Question

12．求證：tan$\frac{π}{7}$tan$\frac{2π}{7}$tan$\frac{3π}{7}$=$\sqrt{7}$．

Answer 1

分析 分別求出cos$\frac{π}{7}cos\frac{2π}{7}cos\frac{3π}{7}$與$sin\frac{π}{7}sin\frac{2π}{7}sin\frac{3π}{7}$的值，由商的關(guān)系求得答案．

解答 證明：∵cos$\frac{π}{7}cos\frac{2π}{7}cos\frac{3π}{7}$=$-cos\frac{π}{7}cos\frac{2π}{7}cos\frac{4π}{7}$
=$\frac{-2sin\frac{π}{7}cos\frac{π}{7}cos\frac{2π}{7}cos\frac{4π}{7}}{2sin\frac{π}{7}}=\frac{-sin\frac{2π}{7}cos\frac{2π}{7}cos\frac{4π}{7}}{2sin\frac{π}{7}}$
=$\frac{-sin\frac{4π}{7}cos\frac{4π}{7}}{4sin\frac{π}{7}}=\frac{-sin\frac{8π}{7}}{8sin\frac{π}{7}}=\frac{1}{8}$，
設(shè)t=$sin\frac{π}{7}sin\frac{2π}{7}sin\frac{3π}{7}$，則${t}^{2}=si{n}^{2}\frac{π}{7}si{n}^{2}\frac{2π}{7}si{n}^{2}\frac{3π}{7}$，
∴$8{t}^{2}=（1-cos\frac{2π}{7}）（1-cos\frac{4π}{7}）（1-cos\frac{6π}{7}）$
=$1-（cos\frac{2π}{7}+cos\frac{4π}{7}+cos\frac{6π}{7}）-$$cos\frac{2π}{7}cos\frac{4π}{7}cos\frac{6π}{7}$
+$（cos\frac{2π}{7}cos\frac{4π}{7}+cos\frac{2π}{7}cos\frac{6π}{7}+cos\frac{4π}{7}cos\frac{6π}{7}）$
=$\frac{3}{8}-（cos\frac{2π}{7}+cos\frac{4π}{7}+cos\frac{6π}{7}）$，
而$cos\frac{2π}{7}+cos\frac{4π}{7}+cos\frac{6π}{7}$=$\frac{2sin\frac{π}{7}cos\frac{2π}{7}+2sin\frac{π}{7}cos\frac{4π}{7}+2sin\frac{π}{7}cos\frac{6π}{7}}{2sin\frac{π}{7}}$
=$\frac{sin\frac{3π}{7}-sin\frac{π}{7}+sin\frac{5π}{7}-sin\frac{3π}{7}+sinπ-sin\frac{5π}{7}}{2sin\frac{π}{7}}$=$\frac{-sin\frac{π}{7}}{2sin\frac{π}{7}}=-\frac{1}{2}$，
∴$8{t}^{2}=\frac{7}{8}$，又t＞0，∴$t=\frac{\sqrt{7}}{8}$，
則tan$\frac{π}{7}$tan$\frac{2π}{7}$tan$\frac{3π}{7}$=$\frac{sin\frac{π}{7}sin\frac{2π}{7}sin\frac{3π}{7}}{cos\frac{π}{7}cos\frac{2π}{7}cos\frac{3π}{7}}=\frac{\frac{\sqrt{7}}{8}}{\frac{1}{8}}=\sqrt{7}$．

點(diǎn)評(píng) 本題考查三角函數(shù)的化簡求值，考查三角函數(shù)的和差化積與積化和差公式的應(yīng)用，考查數(shù)學(xué)轉(zhuǎn)化思想方法，難度較大．