Question

sin(π/4+a)sin(π/4-a)=1/6

要 过程////

Answer 1

sin(π/4-a)=cos[π/2-(π/4-a)]=cos(π+a)
sin(π/4+a)sin(π/4-a)=sin(π/4+a)cos(π/4+a)=sin(π/2+2a)/2
所以sin(π/2+2a)=-cos2a=-1/3
sin2a=2倍根号2/3
sin4a=2sin2acos2a=-4倍根号2/9

Answer 2

积化和差：
sin(45°+α）sin(45°-α)=1/2(cos2α-cos90°)=1/2*cos2α=1/6
故cos2α=1/3.
2α在三四象限，故sin2α=2根2/3.
故sin4α=2sin2αcos2α=4根2/9.

Answer 3

sin(π/4+a)sin(π/4-a)
=sin(π/4+a)sin[π/2-(π/4+a)]
=sin(π/4+a)cos(π/4+a)
=1/2sin(π/2+2a)
=1/2cos2a=1/6
cos2a=1/3 a属于(π/2,π),2a属于(π,2π),
此时sin2a<0 sin2a=-2根号2/3
sin4a=2sin2acos2a=2*1/3*2根号2/3=-4根号2/9

Answer 4

sin(pi/4+a)=cos(pi/4-a) 原式=1/2(2sin(pi/4-a)cos(pi/4-a))=1/2sin(pi/2-2a)=1/6
cos2a=1/3 sin2a=2根号２／３　sin4a=2sin2acos2a=4根号２／９