sin(-π/4-α)+cos(π/4-α)=

显示全部楼层 · 2013-2-11 14:14:28

∵sin-a=-sina cos-a=cosa∴sin(-π/4-α)+cos(π/4-α) =-sin(4\π+a)+cos(a-4\π) =-sin4\πcosa-cos4\πsina+cos4\πcosa+sin4\πsina∵cos4\π=sin4\π=2\√2∴原式=-2\√2cosa-2\√2sina+2\√2cosa+2\√2sina
=0...

千问 · 2013-2-11 14:14:28

sin(-π/4-α)+cos(π/4-α)=-sin(π/4+α)+cos(π/4-α)=-(sinπ/4cosα+cosπ/4sinα)+cosαcosπ/4+sinπ/4sinα=0...

千问 · 2013-2-11 14:14:28

cosα=sin(π/2-α)sin(-π/4-α)+cos(π/4-α)=-sin(π/4+α)+sin[π/2-(π/4-α)]= -sin(π/4+α)+sin(π/4+α)=0...