正弦定理
内容
$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=2r$
r:外接圆半径
证明
锐角三角形
$$\because sinB=\frac{h}{c}, sinC=\frac{h}{b} $$
$$\therefore h=c \cdot sinB=b \cdot sinC$$
$$\therefore \frac{b}{sinB}=\frac{c}{sinC}$$
$$同理, \frac{a}{sinA}=\frac{b}{sinB}$$
$$\therefore \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$$
直角三角形
$$sinB=\frac{b}{c}, sinA=\frac{a}{c}$$
$$c=\frac{b}{sinB}=\frac{a}{sinA}=\frac{c}{sinC} (sinC=1)$$
钝角三角形
$$sinC=\frac{h}{b}, sinB=sin(\pi - B)=\frac{h}{c}(sin \alpha =sin(\pi - \alpha))$$
$$h=b \cdot sinC=c \cdot sinB$$
$$\frac{b}{sinB}=\frac{c}{sinC}$$
$$同理,\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$$
外接圆证明
已知:对于一段弧,其对应的圆周角不变
∠ABC=90°
∠C=∠C'
$$sinC=sinC^{‘}=\frac{c}{2r}$$
$$\frac{c}{sinC}=2r$$
余弦定理
内容
$cosC=\frac{a^2+b^2-c^2}{2ab}$
证明
锐角三角形
设$DC=x$
$$\because h^2=c^2-(a-x)^2$$
$$h^2=b^2-x^2$$
$$\therefore c^2-(a-x)^2=b^2-x^2$$
$$整理得 x=\frac{a^2+b^2-c^2}{2a}$$
$$\therefore cosC=\frac{x}{b}=\frac{a^2+b^2-c^2}{2ab}$$
直角三角形
$$\because a^2+b^2=c^2$$
$$\therefore a^2+b^2-c^2=0$$
$$又\because cosC=0$$
$$\therefore cosC=\frac{a^2+b^2-c^2}{2ab}=0$$
钝角三角形
$$设DC=x$$
$$\because h^2=c^2-(a+x)^2$$
$$h^2=b^2-x^2$$
$$\therefore c^2-(a+x)^2=b^2-x^2$$
$$整理得 x=\frac{c^2-a^2-b^2}{2a}$$
$$\therefore cosC=-cos(\pi - C)=-\frac{x}{b}=-\frac{c^2-a^2-b^2}{2ab}=\frac{a^2+b^2-c^2}{2ab}$$