Question

已知sin[a-(b/2)]=4/5,cos[(a/2)-b]=-12/13 ,且a-b/2和a/2-b分别是第二和第三象限角，求 tan[(a+b)/2]的值

Answer 1

我算出来的答案是63/33

sin((a+b)/2)= sin((a-b/2)-(a/2-b))=sin[a-(b/2)]*cos[(a/2)-b]-sin[(a/2)-b]*cos[a-(b/2)]=4/5*(-12/13)-(-3/5)*(-5/13)=-63/65
同理得，cos((a+b)/2)=(-3/5)*(-12/13)+4/5*(-5/13)=-33/65
tan[(a+b)/2]=63/33

Answer 2

sin[a-(b/2)]=4/5 tan[a-(b/2)]=-4/3
cos[(a/2)-b]=-12/13 tan[(a/2)-b]=5/12
tan[(a+b)/2]=(tan[a-(b/2)]- tan[(a/2)-b])/(1+tan[a-(b/2)]* tan[(a/2)-b])
=(-4/3-5/12)/(1-4/3*5/12)=-63/16

Answer 3

sin[a-(b/2)]=4/5
cos[a-(b/2)]=-3/5
cos[(a/2)-b]=-12/13
sin[(a/2)-b]=-5/13

sin((a+b)/2)=sin((a-b/2)-(a/2-b))=4/5*(-12/13)-(-3/5)*(-5/13)=33/65
cos((a+b)/2)=(-3/5)*(-12/13)+4/5*(-5/13)=16/65
tan=33/16