首页 > 技术文章 > 浅谈三角函数

konjakhzx 2022-05-10 19:35 原文

公式证明不会的可以在评论区发出来,看到后我会发在评论区。


基本知识

如图在\(Rt\triangle_{ABC}\)中,\(\angle C = 90^\circ\)

基本的三角函数 :

\[\sin A = \dfrac{a}{c} \]

\[\cos A = \dfrac{b}{c} \]

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

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

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

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

三角函数中的关系

\[\sin A = \cos (90^\circ-A) \]

\[\cos A = \sin (90^\circ-A) \]

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

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

\[\tan A = \dfrac{\sin A}{\cos A} \]

\[\cot A = \dfrac{\cos A}{\sin A} \]

三角函数常用公式

两角和与差

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

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

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

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

和差化积

\[\sin\alpha+\sin\beta=2\cdot\sin(\dfrac{\alpha+\beta}{2})\cdot\cos(\dfrac{\alpha-\beta}{2}) \]

\[\sin\alpha-\sin\beta=2\cdot\sin(\dfrac{\alpha-\beta}{2})\cdot\cos(\dfrac{\alpha+\beta}{2}) \]

\[\cos\alpha+\cos\beta=2\cdot\cos(\dfrac{\alpha+\beta}{2})\cdot\cos(\dfrac{\alpha-\beta}{2}) \]

\[\cos\alpha-\cos\beta=-2\cdot\sin(\dfrac{\alpha+\beta}{2})\cdot\sin(\dfrac{\alpha-\beta}{2}) \]

二倍角公式

\[\sin 2\alpha = 2\cdot\sin\alpha\cdot\cos\alpha=\dfrac{2}{\tan\alpha+\cot\alpha} \]

\[\cos 2\alpha = \cos^2\alpha-\sin^2\alpha=2\cdot\alpha-1=1-2\cdot\sin^2\alpha \]

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

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

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

半角公式

\[\sin(\dfrac{\alpha}{2})=\pm\sqrt{\dfrac{1-\cos\alpha}{2}} \]

\[\cos(\dfrac{a}{2})=\pm\sqrt{\dfrac{1+\cos\alpha}{2}} \]

\[\tan(\dfrac{\alpha}{2})=\pm\sqrt{\dfrac{1-cos\alpha}{1+cos\alpha}}=\dfrac{\sin\alpha}{1+\cos\alpha}=\dfrac{1-\cos\alpha}{\sin\alpha}=\csc\alpha-\cot\alpha \]

\[\cot(\dfrac{\alpha}{2})=\pm\sqrt{\dfrac{1+\cos\alpha}{1-\cos\alpha}}=\dfrac{1+\cos\alpha}{\sin\alpha}=\dfrac{\sin\alpha}{1-\cos\alpha}=\csc\alpha+\cot\alpha \]

\[\sec(\dfrac{\alpha}{2})=\pm\sqrt{\dfrac{2\cdot\sec\alpha}{\sec\alpha+1}} \]

\[\csc(\dfrac{\alpha}{2})=\pm\sqrt{\dfrac{2\cdot\sec\alpha}{\sec\alpha-1}} \]

三角函数常用定理

正弦定理

设在\(\triangle ABC\)中,\(\angle A\)\(\angle B\)\(\angle C\)的对边边长分别是\(a\)\(b\)\(c\)\(\triangle_{ABC}\)的外接圆的半径长为\(r\),则有:

\[\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c} \]

也可表示为

\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2r \]

正弦定理可以用与求三角形面积:

\[S_{\triangle ABC}=\dfrac{1}{2}\cdot a \cdot b \cdot \sin C=\dfrac{1}{2}\cdot b \cdot c \cdot \sin A=\dfrac{1}{2}\cdot a \cdot c \cdot \sin A \]

余弦定理

设在\(\triangle ABC\)中,\(\angle A\)\(\angle B\)\(\angle C\)的对边边长分别是\(a\)\(b\)\(c\)\(\triangle_{ABC}\),则有:

\[a^2=b^2+c^2-2\cdot b\cdot c\cdot \cos A \]

\[b^2 = a^2 + c^2 - 2 \cdot a \cdot b \cdot \cos B \]

\[c^2 = a^2 + b^2 - 2 \cdot a \cdot b \cdot \cos C \]

也可表示为:

\[\cos A = \dfrac {(b^2 + c^2 - a^2)}{2\cdot b\cdot c} \]

\[\cos B = \dfrac {(a^2 + c^2 - b^2)}{2\cdot a\cdot c} \]

\[\cos C = \dfrac {(a^2 + b^2 - c^2)}{2\cdot a\cdot b} \]

正切定理

设在\(\triangle ABC\)中,\(\angle A\)\(\angle B\)\(\angle C\)的对边边长分别是\(a\)\(b\)\(c\)\(\triangle_{ABC}\),则有:

\[\dfrac{a+b}{a-b}=\dfrac{\tan\dfrac{A+B}{2}}{\tan\dfrac{A-B}{2}} \]

推荐阅读