正弦定理:\(\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}\)
余弦定理:\(c^2=a^2+b^2-2ab\cos\gamma\)
正弦定理推导:
由\(\sin\alpha=\frac{z}{c}\),\(\sin\beta=\frac{y}{c}\)
得\(\frac{z}{\sin\alpha}=\frac{y}{\sin\beta}\)(1)
由\(\sin(\pi-\gamma)=\frac{z}{a}\),\(\sin(\pi-\gamma)=\frac{y}{b}\)
得\(\frac{z}{a}=\frac{y}{b}\)
将(1)左侧乘以\(\frac{a}{z}\),右侧乘以\(\frac{b}{y}\)
得\(\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}\)
同理,得出角\(\alpha,\beta,\gamma\)与边\(a,b,c\)之间的关系为\(\frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}\),即正弦定理。
余弦定理推导:
由图可得三角恒等式:
\(x^2+y^2=b^2\)(1)
\((a+x)^2+y^2=c^2\)(2)
将(1)以\(y^2\)的形式带入(2),可得
\(c^2=(a+x)^2+b^2-x^2=a^2+b^2+2ax\)(3)
由\(\cos(\pi-\gamma)=\frac{x}{b}\),\(\cos(\pi-\gamma)=-\cos\gamma\),可得
\(x=-b\cos\gamma\)(4)
将(4)带入(3)可得
\(c^2=a^2+b^2-2abcos\gamma\),即余弦定理。