Question

已知sinA=12/13,cosB=-3/5,A B均为第二象限角,求cos(A-B),cos(A+B)的值·？

Answer 1

cosA=-5/13 sinB=4/5
cos(A-B)=cosAcosB+sinAsinB=(-5/13)*(-3/5)+(12/13)*(4/5)=63/65
cos(A+B)=cosAcosB-sinAsinB=(-5/13)*(-3/5)-(12/13)*(4/5)=-33/65,5,因为在第二象限，cos<0,sin>0
cosA=-5/13,sinB=4/5
cos(A-B)=cosAcosB+sinAsinB
=-5/13*(-3/5)+12/13*4/5
=63/65
cos(A+B)=cosAcosB-sinAsinB
=-5/13*(-3/5)-12/13*4/5
=-33/65
答案就是这样,2,sinA=12/13,cosB=-3/5,A B均为第二象限角，所以cosA=-5/13，sinB=4/5
所以cos(A-B)=cosAcosB+sinAsinB=63/65,
cos(A+B)=cosAcosB-sinAsinB=-33/65,2,Sin2a+cos2a=1,所以由题得cosa=(1-(12/13)^2)开根号,又因为a.b在第二象限,所以cosa=-5/13,同理可得sinb=4/5,因为cos(a-b)=cosaco *** +sinasin,2,