已知sina+sinb=1/4,cosa+cosb=1/3,求:
(1)cos2a
(2)tanatanb

1)sina+sinb=2sin[(a+b)/2]cos[(a-b)/2]=1/4……(1)
cosa+cosb=2cos[(a+b)/2]cos[(a-b)/2]=1/3……(2)
(1)/(2):
tan[(a+b)/2]=3/4
sin(a+b)=2tan[(a+b)/2]/{1+tan²[(a+b)/2]}=24/25
cos(a+b)=7/25
(1)*(2):
sinacosa+sinacosb+cosasinb+sinbcosb
=(sin2a+sin2b)/2+sin(a+b)
=sin(a+b)cos(a-b)+sin(a+b)
=[cos(a-b)+1]24/25=1/12
cos(a-b)=-263/288
cos2a=cos(a+b+a-b)=cos(a+b)cos(a-b)-sin(a+b)sin(a-b)
带入可计算出cos2a

2)(1)*(1)
sin²a+2sinasinb+sin²b=1/16
2sinasinb=1/16-sin²a-sin²b
`````````=1/16-(1-cos2a)/2-(1-cos2b)/2
`````````=-15/16+(cos2a+cos2b)/2=-15/16+cos(a+b)cos(a-b)……(3)
(2)*(2)
cos²a+2cosacosb+cos²b=1/9
2cosacosb=1/9-(1+cos2a)/2-(1+cos2b)/2
`````````=-8/9-cos(a+b)cos(a-b)……(4)
(3)/(4)
tanatanb=[-15/16+cos(a+b)cos(a-b)]/[-8/9-cos(a+b)cos(a-b)]