1、
cos[(π/4)-α]=12/13
===> cos[(π/2)-2α]=2cos²[(π/4)-α]-1
===> cos[(π/2)-2α]=2*(12/13)²-1=119/169
===> sin2α=119/169
===> cos2α=√(1-sin²2α)=120/169
因?yàn)閇(π/4)-α]+[(π/4)+α]=π/2
所以,sin[(π/4)+α]=cos[(π/4)-α]=12/13
則,原式=(120/169)/(12/13)=10/13
2、
sin[α+(π/3)]=-4/5
則,cos[α+(π/3)]=-3/5
所以,cos[2α+(2π/3)]=2cos²[α+(π/3)]-1=2*(-3/5)²-1=-7/25
那么,sin[2α+(2π/3)]=-√[1-cos²(2α+2π/3)]=-24/25
則,cos[2α+(5π/6)]=cos[2α+(2π/3)+(π/6)]
=cos[2α+(2π/3)]cos(π/6)-sin[2α+(2π/3)]sin(π/6)
=(-7/25)*(√3/2)-(-24/25)*(1/2)
=(24-7√3)/50