已知tan a/2=2,6sin(π-α)+sin(π/2+α)/3cos(3π/2+α)-cos(-α)=
问题描述：
已知tan a/2=2,6sin(π-α)+sin(π/2+α)/3cos(3π/2+α)-cos(-α)=
答
tan a/2=sina/(1+cosa)=2
sina=2(1+cosa)
sin²a=4(1+cosa)²
1-cos²a=4+8cosa+4cos²a
5cos²a+8cosa+3=0
(5cosa+3)(cosa+1)=0
cosa=-3/5或cosa=-1(舍去)
6sin(π-α)+sin(π/2+α)/3cos(3π/2+α)-cos(-α)
=(6sinα+cosα)/(3sinα-cosα)
=(12+13cosα)/(6+5cosα)
=(12-39/5)/(6-3)
=7/5