物理

[TINY NSD]三角函数续集3

${3.正弦定理和余弦定理}$

${3.1~\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}}$

${[证明]由于形式对称，只需证a\sin B=b\sin A}$

${若A, B均为锐角，显然a\sin B，b\sin A均表示c边上的高，显然相等}$

${若A, B中有直角，不妨设B=\frac{\pi}{2}，则b\sin A=a=a\sin B，也相等}$

${若A, B中有钝角，不妨设B \gt \frac{\pi}{2}}$

${则a\sin (\pi-B)，b\sin A表示c边上的高，必然相等}$

${又\sin (\pi-B)=\sin B，故a\sin B=b\sin A}$

${综上，总有a\sin B=b\sin A，又\sin A, \sin B \gt 0，从而\frac{a}{\sin A}=\frac{b}{\sin B}}$

${故\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}}$

${3.2~a^2+b^2-c^2=2ab\cos C，b^2+c^2-a^2=2bc\cos A，c^2+a^2-b^2=2ca\cos B}$

${[证明]由于形式对称，只需要证a^2+b^2-c^2=2ab\cos C}$

而${2ab\cos C=2\mathbf{a} \cdot \mathbf{b}=\mathbf{a}^2+\mathbf{b}^2-(\mathbf{a}-\mathbf{b})^2}$

${=\mathbf{a}^2+\mathbf{b}^2-\mathbf{c}^2=a^2+b^2-c^2}$

${其中\mathbf{a}=\vec{CB}，\mathbf{b}=\vec{CA}，\mathbf{c}=\vec{AB}}$

${同理可证另外两个等式}$

${3.3~a=b\cos C+c\cos B，b=c\cos A+a\cos C，c=a\cos B+b\cos A}$

${[证明]由3.2，b\cos C+c\cos B=\frac{2ab\cos C+2ac\cos B}{2a}=\frac{(a^2+b^2-c^2)+(a^2+c^2-b^2)}{2a}={2a^2}{2a}=a}$

${同理可证另外两个等式}$

${3.4~\forall x, y, z \in \mathbb{R}，x^2+y^2+z^2 \ge 2(xy\cos C+yz\cos A+zx\cos B)}$

${[证明]固定x, y，令f(z)=z^2-2z(y\cos A+x\cos B)+(x^2+y^2-2xy\cos C)}$

${则f(z)取最小值当且仅当z=y\cos A+x\cos B}$

${故x^2+y^2+z^2-2(xy\cos C+yz\cos A+zx\cos B) \ge –(y\cos A+x\cos B)^2+(x^2+y^2-2xy\cos C)}$

${=x^2\sin^2 B+y^2\sin^2 A-2xy[\cos A\cos B-\cos(A_{}+B)]}$

${=x^2\sin^2 B+y^2\sin^2 A-2xy\sin A\sin B=(x\sin B-y\sin A)^2 \ge 0}$

${取等当且仅当x\sin B=y\sin A且z=y\cos A+x\cos B}$

${由3.2，易知此时必有\frac{x}{a}=\frac{y}{b}=\frac{z}{c}}$

${4.三角形中顶点到对边的连线}$

${4.1~b^2 \cdot BD+c^2 \cdot CD=a \cdot(AD^2+BD \cdot CD)}$

${[证明]设BD=\lambda a, CD=(1-\lambda)a}$

${则\vec{AD}=\lambda\mathbf{b}+(1-\lambda) \mathbf{c}，其中\mathbf{b}=\vec{AC}，\mathbf{c}=\vec{AB}}$

${故a \cdot (AD^2+BD\cdot CD)=a \cdot [\lambda \mathbf{b}+(1-\lambda) \mathbf{c}^2+\lambda(1-\lambda)a^2]}$

${=a \cdot [\lambda^2 \mathbf{b}^2+(1-\lambda)^2\mathbf{c}^2+2\lambda (1-\lambda) \mathbf{b} \cdot \mathbf{c}+\lambda(1-\lambda)a^2]}$

${=a \cdot [\lambda^2 b^2+(1-\lambda)^2 c^2+\lambda(1-\lambda)(b^2+c^2-a^2)+\lambda(1-\lambda)a^2]}$

${=a \cdot [\lambda b^2+(1-\lambda) c^2]=\lambda a \cdot b^2+(1-\lambda)a \cdot c^2}$

${=BD \cdot b^2+CD \cdot c^2，即证}$

${4.2~\frac{\sin \angle BAD}{b}+\frac{\sin\angle CAD}{c}=\frac{\sin A}{AD}}$

${[证明]\frac{\sin\angle BAD}{b}+\frac{\sin \angle CAD}{c}=\frac{b\sin \angle CAD+c\sin \angle BAD}{bc}}$

${=\frac{CD\sin \angle ADC+BD\sin \angle ADB}{bc}=\frac{a}{c} \cdot \frac{\sin \angle ADC}{b}}$

${=\frac{\sin A}{\sin C} \cdot \frac{\sin C}{AD}=\frac{\sin A}{AD}，即证}$

${5.三角形中的二倍角}$

${a^2=b(b+c) \Leftrightarrow b=\frac{c}{1+2\cos A} \Leftrightarrow A=2B}$

${[证明]只需证① \Rightarrow ② \Rightarrow ③ \Rightarrow ①}$

${① \Rightarrow ②：由3.2，b^2+c^2-a^2=2bc\cos A}$

${又a^2-b^2=b(b+c)-b^2=bc，有c^2-bc=2bc\cos A}$

${而c \ne 0，从而b=\frac{c^2}{c+2c\cos A}=\frac{c}{1+2\cos A}}$

${② \Rightarrow ③：由②，c=b(1+2\cos A)}$

${从而\sin A \cos B+\sin B \cos A=\sin (A+B)=\sin C=\sin B+2\sin B \cos A}$

${\sin B=\sin A \cos B+\sin B \cos A-2\sin B \cos A=\sin A \cos B-\sin B \cos A=\sin (A-B)}$

${由三角形角的取值范围，必有B=A-B或B+(A-B)=\pi，后者需A=\pi，矛盾}$

${故B=A-B，即A=2B}$

${③ \Rightarrow ①：设AD为A的内角平分线}$

${则\angle CAD=\frac{A}{2}=B，故\triangle DCA \backsim \triangle ACB}$

${从而CD=\frac{AC^2}{BC}=\frac{b^2}{a}，BD=a-\frac{b^2}{a}}$

${由角平分线定理，\frac{b}{c}=\frac{CD}{BD}=\frac{\frac{b^2}{a}}{a-\frac{b^2}{a}}=\frac{b^2}{a^2-b^2}}$

${由b \ne 0，整理得a^2-b^2=bc，即a^2=b(b+c)}$

${综上，①，②，③等价，原命题成立}$