第1个回答 2013-04-23
【参考答案】
（1）∵a、b∈(3π/4, π)
∴a+b∈(3π/2, 2π)
∵sin(a+b)=-3/5
∴cos(a+b)=4/5
∴sin2(a+b)=2sin(a+b)cos(a+b)=-24/25
(2)∵a、b∈(3π/4, π)
∴a+ π/4∈(π，5π/4)
b- π/4∈(π/2, 3π/4)
又∵sin(b -π/4)=(3/10)√10
∴cos(b- π/4)=-√10/10
∴sin(a+b)=-3/5=sin[(a+ π/4)+(b- π/4)]
=(-√10/10)sin(a+ π/4)+(3/10)√10cos(a+ π/4)
将上式与sin²(a+ π/4)+cos²(a+ π/4)=1联立，
解得 cos(a+ π/4)=(-13/50)√10或(-1/10)√10
sin2a=－cos[2(a＋π/4)]=－[2cos²(a＋π/4)－1]=－44/125
（1）∵a、b∈(3π/4, π)
∴a+b∈(3π/2, 2π)
∵sin(a+b)=-3/5
∴cos(a+b)=4/5
∴sin2(a+b)=2sin(a+b)cos(a+b)=-24/25
(2)∵a、b∈(3π/4, π)
∴a+ π/4∈(π，5π/4)
b- π/4∈(π/2, 3π/4)
又∵sin(b -π/4)=(3/10)√10
∴cos(b- π/4)=-√10/10
∴sin(a+b)=-3/5=sin[(a+ π/4)+(b- π/4)]
=(-√10/10)sin(a+ π/4)+(3/10)√10cos(a+ π/4)
将上式与sin²(a+ π/4)+cos²(a+ π/4)=1联立，
解得 cos(a+ π/4)=(-13/50)√10或(-1/10)√10
sin2a=－cos[2(a＋π/4)]=－[2cos²(a＋π/4)－1]=－44/125
第2个回答 2013-04-23
(1)a∈(3π/4,π),b∈(3π/4,π),a+b∈(3π/2,2π), 即在第四象限
sin(a+b)=-3/5, cos(a+b)=√[1-sin²(a+b)]=4/5
sin2(a+b)=2sin(a+b)cos(a+b)=2*(-3/5)*(4/5)=-24/25
(2)b-π/4∈(π/2,3π/4),在第二象限
sin(b-π/4)=3√10/10, cos(b-π/4)=-1/10
sin(a+b)=sin[(a+π/4)+(b-π/4)]=sin(a+π/4)cos(b-π/4)+cos(a+π/4)sin(b-π/4)
=(3√10cos(a+π/4)-sin(a+π/4))/10=-3/5
3√10cos(a+π/4)-sin(a+π/4)=-6 .....(1)
cos(a+b)=cos[(a+π/4)+(b-π/4)]=cos(a+π/4)cos(b-π/4)-sin(a+π/4)sin(b-π/4)
=(-3√10sin(a+π/4)-cos(a+π/4))/10=4/5
3√10sin(a+π/4)+cos(a+π/4)=-8 ....(2)
3√10(1)+(2): 91cos(a+π/4)=-8-18√10, cos(a+π/4)=-(8+18√10)/91 ....(3)
3√10(2)-(1): 91sin(a+π/4)=6-24√10, sin(a+π/4)=(6-24√10)/91 ....(4)
由(3): cosacosπ/4-sinasinπ/4=-(8+18√10)/91
cosa-sina=-(8+18√10)√2/91 ....(5)
由(4): sinacosπ/4+cosasinπ/4=(6-24√10)/91
sina+cosa=(6-24√10)√2/91 ....(6)
(5)+(6): cosa=-(1+21√10)√2/91
(6)-(5): sina=(7-3√10)√2/91
sin2a=2sinacosa=-2*(7-3√10)√2/91(1+21√10)√2/91
=4(623+150√10)/8281
怎么这么繁,是不是有错?
