Question

已知π/4<a<b<π/2,且sin(a+b)=4/5,cos(a-b)=12/13

1、用a+b,a-b表示2a；
2、求cos2a,sin2a,tan2a的值。
【需过程】

Answer 1

2a=a+b+a-b;
cos2a=cos[(a+b)+(a-b)]
=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
=3/5*12/13-4/5*5/13
=16/65;
sin(2a)=sin(a+b)cos(a-b)+sin(a-b)*cos(a+b)
=4/5*12/13+3/5*5/13
=(48+15)/65=63/65
tan2a=sin2a/cos2a=63/16;