sin(a+b)=-3/5,sin(b-π/4)=12/13,求cos(a+π/4)值

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/11 17:16:27
sin(a+b)=-3/5,sin(b-π/4)=12/13,求cos(a+π/4)值

sin(a+b)=-3/5,sin(b-π/4)=12/13,求cos(a+π/4)值
sin(a+b)=-3/5,sin(b-π/4)=12/13,求cos(a+π/4)值

sin(a+b)=-3/5,sin(b-π/4)=12/13,求cos(a+π/4)值
cos(a+π/4)
=cos[(a+b)-(b-π/4)]
=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
又sin(a+b)=-3/5,sin(b-π/4)=12/13,
所以cos(a+b)=4/5(a+b为第四象限角),cos(a+b)=4/5(a+b为第四象限角)
cos(b-π/4)=5/13(b-π/4为第一象限角),cos(b-π/4)=-5/13(b-π/4为第二象限角),
所以cos(a+π/4)=-16/65,-56/65

你好,cosA=4/5,cosB=5/13
A,B为第一象限角,sinA=3/5,sinB=12/13
cosC
=cos(派-(A+B))
=cos派cos(A+B)-sin派sin(A+B)
=-cos(A+B)
=-cosAcosB+sinAsinB
=-20/65+36/65
=16/65 15939希望对你有帮助!

sin(a+b)=-3/5
则cos(a+b)=4/5或(-4/5)
sin(b-π/4)=12/13
则cos(b-π/4)=5/13或(-5/13)

所以
cos(a+π/4)
=cos[(a+b)-(b-π/4)]
=cos(a+b)*cos(b-π/4)+sin(a+b)*sin(b-π/4)
=cos...

全部展开

sin(a+b)=-3/5
则cos(a+b)=4/5或(-4/5)
sin(b-π/4)=12/13
则cos(b-π/4)=5/13或(-5/13)

所以
cos(a+π/4)
=cos[(a+b)-(b-π/4)]
=cos(a+b)*cos(b-π/4)+sin(a+b)*sin(b-π/4)
=cos(a+b)*cos(b-π/4)-36/65
当cos(a+b)与cos(b-π/4)同号时(同正或同负)
cos(a+b)*cos(b-π/4)=20/65
cos(a+π/4)=-16/65
当cos(a+b)与cos(b-π/4)异号时(一正一负)
cos(a+b)*cos(b-π/4)=-20/65
cos(a+π/4)=-56/65

收起

sin(a+b)=-3/5
则cos(a+b)=4/5 或cos(a+b)=-4/5
sin(b-π/4)=12/13
则cos(b-π/4)=5/13 或cos(b-π/4)=-5/13
所以cos(a+π/4)
=cos[(a+b)-(b-π/4)]
=cos(a+b)*cos(b-π/4)+sin(a+b)*sin(b-π/4) <...

全部展开

sin(a+b)=-3/5
则cos(a+b)=4/5 或cos(a+b)=-4/5
sin(b-π/4)=12/13
则cos(b-π/4)=5/13 或cos(b-π/4)=-5/13
所以cos(a+π/4)
=cos[(a+b)-(b-π/4)]
=cos(a+b)*cos(b-π/4)+sin(a+b)*sin(b-π/4)
=(4/5)(5/13)-(-3/5)(12/13)
=56/65 ,还有三种答案。(因为你的条件不完善,所以答案不唯一)

收起

sin(a+b)=-3/5,cos(a+b)=4/5,或-4/5
sin(b-π/4)=12/13,cos(b-π/4)=5/13,或-5/13
cos(a+π/4)=cos[(a+b)-(b-π/4)]=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
cos(a+b)与cos(b-π/4)同号时(同正或同负)
=20/65-36/65=-16/65
cos(a+b)与cos(b-π/4)异号时(一正一负)
=-20/65-36/65=-56/65

cos(a+π/4)=cos[(a+b)-(b-π/4)]
=cos(a+b)cos(b-π/4)+sin(a+b)sin(b-π/4)
再利用平方关系,求出cos(a+b),与cos(b-π/4)的值就可以了
注意:cos(a+b),与cos(b-π/4)应该有两个值,一正一负
sin(a+b)的平方+cos(a+b)的平方=1,故cos(a+b)=正负4/5