Question

sina=3/5,sin(a+B)=5/13(a,B为第一象限角)则cosB= 要过程，谢谢

Answer 1

sina=3/5
cosa=4/5
sin(a+b)=5/13
cos(a+b)=+(-)12/13
cosb=cos(a+b)*cosa-sin(a+b)*sina
=+(-)12/13*4*5-5/13*3/5
=(-15+(-)48)/65
∵b在第一象限，所以cosb大于0
所以cosb=33/65

Answer 2

cosa=4/5>0
2k1pai<a<2k1pai+pai/2
2k2pai<b<2k2pai+pai/2,k1,k2:Z
(k1+k2)2pai<a+b<(k1+k2)*2pai+pai
k=k1+k2:Z
2kpai<a+b<2kpai+pai
a+b在一，二象限
cos(a+b)=12/13orcos(a+b)=-12/13
1.cos(a+b)=12/13
cosb=cos((a+b)-a)=cos(a+b)cosa+sin(a+b)sina
=12/13*4/5+5/13*3/5=48/65+15/65=63/65
2.cos(a+b)=-12/13
cosb=cos(a+b)-a)=cos(a+b)cosa+sin(a+b)sina
=-12/13*4/5+5/13*3/5=15/65-48/65=-33/65
所以cosb=63/65orcosb=-33/65
两解

Answer 3

a,B为第一象限角 sina=3/5 cosa=4/5 sin(a+B)=5/13 0<a+B< π
cos(a+B)=12/13 或cos(a+B)=-12/13
cosB=cos[(a+B)-a]=cos(a+B)xcosa+ sin(a+B)xsina=12/13 x4/5 +5/13x3/5=63/65
或cosB=cos[(a+B)-a]=cos(a+B)xcosa+ sin(a+B)xsina=-12/13 x4/5 +5/13x3/5=-33/65