aligned}\tan(\theta_1-\theta_2)&=\tan[\theta_1+(-\theta_2)]\\&=\frac{\tan\theta_1+\tan(-\theta_2)}{1-\tan\theta_1\tan(-\theta_2)}\\&=\frac{\tan\theta_1-\tan\theta_2}{1+\tan\theta_1\tan\theta_2}\end{aligned} tan(θ1−θ2)=tan[θ1+(−θ2)]=1−tanθ1tan(−θ2)tanθ1+tan(−θ2)=1+tanθ1tanθ2tanθ1−tanθ2
倍角公式
由和差公式可推出
- sin 2 θ = 2 sin θ cos θ = 2 sin θ cos θ sin 2 θ + cos 2 θ = 2 tan θ 1 + tan 2 θ \begin{aligned}\sin 2\theta&=2\sin\theta\cos\theta\\&=\frac{2\sin\theta\cos\theta}{\sin^2\theta+\cos^2\theta}\\&=\frac{2\tan\theta}{1+\tan^2\theta}\end{aligned} sin2θ=2sinθcosθ=sin2θ+cos2θ2sinθcosθ=1+tan2θ2tanθ
- cos 2 θ = cos 2 θ − sin 2 θ = 1 − tan 2 θ 1 + tan 2 θ = 2 cos 2 θ − 1 = 1 − 2 sin 2 θ \begin{aligned}\cos 2\theta&=\cos^2\theta-\sin^2\theta\\&=\frac{1-\tan^2\theta}{1+\tan^2\theta}\\&=2\cos^2 \theta-1\\&=1-2\sin^2\theta\end{aligned} cos2θ=cos2θ−sin2θ=1+tan2θ1−tan2θ=2cos2θ−1=1−2sin2θ
- tan 2 θ = 2 tan θ 1 − tan 2 θ \tan 2\theta=\frac{2\tan \theta}{1-\tan^2 \theta} tan2θ=1−tan2θ2tanθ
降幂公式
由倍角公式容易推出
- sin 2 θ = 1 − cos 2 θ 2 \sin^2\theta=\frac{1-\cos2\theta}{2} sin2θ=21−cos2θ
- cos 2 θ = cos 2 θ + 1 2 \cos^2\theta=\frac{\cos2\theta+1}{2} cos2θ=2cos2θ+1
由以上两个公式容易推出
- tan 2 θ = 1 − cos 2 θ 1 + cos 2 θ \tan^2\theta=\frac{1-\cos2\theta}{1+\cos2\theta} tan2θ=1+cos2θ1−cos2θ
半角公式
由倍角公式可推出
tan θ 2 = 1 − cos θ sin θ = sin θ 1 + cos θ \tan\frac{\theta}{2}=\frac{1-\cos \theta}{\sin \theta}=\frac{\sin \theta}{1+\cos \theta} tan2θ=sinθ1−cosθ=1+cosθsinθ
∵ cos θ = 1 − 2 sin 2 θ 2 \because \cos \theta=1-2\sin^2\frac{\theta}{2} ∵cosθ=1−2sin22θ
∴ 1 − cos θ = 2 sin 2 θ 2 \therefore 1-\cos \theta=2\sin^2\frac{\theta}{2} ∴1−cosθ=2sin22θ
∵ sin θ = 2 sin θ 2 cos θ 2 \because \sin \theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2} ∵sinθ=2sin2θcos2θ
∴ \therefore ∴
1 − cos θ sin θ = 2 sin 2 θ 2 2 sin θ 2 cos θ 2 = sin θ 2 cos θ 2 = tan θ 2 \begin{aligned}\frac{1-\cos \theta}{\sin \theta}&=\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\\&=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\\&=\tan\frac{\theta}{2}\end{aligned} sinθ1−cosθ=2sin2θcos2θ2sin22θ=cos2θsin2θ=tan2θ
∵ cos θ = 2 cos 2 θ 2 − 1 \because \cos \theta=2\cos^2\frac{\theta}{2}-1 ∵cosθ=2cos22θ−1
∴ 1 + cos θ = 2 cos 2 θ 2 \therefore 1+\cos \theta=2\cos^2\frac{\theta}{2} ∴1+cosθ=2cos22θ
∴ \therefore ∴
sin θ 1 + cos θ = 2 sin θ 2 cos θ 2 2 cos 2 θ 2 = sin θ 2 cos θ 2 = tan θ 2 \begin{aligned}\frac{\sin \theta}{1+\cos \theta}&=\frac{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}\\&=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}\\&=\tan\frac{\theta}{2}\end{aligned} 1+cosθsinθ=2cos22θ2sin2θcos2θ=cos2θsin2θ=tan2θ
积化和差公式
