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