:已知tan(π+α)=2,求3cos(π-α)-2cos(π/2+α)/4cos(-α)+sin(2π-α)的值:已知tan(π+α)=2,求3cos(π-α)-2cos(π/2+α)/4cos(-

题目

:已知tan(π+α)=2,求3cos(π-α)-2cos(π/2+α)/4cos(-α)+sin(2π-α)的值:已知tan(π+α)=2,求3cos(π-α)-2cos(π/2+α)/4cos(-α)+sin(

答案

tan(π+α)=2,即有tana=2,tana=2
3cos(π-α)-2cos(π/2+α)/4cos(-α)+sin(2π-α)
=[-3cosa+2sina]/[4cosa-sina]
=[-3+2tana]/[4-tana],(分子分母同除以cosa)
=(-3+4)/(4-2)
=1/2