- sin θ 1 sin θ 2 = − 1 2 [ cos ( θ 1 + θ 2 ) − cos ( θ 1 − θ 2 ) ] \sin\theta_1\sin\theta_2=-\frac{1}{2}[\cos(\theta_1+\theta_2)-\cos(\theta_1-\theta_2)] sinθ1sinθ2=−21[cos(θ1+θ2)−cos(θ1−θ2)]
- sin θ 1 cos θ 2 = 1 2 [ sin ( θ 1 + θ 2 ) + sin ( θ 1 − θ 2 ) ] \sin\theta_1\cos\theta_2=\frac{1}{2}[\sin(\theta_1+\theta_2)+\sin(\theta_1-\theta_2)] sinθ1cosθ2=21[sin(θ1+θ2)+sin(θ1−θ2)]
- cos θ 1 sin θ 2 = 1 2 [ sin ( θ 1 + θ 2 ) − sin ( θ 1 − θ 2 ) ] \cos\theta_1\sin\theta_2=\frac{1}{2}[\sin(\theta_1+\theta_2)-\sin(\theta_1-\theta_2)] cosθ1sinθ2=21[sin(θ1+θ2)−sin(θ1−θ2)]
- cos θ 1 cos θ 2 = 1 2 [ cos ( θ 1 + θ 2 ) + cos ( θ 1 − θ 2 ) ] \cos\theta_1\cos\theta_2=\frac{1}{2}[\cos(\theta_1+\theta_2)+\cos(\theta_1-\theta_2)] cosθ1cosθ2=21[cos(θ1+θ2)+cos(θ1−θ2)]
由和差公式得
{ sin ( θ 1 + θ 2 ) = sin θ 1 cos θ 2 + cos θ 1 sin θ 2 ( 1 ) sin ( θ 1 − θ 2 ) = sin θ 1 cos θ 2 − cos θ 1 sin θ 2 ( 2 ) cos ( θ 1 + θ 2 ) = cos θ 1 cos θ 2 − sin θ 1 sin θ 2 ( 3 ) cos ( θ 1 − θ 2 ) = cos θ 1 cos θ 2 + sin θ 1 sin θ 2 ( 4 ) \begin{cases} \sin(\theta_1+\theta_2)=\sin\theta_1\cos\theta_2+\cos\theta_1\sin\theta_2&&&&&(1)\\ \sin(\theta_1-\theta_2)=\sin\theta_1\cos\theta_2-\cos\theta_1\sin\theta_2&&&&&(2)\\ \cos(\theta_1+\theta_2)=\cos\theta_1\cos\theta_2-\sin\theta_1\sin\theta_2&&&&&(3)\\ \cos(\theta_1-\theta_2)=\cos\theta_1\cos\theta_2+\sin\theta_1\sin\theta_2&&&&&(4)\\ \end{cases} ⎩ ⎨ ⎧sin(θ1+θ2)=sinθ1cosθ2+cosθ1sinθ2sin(θ1−θ2)=sinθ1cosθ2−cosθ1sinθ2cos(θ1+θ2)=cosθ1cosθ2−sinθ1sinθ2cos(θ1−θ2)=cosθ1cosθ2+sinθ1sinθ2(1)(2)(3)(4)
式 ( 4 ) − (4)- (4)− 式 ( 3 ) (3) (3) 得
sin θ 1 sin θ 2 = 1 2 [ cos ( θ 1 − θ 2 ) − cos ( θ 1 + θ 2 ) ] \sin\theta_1\sin\theta_2=\frac{1}{2}[\cos(\theta_1-\theta_2)-\cos(\theta_1+\theta_2)] sinθ1sinθ2=21[cos(θ1−θ2)−cos(θ1+θ2)]
式 ( 4 ) + (4)+ (4)+ 式 ( 3 ) (3) (3) 得
cos θ 1 cos θ 2 = 1 2 [ cos ( θ 1 + θ 2 ) + cos ( θ 1 − θ 2 ) ] \cos\theta_1\cos\theta_2=\frac{1}{2}[\cos(\theta_1+\theta_2)+\cos(\theta_1-\theta_2)] cosθ1cosθ2=21[cos(θ1+θ2)+cos(θ1−θ2)]
式 ( 1 ) + (1)+ (1)+ 式 ( 2 ) (2) (2) 得
sin θ 1 cos θ 2 = 1 2 [ sin ( θ 1 + θ 2 ) + sin ( θ 1 − θ 2 ) ] \sin\theta_1\cos\theta_2=\frac{1}{2}[\sin(\theta_1+\theta_2)+\sin(\theta_1-\theta_2)] sinθ1cosθ2=21[sin(θ1+θ2)+sin(θ1−θ2)]
式 ( 1 ) − (1)- (1)− 式 ( 2 ) (2) (2) 得
cos θ 1 sin θ 2 = 1 2 [ sin ( θ 1 + θ 2 ) − sin ( θ 1 − θ 2 ) ] \cos\theta_1\sin\theta_2=\frac{1}{2}[\sin(\theta_1+\theta_2)-\sin(\theta_1-\theta_2)] cosθ1sinθ2=21[sin(θ1+θ2)−sin(θ1−θ2)]
和差化积公式
- sin θ 1 + sin θ 2 = 2 sin ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) \sin\theta_1+\sin\theta_2=2\sin(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2}) sinθ1+sinθ2=2sin(2θ1+θ2)cos(2θ1−θ2)
- sin θ 1 − sin θ 2 = 2 cos ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) \sin\theta_1-\sin\theta_2=2\cos(\frac{\theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2}) sinθ1−sinθ2=2cos(2θ1+θ2)sin(2θ1−θ2)
