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: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(

分类: 作业答案 • 2022-05-30 22:56:37
问题描述：

: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

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

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