「数学」三角函数公式以及部分证明

定义

\(Rt\triangle ABC\)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:

\[\tan A = \frac{a}{b} \]

级数表示:\(\tan (x)=i+2 i \sum_{k=1}^{\infty}(-1)^{k} q^{2 k} \color{gray}\textrm{ for } q=e^{i x}\)

余切:

\[\cot A = \frac{b}{a} \]

级数表示:\(\cot (x)=-i-2 i \sum_{k=1}^{\infty} q^{2 k} \color{gray}\text { for } q=e^{i x}\)

正割:

\[\sec A = \frac{c}{b} \]

级数表示:\(\sec (x)=-2 \sum_{k=1}^{\infty}(-1)^{k} q^{-1+2 k} \color{gray}\text { for } q=e^{i x}\)

余割:

\[\csc A = \frac{c}{a} \]

级数表示:\(\csc (x)=-2 i \sum_{k=1}^{\infty} q^{-1+2 k} \color{gray}\text { for } q=e^{i x}\)

诱导公式

链接

关系 & 定理 & 公式

倒数关系

\[\cos \alpha \cdot \sec \alpha = 1 \]

\[\sin \alpha \cdot \csc \alpha = 1 \]

\[\tan \alpha \cdot \cot \alpha = 1 \]

平方关系

\[1 + \tan ^ 2 \alpha = \sec ^ 2 \alpha \]

\[1 + \cot ^ 2 \alpha = \csc ^ 2 \alpha \]

\[\sin^2 \alpha + cos ^ 2 \alpha = 1 \]

商的关系

\[\frac{\sin \alpha}{\cos \alpha} = \frac{\sec \alpha}{\csc \alpha} = \tan \alpha \]

\[\frac{\cos \alpha}{\sin \alpha} = \frac{\csc \alpha}{\sec \alpha} = \cot \alpha \]

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 \(\triangle ABC\) 中可得 \(\sin A = \frac{h}{b}\)\(\sin B = \frac{h}{a}\)

\[\therefore h = \sin A \times b, h = \sin B \times a \\\\ \therefore \sin A \times b = \sin B \times a \\\\ \therefore \frac{\sin A}{a} = \frac{\sin B}{b} \\\\ \therefore \frac{a}{\sin A} = \frac{b}{\sin B} \\\\ \textrm{同理:} \frac{a}{\sin A} = \frac{c}{\sin C} \\\\ \therefore \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \]

如图, \(\triangle CDB\) 中线段 \(CD\) 经过圆心,所以 \(\angle CBD = 90 ^ \circ\)\(CD = 2R\)

\[\therefore \sin A = \sin D = \frac{CB}{CD} = \frac{a}{2R} \\\\ \therefore \frac{a}{\sin A} = 2R \\\\ \textrm{同理:} \frac{b}{\sin B} = 2R, \frac{c}{\sin C} = 2R \\\\ \therefore \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R = D \]

余弦定理

\[a ^ 2 = b ^ 2 + c ^ 2 - 2bc\cos A,\ b ^ 2 = a ^ 2 + c ^ 2 - 2ac\cos B,\ c ^ 2 = a ^ 2 + b ^ 2 - 2ab\cos C \\\\ \rm{或} \\\\ \cos A = \frac{b ^ 2 + c ^ 2 - a ^ 2}{2bc},\ \cos B = \frac{a ^ 2 + c ^ 2 - b ^ 2}{2ac},\ \cos C = \frac{a ^ 2 + b ^ 2 - c ^ 2}{2ab} \]

证明:

如图,在 \(\triangle ABC\) 中,令\(\vec{AB} = \vec{c}, \vec{CB} = \vec{a}, \vec{CA} = \vec{b}\)

\[\therefore \vec{c} = \vec{AB} = \vec{CB} - \vec{CA} = \vec{a} - \vec{b} \\\\ \therefore (\vec{c}) ^ 2 = (\vec{a} - \vec{b}) ^ 2 = \vec{a} ^ 2 + \vec{b} ^ 2 - 2 \vec{a} \cdot \vec{b} \\\\ \therefore |\vec{c}| ^ 2 = |\vec{a}| ^ 2 + |\vec{b}| ^ 2 - 2 |\vec{a}| \cdot |\vec{b}| \cdot \cos C \\\\ \therefore c ^ 2 = a ^ 2 + b ^ 2 - 2ab\cos C \\\\ 同理:\cos A = \frac{b ^ 2 + c ^ 2 - a ^ 2}{2bc},\ \cos B = \frac{a ^ 2 + c ^ 2 - b ^ 2}{2ac} \]

和角公式

\[\sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \]

\[\cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \]

\[\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

差角公式

\[\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \]

\[\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \]

\[\tan(\alpha - \beta) = \frac{\tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \]

