sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
问题描述：
sin(-1200°)-cos1290°+cos(-120)°+cos(-1020)°-sin(-1050°)+tan855°
答
sin(-1200°)=-sin(3*360° +120° )=-sin120°=-1/2cos1290°=cos(4*360°-150°)=cos150°=-√3/2cos(-120)°=cos120°=-1/2cos(-1020)°=cos1020°=cos(3*360°-60°）=1/2sin(-1050°)=sin(3*360°-30°）=-sin3...