求sin(2a+pai/3)已知:角a的顶点与直角坐标系的原点重合,始边在x轴的正半轴上,始边经过点P(1,2),求(2

a∈(0,π/2),tana=1/2,则tan2a=[2tana]/[1－tan²a]=4/3；sin(2a＋π/3)=(1/2)sin2a＋(√3/2)cos2a.sin2a=[2tana]/[1＋tan²a]=4/5,cos2a=[1－tan²a]/[1＋tan²a]=3/5,则sin(2a＋π/3)=(4＋3√3)/10

你说的很对,来看看没有回答的问题,看到了你的这道题
非常容易,怎么没人应答呢!
tanatan(a+π/3)=2,求cos(2a+π/3)
cos(2a+π/3)=cosacos(a+π/3)-sinasin(a+π/3)
=cosacos(a+π/3)[1-tanatan(a+π/3)
=-cosacos(a+π/3)
2cos(2a+π/3)=-2cosacos(a+π/3)【积化和差】
=-cos(2a+π/3)-cosπ/3
cos(2a+π/3)=(-1/2)/3=-1/6