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问题: 三角函数,求解题过程

解答:

积化和差:
cos(A+B)+cos(A-B)=(cosAcosB-sinAsinB)+(cosAcosB+sinAsinB)
=2cosAcosB
--->cosAcosB=(1/2)[cos(A+B)+cos(A-B)]
同样可以得到sinAsinB=(-1/2)[cos(A+B)-cos(A-B)]
反过来就是和差化积

原式=cos66cos6*(cos78cos42)
=(1/2)[cos(66+6)+cos(66-6)]*(1/2)[cos(78+42)+cos(78-42)]
=(1/4)(cos72+cos60)(cos120+cos36)
=(1/4)(1/2+cos72)(-1/2+cos36)
=(1/4)[(-1/4)+(cos36-cos72)/2+cos72cos36]
=(1/4)[(-1/4)+(-2)sin54sin(-18)/2+cos72cos36]
=-1/16+(1/4)(cos36cos72+cos72cos36)
=-1/16+(1/2)cos36cos72
=-1/16+(4sin36cos36cos72)/(8sin36) 两次使用sin2A=2sinAcosA
=-1/16+sin144/(8sin36)
=-1/16+sin36/(8sin36)
=-1/16+1/8
=1/16.