- cos θ 1 + cos θ 2 = 2 cos ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) \cos\theta_1+\cos\theta_2=2\cos(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2}) cosθ1+cosθ2=2cos(2θ1+θ2)cos(2θ1−θ2)
- cos θ 1 − cos θ 2 = − 2 sin ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) \cos\theta_1-\cos\theta_2=-2\sin(\frac{\theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2}) cosθ1−cosθ2=−2sin(2θ1+θ2)sin(2θ1−θ2)
∵ \because ∵
θ 1 = θ 1 + θ 2 2 + θ 1 − θ 2 2 , θ 2 = θ 1 + θ 2 2 − θ 1 − θ 2 2 \theta_1=\frac{\theta_1+\theta_2}{2}+\frac{\theta_1-\theta_2}{2},\ \theta_2=\frac{\theta_1+\theta_2}{2}-\frac{\theta_1-\theta_2}{2} θ1=2θ1+θ2+2θ1−θ2, θ2=2θ1+θ2−2θ1−θ2
∴ \therefore ∴
sin θ 1 = sin ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) + cos ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) sin θ 2 = sin ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) − cos ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) cos θ 1 = cos ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) − sin ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) cos θ 2 = cos ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) + sin ( θ 1 + θ 2 2 ) sin ( θ 1 − θ 2 2 ) \begin{aligned}&\sin\theta_1=\sin(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})+\cos(\frac{\theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2})\\&\sin\theta_2=\sin(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})-\cos(\frac{\theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2})\\&\cos\theta_1=\cos(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})-\sin(\frac{\theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2})\\&\cos\theta_2=\cos(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2})+\sin(\frac{\theta_1+\theta_2}{2})\sin(\frac{\theta_1-\theta_2}{2})\end{aligned} sinθ1=sin(2θ1+θ2)cos(2θ1−θ2)+cos(2θ1+θ2)sin(2θ1−θ2)sinθ2=sin(2θ1+θ2)cos(2θ1−θ2)−cos(2θ1+θ2)sin(2θ1−θ2)cosθ1=cos(2θ1+θ2)cos(2θ1−θ2)−sin(2θ1+θ2)sin(2θ1−θ2)cosθ2=cos(2θ1+θ2)cos(2θ1−θ2)+sin(2θ1+θ2)sin(2θ1−θ2)
∴ \therefore ∴
sin θ 1 + sin θ 2 = 2 sin ( θ 1 + θ 2 2 ) cos ( θ 1 − θ 2 2 ) \sin\theta_1+\sin\theta_2=2\sin(\frac{\theta_1+\theta_2}{2})\cos(\frac{\theta_1-\theta_2}{2}) sinθ1+sinθ2=2sin(2θ1+θ2)cos(2θ1−θ2)
其他同理可证
辅助角公式
a sin θ + b cos θ = a 2 + b 2 sin ( θ + φ ) a\sin\theta+b\cos\theta=\sqrt{a^2+b^2}\sin(\theta+\varphi) asinθ+bcosθ=a2+b2sin(θ+φ)
其中
sin ( φ ) = b a 2 + b 2 , cos ( φ ) = a a 2 + b 2 \sin(\varphi)=\frac{b}{\sqrt{a^2+b^2}},~\cos(\varphi)=\frac{a}{\sqrt{a^2+b^2}} sin(φ)=a2+b2b, cos(φ)=a2+b2a
参考
[1] 维基百科三角函数https://zh.wikipedia.org/wiki/三角函数
[2] 百度百科三角函数公式https://baike.baidu.com/item/三角函数公式/4374733