Question

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

要求详细的解题过程，在解题过程中求出的几个值正负问题的解释

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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）

Answer 5

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！

Answer 6

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