sin(a+b)=-3/5, cos(a+b)=√[1-sin²(a+b)]=4/5
sin2(a+b)=2sin(a+b)cos(a+b)=2*(-3/5)*(4/5)=-24/25
(2)b-π/4∈(π/2,3π/4),在第二象限
sin(b-π/4)=3√10/10, cos(b-π/4)=-1/10
sin(a+b)=sin[(a+π/4)+(b-π/4)]=sin(a+π/4)cos(b-π/4)+cos(a+π/4)sin(b-π/4)
=(3√10cos(a+π/4)-sin(a+π/4))/10=-3/5
3√10cos(a+π/4)-sin(a+π/4)=-6 .....(1)
cos(a+b)=cos[(a+π/4)+(b-π/4)]=cos(a+π/4)cos(b-π/4)-sin(a+π/4)sin(b-π/4)
=(-3√10sin(a+π/4)-cos(a+π/4))/10=4/5
3√10sin(a+π/4)+cos(a+π/4)=-8 ....(2)
3√10(1)+(2): 91cos(a+π/4)=-8-18√10, cos(a+π/4)=-(8+18√10)/91 ....(3)
3√10(2)-(1): 91sin(a+π/4)=6-24√10, sin(a+π/4)=(6-24√10)/91 ....(4)
由(3): cosacosπ/4-sinasinπ/4=-(8+18√10)/91
cosa-sina=-(8+18√10)√2/91 ....(5)
由(4): sinacosπ/4+cosasinπ/4=(6-24√10)/91
sina+cosa=(6-24√10)√2/91 ....(6)
(5)+(6): cosa=-(1+21√10)√2/91
(6)-(5): sina=(7-3√10)√2/91
sin2a=2sinacosa=-2*(7-3√10)√2/91(1+21√10)√2/91
=4(623+150√10)/8281
怎么这么繁,是不是有错?
第3个回答 2013-04-23
a、b∈(3π/4，π)
则：a＋b∈(3π/2，2π)
sin(a＋b)=－3/5，则：cos(a＋b)=4/5
【1】
sin2(a＋b)=2sin(a＋b)cos(a＋b)=－24/25
【2】
sin(b－π/4)=(3√10)/10、b－π/4∈(π/2，3π/4)
则：cos(b－π/4)=－√10/10
cos(a＋π/4)=cos[(a＋b)－(b－π/4)]=cos(a＋b)cos(b－π/4)＋sin(a＋b)sin(b－π/4)
=－(13√10)/50
sin2a=－cos[2a＋π/2]=－cos[2(a＋π/4)]=－[2cos²(a＋π/4)－1]=－44/125本回答被提问者采纳
则：a＋b∈(3π/2，2π)
sin(a＋b)=－3/5，则：cos(a＋b)=4/5
【1】
sin2(a＋b)=2sin(a＋b)cos(a＋b)=－24/25
【2】
sin(b－π/4)=(3√10)/10、b－π/4∈(π/2，3π/4)
则：cos(b－π/4)=－√10/10
cos(a＋π/4)=cos[(a＋b)－(b－π/4)]=cos(a＋b)cos(b－π/4)＋sin(a＋b)sin(b－π/4)
=－(13√10)/50
sin2a=－cos[2a＋π/2]=－cos[2(a＋π/4)]=－[2cos²(a＋π/4)－1]=－44/125本回答被提问者采纳
第4个回答 2013-04-23
∵a∈(3π/4,π),b∈(3π/4,π)
∴(a+b)∈(3π/2,2π)
∴cos(a+b)>0
∴cos(a+b)=4/5
∴sin2(a+b)=-24/25
sin(b-π/4)=sin(b+π/4-π/2)=-cos(b+π/4)=3√10/10
sin(b－π/4)=(3√10)/10、b－π/4∈(π/2，3π/4)
则：cos(b－π/4)=－√10/10
cos(a＋π/4)=cos[(a＋b)－(b－π/4)]=cos(a＋b)cos(b－π/4)＋sin(a＋b)sin(b－π/4)
=－(13√10)/50
sin2a=－cos[2a＋π/2]=－cos[2(a＋π/4)]=－[2cos²(a＋π/4)－1]=－44/125
∴(a+b)∈(3π/2,2π)
∴cos(a+b)>0
∴cos(a+b)=4/5
∴sin2(a+b)=-24/25
sin(b-π/4)=sin(b+π/4-π/2)=-cos(b+π/4)=3√10/10
sin(b－π/4)=(3√10)/10、b－π/4∈(π/2，3π/4)
则：cos(b－π/4)=－√10/10
cos(a＋π/4)=cos[(a＋b)－(b－π/4)]=cos(a＋b)cos(b－π/4)＋sin(a＋b)sin(b－π/4)
=－(13√10)/50
sin2a=－cos[2a＋π/2]=－cos[2(a＋π/4)]=－[2cos²(a＋π/4)－1]=－44/125
第5个回答 2013-04-23
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