若tanx=-2分之一,求2sin^2x+sinx乘cosx-3cos^x
问题描述：
若tanx=-2分之一,求2sin^2x+sinx乘cosx-3cos^x
答
解:
2sin^2x+sinx乘cosx-3cos^x
=[2tan^2x+tanx-3]/[tan^2x+1]
=[1/2-1/2-3]/[1/4+1]
=-3/5/4
=-12/5
答
标准做法
2sin^2x+sinx乘cosx-3cos^x
=(2sin^2x+sinx乘cosx-3cos^x)/(sin²x+cos²x)
=(2tan²x+tanx-3)/(tan²x+1)
=(1/2-1/2-3)/(1/4+1)
=-12/5