Question

sina+sinb=1/4,cosa+cosb=1/3.求sin(a+b),cos(a-b).要详细的过程！~急求，速度！！！~

sina+sinb=1/4,cosa+cosb=1/3.求sin(a+b),cos(a-b).

要详细的过程！~急求，速度！！！~

Answer 1

sina+sinb=2sin[(a+b)/2]cos[(a-b)/2]=1/4;
cosa+cosb=2cos[(a+b)/2]cos[(a-b)/2]=1/3;
上下两返棚者式相除,得
tan[(a+b)/2]=3/4;
sin(a+b)=2sin[(a+b)/2]cos[(a+b)/2]
=2sin[(a+b)/2]cos[(a+b)/2]/{sin[(a+b)/2]^2+cos[(a+b)/和粗2]^2}
=2tan[(a+b)/2]/(tan[(a+b)/2]^2+1)
=2*3/4/漏薯[(3/4)^2+1]
=24/25

(sina+sinb)^2=(sina)^2+(sinb)^2+2*sinasinb=1/16
(cosa+cosb)^2=(cosa)^2+(cosb)^2+2*cosacosb=1/9
以上两式相加，得1+1+2*（cosacosb+sinasinb）=25/144
cos(a-b)=cosacosb+sinasinb=-263/288