Question

已知sina+sinb=1/4，cosa+cosb=1/3，则tan(a+b)的值为

Answer 1

(sina+sinb)(cosa+cosb)=1/12，sinacosa+sinbcosb+sin(a+b)=1/12，sin(a+b)cos(a-b)+sin(a+b)=1/12；(sina+sinb)²+(cosa+cosb)²=1/16+1/9，2cos(a-b)+2=25/16*9，cos(a-b)=-263/288，代入sin(a+b)cos(a-b)+sin(a+b)=1/12，得：sin(a+b)=24/25；
(cosa+cosb)²-(sina+sinb)²=1/9-1/16，2cos(a+b)+2cos(a+b)cos(a-b)=7/16*9，得：cos(a+b)=7/25，则tan(a+b)=24/7

Answer 2

sinA+sinB=1/4(1)
cosA+cosB=1/3(2)
由（1）^2-（2）^2并整理得
2sinAsinB-2cosAcosB=25/144
即2cos(A+B)=-25/144
解得cos(A+B）=-25/288
根据[sin(A+B)]^2+[cos(A+B)]^2=1得
sin(A+B)=±√13
tan(A+B)=sin(A+B)/cos(A+B)
=±√13/(-25/288)
=±288√13/25

Answer 3

(sina+sinb)(cosa+cosb)=1/12，sinacosa+sinbcosb+sin(a+b)=1/12，sin(a+b)cos(a-b)+sin(a+b)=1/12；(sina+sinb)²+(cosa+cosb)²=1/16+1/9，2cos(a-b)+2=25/16*9，cos(a-b)=-263/288，代入sin(a+b)cos(a-b)+sin(a+b)=1/12，得：sin(a+b)=24/25；
(cosa+cosb)²-(sina+sinb)²=1/9-1/16，2cos(a+b)+2cos(a+b)cos(a-b)=7/16*9，得：cos(a+b)=7/25，则tan(a+b)=24/7

Answer 4

2sin(a+b)/2cos(a-b)/2=1/4

cosa+cosb=1/3
2cos(a+b)/2cos(a-b)/2=1/3

tan(a+b)/2=3/4
tan(a+b)=2tan(a+b)/2 /1-tan(a+b)/2平方
=24/7

Answer 5

2sin(a+b)/2cos(a-b)/2=1/4
cosa+cosb=1/3
2cos(a+b)/2cos(a-b)/2=1/3
tan(a+b)/2=3/4
tan(a+b)=2tan(a+b)/2 /1-tan(a+b)/2平方
=24/7