定义
![](https://s2.ax1x.com/2019/12/15/QhClnK.png)
在(Rt riangle ABC)中,如下有六个三角函数的定义:
正弦:
[sin A = frac{a}{c}
]
级数表示:(sin (x)==sum_{k=0}^{infty} frac{(-1)^{k} x^{1+2k}}{(1+2k)!})
余弦:
[cos A = frac{b}{c}
]
级数表示:(cos (x)=sum_{k=0}^{infty} frac{(-1)^{k} x^{2 k}}{(2 k) !})
正切:
[ an A = frac{a}{b}
]
级数表示:( an (x)=i+2 i sum_{k=1}^{infty}(-1)^{k} q^{2 k} color{gray} extrm{ for } q=e^{i x})
余切:
[cot A = frac{b}{a}
]
级数表示:(cot (x)=-i-2 i sum_{k=1}^{infty} q^{2 k} color{gray} ext { for } q=e^{i x})
正割:
[sec A = frac{c}{b}
]
级数表示:(sec (x)=-2 sum_{k=1}^{infty}(-1)^{k} q^{-1+2 k} color{gray} ext { for } q=e^{i x})
余割:
[csc A = frac{c}{a}
]
级数表示:(csc (x)=-2 i sum_{k=1}^{infty} q^{-1+2 k} color{gray} ext { for } q=e^{i x})
诱导公式
链接
关系 & 定理 & 公式
倒数关系
[cos alpha cdot sec alpha = 1
]
[sin alpha cdot csc alpha = 1
]
[ an alpha cdot cot alpha = 1
]
平方关系
[1 + an ^ 2 alpha = sec ^ 2 alpha
]
[1 + cot ^ 2 alpha = csc ^ 2 alpha
]
[sin^2 alpha + cos ^ 2 alpha = 1
]
商的关系
[frac{sin alpha}{cos alpha} = frac{sec alpha}{csc alpha} = an alpha
]
[frac{cos alpha}{sin alpha} = frac{csc alpha}{sec alpha} = cot alpha
]
正弦定理
[frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} = 2R = D
]
(R) 为三角形外切圆半径,(D) 为三角形外切圆直径。
证明:
![](https://s2.ax1x.com/2019/12/25/lFqL0x.png)
如图在 ( riangle ABC) 中可得 (sin A = frac{h}{b}) 和 (sin B = frac{h}{a}) 。
[ herefore h = sin A imes b, h = sin B imes a \\ herefore sin A imes b = sin B imes a \\ herefore frac{sin A}{a} = frac{sin B}{b} \\ herefore frac{a}{sin A} = frac{b}{sin B} \\ extrm{同理:} frac{a}{sin A} = frac{c}{sin C} \\ herefore frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C}
]
![](https://s2.ax1x.com/2019/12/29/lKnCNQ.png)
如图, ( riangle CDB) 中线段 (CD) 经过圆心,所以 (angle CBD = 90 ^ circ) , (CD = 2R)。
[ herefore sin A = sin D = frac{CB}{CD} = frac{a}{2R} \\ herefore frac{a}{sin A} = 2R \\ extrm{同理:} frac{b}{sin B} = 2R, frac{c}{sin C} = 2R \\ herefore frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C} = 2R = D
]
余弦定理
[a ^ 2 = b ^ 2 + c ^ 2 - 2bccos A, b ^ 2 = a ^ 2 + c ^ 2 - 2accos B, c ^ 2 = a ^ 2 + b ^ 2 - 2abcos C \\
m{或} \\ cos A = frac{b ^ 2 + c ^ 2 - a ^ 2}{2bc}, cos B = frac{a ^ 2 + c ^ 2 - b ^ 2}{2ac}, cos C = frac{a ^ 2 + b ^ 2 - c ^ 2}{2ab}
]
证明:
![](https://s2.ax1x.com/2020/01/16/lxPR10.png)
如图,在 ( riangle ABC) 中,令(vec{AB} = vec{c}, vec{CB} = vec{a}, vec{CA} = vec{b})。
[ herefore vec{c} = vec{AB} = vec{CB} - vec{CA} = vec{a} - vec{b} \\ herefore (vec{c}) ^ 2 = (vec{a} - vec{b}) ^ 2 = vec{a} ^ 2 + vec{b} ^ 2 - 2 vec{a} cdot vec{b} \\ herefore |vec{c}| ^ 2 = |vec{a}| ^ 2 + |vec{b}| ^ 2 - 2 |vec{a}| cdot |vec{b}| cdot cos C \\ herefore c ^ 2 = a ^ 2 + b ^ 2 - 2abcos C \\ 同理:cos A = frac{b ^ 2 + c ^ 2 - a ^ 2}{2bc}, cos B = frac{a ^ 2 + c ^ 2 - b ^ 2}{2ac}
]
和角公式
