Question

14．在△ABC中，a，b，c是角A，B，C所對(duì)應(yīng)邊，且a，b，c成等比數(shù)列，則sinA（$\frac{1}{tanA}$+$\frac{1}{tanB}$）的取值范圍是（$\frac{\sqrt{5}-1}{2}$，$\frac{\sqrt{5}+1}{2}$）．

Answer 1

分析 設(shè)a，b，c分別為a，aq，aq²．則有$\left\{\begin{array}{l}{a+aq＞a{q}^{2}}\\{a+a{q}^{2}＞aq}\\{aq+a{q}^{2}＞a}\end{array}\right.$⇒$\left\{\begin{array}{l}{{q}^{2}+q-1＞0…①}\\{{q}^{2}-q+1＞0…②}\\{{q}^{2}+q-1＞0…③}\end{array}\right.$⇒$\frac{\sqrt{5}-1}{2}＜q＜\frac{\sqrt{5}+1}{2}$．化簡(jiǎn)sinA（$\frac{1}{tanA}$+$\frac{1}{tanB}$）=q即可

解答 解：∵△ABC中，∠A，∠B，∠C所對(duì)的邊分別為a，b，c，
∵a，b，c成等比數(shù)列，sin²B=sinAsinB
設(shè)a，b，c分別為a，aq，aq²．
則有$\left\{\begin{array}{l}{a+aq＞a{q}^{2}}\\{a+a{q}^{2}＞aq}\\{aq+a{q}^{2}＞a}\end{array}\right.$⇒$\left\{\begin{array}{l}{{q}^{2}+q-1＞0…①}\\{{q}^{2}-q+1＞0…②}\\{{q}^{2}+q-1＞0…③}\end{array}\right.$⇒$\frac{\sqrt{5}-1}{2}＜q＜\frac{\sqrt{5}+1}{2}$．
 sinA（$\frac{1}{tanA}+\frac{1}{tanB}$）=sinA（$\frac{cosA}{sinA}+\frac{cosB}{sinB}$）=sinA$•\frac{sin（A+B）}{sinAsinB}$
=$\frac{sinAsinC}{sinAsinB}=\frac{si{n}^{2}B}{sinAsinB}=\frac{sinB}{sinA}=\frac{a}=q$
∴sinA（$\frac{1}{tanA}$+$\frac{1}{tanB}$）的取值范圍是：（$\frac{\sqrt{5}-1}{2}$，$\frac{\sqrt{5}+1}{2}$）

點(diǎn)評(píng) 本題主要考查同角三角函數(shù)的基本關(guān)系、等比中項(xiàng)，及三角形三邊的數(shù)量關(guān)系，屬于中檔題