几道三角函数题 1.cos(x+20)cos(x-40)+cos(x-70)sin(x-40) 2.sin347cos148+sin77cos58

几道三角函数题 1.cos(x+20)cos(x-40)+cos(x-70)sin(x-40) 2.sin347cos148+sin77cos58

1.cos(x+20)cos(x-40)+cos(x-70)sin(x-40)
=cos(x+20)cos(x-40)+sin(x+20)sin(x-40)
=cos[(x+20)-(x-40)]=cos60=1/2
2.sin347cos148+sin77cos58
=sin(270+77)cos(90+58)+sin77cos58
=cos77son58+sin77cos58
=sin(77+58)=sin135=cos45=二分之根号二
3.cos(a-b/2)=-1/9,sin[(a/2)-b]=2/3,π/2sin(a-b/2)=(4X5^0.5)/9,cos[(a/2)-b]=5^0.5/3
cos[(a+b)/2]=cos[(a-b/2)-(a/2-b)]=cos(a-b/2)cos[(a/2)-b]+sin(a-b/2)sin[(a/2)-b]
=-1/9 X 5^0.5/3 + 4X5^0.5)/9 X 2/3=7X5^0.5/27
4.y=sin[(π/3)-2a]+cos[(π/3+2a)]=sin(π/3)cos2a-coa(π/3)sin2a+cos(π/3)cos2a-sin(π/3)sin2a
=(1+3^0.5)/2(cos2a-sin2a)
=(2^0.5+6 ^0.5)/2cos(2a+π/4)
所以最大值为(2^0.5+6 ^0.5)/2，周期为T=2π/2=π
单调递减区间满足：2kπ<2a+π/4<π/2+2kπ,k=0,±1,...
kπ-π/8所以0≤a<π/8或者7π/8所以
单调递减区间为[0,π/8),(7π/8,π]追问还有后面的两道嘞..........追答四道题而已吧，不是已经回答完了吗

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