Question

高一数学题

已知cos(a+b)=4/5,sin(a-b)=3/5,3π/2<a+b<2π，π/2<a-b<π，求cos2b的值。
写明步骤啊~~~求高人指点，谢谢！！！！

Answer 1

解：
因为：3π/2<a+b<2π,π/2<a-b<π
所以：sin(a+b)<0
sin(a-b)>0

由：(sinx)^2+(cosx)^2=1
所以：sin(a+b)=-3/5,sin(a-b)=3/5
cos2b=cos[(a+b)-(a-b)]
=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
=-16/25-9/25
=-1

Answer 2

cos(a+b)=4/5, sin(a+b)=-3/5
cos(a-b)=-4/5 sin(a-b)=3/5
cos2b
=cos[(a+b)-(a-b)]
=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)
=(-16/25)+(-9/25)
=-1

Answer 3

sin(a+b)= - 3/5 cos(a-b)= -4/5
cos2b=cos((a+b)-(a-b))=cos(a+b)cos(a-b)+sin(a+b)sin(a-b)=-1