Question

1．若tanα=2tan$\frac{π}{5}$，則$\frac{{cos（α-\frac{3π}{10}）}}{{sin（α-\frac{π}{5}）}}$=（　�。�

• A． 1
• B． 2
• C． 3
• D． 4

Answer 1

分析 直接利用兩角和與差的三角函數(shù)化簡所求表達(dá)式，利用同角三角函數(shù)的基本關(guān)系式結(jié)合已知條件以及積化和差個數(shù)化簡求解即可．

解答 解：tanα=2tan$\frac{π}{5}$，則$\frac{{cos（α-\frac{3π}{10}）}}{{sin（α-\frac{π}{5}）}}$=$\frac{cosαcos\frac{3π}{10}+sinαsin\frac{3π}{10}}{sinαcos\frac{π}{5}-cosαsin\frac{π}{5}}$=$\frac{cos\frac{3π}{10}+tanαsin\frac{3π}{10}}{tanαcos\frac{π}{5}-sin\frac{π}{5}}$
=$\frac{cos\frac{3π}{10}+2tan\frac{π}{5}sin\frac{3π}{10}}{2tan\frac{π}{5}cos\frac{π}{5}-sin\frac{π}{5}}$=$\frac{cos\frac{3π}{10}+2\frac{sin\frac{π}{5}}{cos\frac{π}{5}}sin\frac{3π}{10}}{2\frac{sin\frac{π}{5}}{cos\frac{π}{5}}cos\frac{π}{5}-sin\frac{π}{5}}$=$\frac{cos\frac{π}{5}cos\frac{3π}{10}+2sin\frac{π}{5}sin\frac{3π}{10}}{2sin\frac{π}{5}cos\frac{π}{5}-cos\frac{π}{5}sin\frac{π}{5}}$=$\frac{cos（\frac{π}{5}-\frac{3π}{10}）+sin\frac{π}{5}sin\frac{3π}{10}}{sin\frac{π}{5}cos\frac{π}{5}+sin（\frac{π}{5}-\frac{π}{5}）}$=$\frac{cos\frac{π}{10}+sin\frac{π}{5}sin\frac{3π}{10}}{sin\frac{π}{5}cos\frac{π}{5}}$=$\frac{cos\frac{π}{10}-\frac{1}{2}[cos（\frac{π}{5}+\frac{3π}{10}）-cos（\frac{π}{5}-\frac{3π}{10}）]}{\frac{1}{2}sin\frac{2π}{5}}$=$\frac{cos\frac{π}{10}+\frac{1}{2}cos\frac{π}{10}}{\frac{1}{2}sin\frac{2π}{5}}$=$\frac{3cos\frac{π}{10}}{sin\frac{2π}{5}}$=$\frac{3cos\frac{π}{10}}{sin（\frac{π}{2}-\frac{π}{10}）}$=$\frac{3cos\frac{π}{10}}{cos\frac{π}{10}}$=3．
故答案為：3．

點評 本題考查兩角和與差的三角函數(shù)，積化和差以及誘導(dǎo)公式的應(yīng)用，考查計算能力．