「数学」三角函数公式以及部分证明

定义

(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:

如图在 ( riangle ABC) 中可得 (sin A = frac{h}{b})(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} ]

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:

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:

(alpha) 弧度 (sin) (cos) ( an)
(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)
原文地址:https://www.cnblogs.com/shenxiaohuang/p/12579303.html