tanx=2 x∈(π,3π/2),求[sin(π-α)+2sin(3π/2+α)]/[cos(3π-α)+1]的值.

问题描述:

tanx=2 x∈(π,3π/2),求[sin(π-α)+2sin(3π/2+α)]/[cos(3π-α)+1]的值.

[sin(π-α)+2sin(3π/2+α)]/[cos(3π-α)+1]=[sinα-2cosα]/[-cosα+1]=[sinα/cosα-2cosα/cosα]/[-cosα/cosα+1/cosα]=[tanα-2]/[-1+1/cosα]=[2-2]/[-1+1/cosα]=0/[-1+1/cosα]=0