高一数学三角函数请详细解答,谢谢! (20 17:30:51)

显示全部楼层 · 2009-8-20 17:35:01

[cos2a] / sin(a-π/4)=-√2/2 [cos2a] / sin(a-π/4)=（cosa-sina)(cosa+sina)/[√2(sina-cosa)/2]=-(cosa+sina)/(√2/2)=-√2/2所以sina+cosa=1/2...

千问 · 2009-8-20 17:35:01

解:由cos2α/sin(α-π/4)=-(√2)/2, 且 cos2α=cos2α-sin2α sin(α-π/4)=(√2)/2*(sinα-cosα) 则: cos2α/sin(α-π/4)=cos2α-sin2α/((√2)/2*(sinα-cosα))=-(√2)/2 ...

千问 · 2009-8-20 17:35:01

cos2α/sin(α-π/4)=-（根号2）/2 =>cos2α=-（根号2）/2 *sin(α-π/4)=>cos2α=1/2 [cosa-sina]=>(cosa-sina)(sina+cosa)=1/2(cosa-sina)=>(cosa-sina)(sina+cosa-1/2)=0=>sina+cosa=1/2,因为sina-c...

千问 · 2009-8-20 17:35:01

cos2α/sin(α-π/4)=[(cosα)^2-(sinα)^2]/[√2sinα/2+√2cosα/2]=√2（cosα-sinα）=-√2/2cosα-sinα=-1/2(sinα+cosα)^2=(cosα)^2+(sinα)^2+2sinαcosα(sinα-cosα)^2=(cosα)^2+(sinα)^2-2sinαcosα...

千问 · 2009-8-20 17:35:01

由cos2α/sin(α-π/4)=-(√2)/2, cos2α=cos2α-sin2α sin(α-π/4)=(√2)/2*(sinα-cosα) cos2α/sin(α-π/4)=cos2α-sin2α/((√2)/2*(sinα-cosα))=-(√2)/2 等式左边同...

高一 数学 三角函数 请详细解答,谢谢! (20 17:30:51)

高一数学三角函数请详细解答,谢谢! (20 17:30:51)