和差化积

\[\sin \alpha+\sin \beta=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right) \]

\[\sin \alpha-\sin \beta=2 \sin \left(\frac{\alpha-\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right) \]

\[\cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right) \]

\[\cos \alpha-\cos \beta=-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right) \]

积化和差

\[\cos \alpha \sin \beta=\frac{1}{2}[\sin (\alpha+\beta)-\sin (\alpha-\beta)] \]

\[\sin \alpha \cos \beta=\frac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)] \]

\[\cos \alpha \cos \beta=\frac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)] \]

\[\sin \alpha \sin \beta=-\frac{1}{2}[\cos (\alpha+\beta)-\cos (\alpha-\beta)] \]

倍角公式

\[\sin 2 \alpha = 2 \sin \alpha \cos \alpha \]

\[\cos 2 \alpha = \cos ^ 2 \alpha - \sin ^ 2 \alpha \]

\[\tan 2 \alpha = \frac{2 \tan \alpha}{1 - \tan ^ 2 \alpha} \]

\[\cot 2 \alpha=\frac{\cot ^{2} \alpha-1}{2 \cot \alpha} \]

\[\sec 2 \alpha=\frac{\sec ^{2} \alpha}{1-\tan ^{2} \alpha} \]

\[\csc 2 \alpha=\frac{1}{2} \sec \alpha \csc \alpha \]

半角公式

\[\sin \left(\frac{\alpha}{2}\right) = \sqrt{\frac{1-\cos \alpha}{2}} \]

\[\cos \left(\frac{\alpha}{2}\right) = \sqrt{\frac{1+\cos \alpha}{2}} \]

\[\tan \left(\frac{\alpha}{2}\right) = \csc \alpha-\cot \alpha \]

\[\cot \left(\frac{\alpha}{2}\right) = \csc \alpha+\cot \alpha \]

\[\sec \left(\frac{\alpha}{2}\right) = \sqrt{\frac{2 \sec \alpha}{\sec \alpha+1}} \]

\[\csc \left(\frac{\alpha}{2}\right) = \sqrt{\frac{2 \sec \alpha}{\sec \alpha-1}} \]

Attachment

常用三角函数值对照表:

\(\alpha\) 弧度 \(\sin\) \(\cos\) \(\tan\)
\(0^\circ\) \(0\) \(0\) \(1\) \(0\)
\(15^\circ\) \(\frac{\pi}{12}\) \(\frac{\sqrt{6} - \sqrt{2}}{4}\) \(\frac{\sqrt{6} + \sqrt{2}}{4}\) \(2 - \sqrt{3}\)
\(22.5^\circ\) \(\frac{\pi}{8}\) \(\frac{\sqrt{2 - \sqrt{2}}}{2}\) \(\frac{\sqrt{2 + \sqrt{2}}}{2}\) \(-1 + \sqrt{2}\)
\(30^\circ\) \(\frac{\pi}{6}\) \(\frac{1}{2}\) \(\frac{\sqrt{3}}{2}\) \(\frac{\sqrt{3}}{3}\)
\(45^\circ\) \(\frac{\pi}{4}\) \(\frac{\sqrt{2}}{2}\) \(\frac{\sqrt{2}}{2}\) \(1\)
\(60^\circ\) \(\frac{\pi}{3}\) \(\frac{\sqrt{3}}{2}\) \(\frac{1}{2}\) \(\sqrt{3}\)
\(75^\circ\) \(\frac{5\pi}{12}\) \(\frac{\sqrt{6} + \sqrt{2}}{4}\) \(\frac{\sqrt{6} - \sqrt{2}}{4}\) \(2 + \sqrt{3}\)
\(90^\circ\) \(\frac{\pi}{2}\) \(1\) \(0\) \(\rm{无}\)
\(120^\circ\) \(\frac{2\pi}{3}\) \(\frac{\sqrt{3}}{2}\) \(-\frac{1}{2}\) \(-\sqrt{3}\)
\(135^\circ\) \(\frac{3\pi}{4}\) \(\frac{\sqrt{2}}{2}\) \(-\frac{\sqrt{2}}{2}\) \(-1\)
\(150^\circ\) \(\frac{5\pi}{6}\) \(\frac{1}{2}\) \(-\frac{\sqrt{3}}{2}\) \(\frac{\sqrt{3}}{3}\)
\(180^\circ\) \(\pi\) \(0\) \(-1\) \(0\)
\(270^\circ\) \(\frac{3\pi}{2}\) \(-1\) \(0\) \(\rm{无}\)
\(360^\circ\) \(2\pi\) \(0\) \(1\) \(0\)
posted @ 2020-02-26 09:27  XiaoHuang666  阅读(1126)  评论(0编辑  收藏  举报