[sin(alpha + eta) = sin alpha cos eta + cos alpha sin eta
]
[cos(alpha + eta) = cos alpha cos eta - sin alpha sin eta
]
[ an(alpha + eta) = frac{ an alpha + an eta}{1 - an alpha an eta}
]
差角公式
[sin(alpha - eta) = sin alpha cos eta - cos alpha sin eta
]
[cos(alpha - eta) = cos alpha cos eta + sin alpha sin eta
]
[ an(alpha - eta) = frac{ an alpha - an eta}{1 + an alpha an eta}
]
和差化积
[sin alpha+sin eta=2 sin left(frac{alpha+eta}{2}
ight) cos left(frac{alpha-eta}{2}
ight)
]
[sin alpha-sin eta=2 sin left(frac{alpha-eta}{2}
ight) cos left(frac{alpha+eta}{2}
ight)
]
[cos alpha+cos eta=2 cos left(frac{alpha+eta}{2}
ight) cos left(frac{alpha-eta}{2}
ight)
]
[cos alpha-cos eta=-2 sin left(frac{alpha+eta}{2}
ight) sin left(frac{alpha-eta}{2}
ight)
]
积化和差
[cos alpha sin eta=frac{1}{2}[sin (alpha+eta)-sin (alpha-eta)]
]
[sin alpha cos eta=frac{1}{2}[sin (alpha+eta)+sin (alpha-eta)]
]
[cos alpha cos eta=frac{1}{2}[cos (alpha+eta)+cos (alpha-eta)]
]
[sin alpha sin eta=-frac{1}{2}[cos (alpha+eta)-cos (alpha-eta)]
]
倍角公式
[sin 2 alpha = 2 sin alpha cos alpha
]
[cos 2 alpha = cos ^ 2 alpha - sin ^ 2 alpha
]
[ an 2 alpha = frac{2 an alpha}{1 - an ^ 2 alpha}
]
[cot 2 alpha=frac{cot ^{2} alpha-1}{2 cot alpha}
]
[sec 2 alpha=frac{sec ^{2} alpha}{1- an ^{2} alpha}
]
[csc 2 alpha=frac{1}{2} sec alpha csc alpha
]
半角公式
[sin left(frac{alpha}{2}
ight) = sqrt{frac{1-cos alpha}{2}}
]
[cos left(frac{alpha}{2}
ight) = sqrt{frac{1+cos alpha}{2}}
]
[ an left(frac{alpha}{2}
ight) = csc alpha-cot alpha
]
[cot left(frac{alpha}{2}
ight) = csc alpha+cot alpha
]
[sec left(frac{alpha}{2}
ight) = sqrt{frac{2 sec alpha}{sec alpha+1}}
]
[csc left(frac{alpha}{2}
ight) = sqrt{frac{2 sec alpha}{sec alpha-1}}
]
Attachment
常用三角函数值对照表:
角(alpha) |
弧度 |
(sin)值 |
(cos)值 |
( an)值 |
(0^circ) |
(0) |
(0) |
(1) |
(0) |
(15^circ) |
(frac{pi}{12}) |
(frac{sqrt{6} - sqrt{2}}{4}) |
(frac{sqrt{6} + sqrt{2}}{4}) |
(2 - sqrt{3}) |
(22.5^circ) |
(frac{pi}{8}) |
(frac{sqrt{2 - sqrt{2}}}{2}) |
(frac{sqrt{2 + sqrt{2}}}{2}) |
(-1 + sqrt{2}) |
(30^circ) |
(frac{pi}{6}) |
(frac{1}{2}) |
(frac{sqrt{3}}{2}) |
(frac{sqrt{3}}{3}) |
(45^circ) |
(frac{pi}{4}) |
(frac{sqrt{2}}{2}) |
(frac{sqrt{2}}{2}) |
(1) |
(60^circ) |
(frac{pi}{3}) |
(frac{sqrt{3}}{2}) |
(frac{1}{2}) |
(sqrt{3}) |
(75^circ) |
(frac{5pi}{12}) |
(frac{sqrt{6} + sqrt{2}}{4}) |
(frac{sqrt{6} - sqrt{2}}{4}) |
(2 + sqrt{3}) |
(90^circ) |
(frac{pi}{2}) |
(1) |
(0) |
(
m{无}) |
(120^circ) |
(frac{2pi}{3}) |
(frac{sqrt{3}}{2}) |
(-frac{1}{2}) |
(-sqrt{3}) |
(135^circ) |
(frac{3pi}{4}) |
(frac{sqrt{2}}{2}) |
(-frac{sqrt{2}}{2}) |
(-1) |
(150^circ) |
(frac{5pi}{6}) |
(frac{1}{2}) |
(-frac{sqrt{3}}{2}) |
(frac{sqrt{3}}{3}) |
(180^circ) |
(pi) |
(0) |
(-1) |
(0) |
(270^circ) |
(frac{3pi}{2}) |
(-1) |
(0) |
(
m{无}) |
(360^circ) |
(2pi) |
(0) |
(1) |
(0) |