Question

3cos(2a+b)+5cosb=0，则tan(a+b)tana=?

Answer 1

3cos(2a+b)+5cosb=0
3cos2acosb-3sin2asinb=-5cosb
cos2acosb-sin2asinb=-5/3cosb
2(cosa)^2 cosb-cosb-sin2asinb=-5/3cosb
2(cosa)^2 cosb-sin2asinb=-2/3cosb
(cosa)^2 cosb-0.5sin2asinb=-1/3cosb (1)
[1-(sina)^2] cosb-0.5sin2asinb=-1/3cosb
sina)^2cosb+0.5sin2asinb=4/3cosb (2)
tan(a+b)tana=(2)/(1)=-4
望采纳呵呵

Answer 2

3cos(2a+b)+5cosb
=3cos[(a+b)+a]+5cos[(a+b)-a]
=3[cos(a+b)cosa-sin(a+b)sina]+5[cos(a+b)cosa+sin(a+b)sina]
=8cos(a+b)cosa+2sin(a+b)sina
=0
8cos(a+b)cosa=-2sin(a+b)sina
tan(a+b)tana=[sin(a+b)sina]/[cos(a+b)cosa]=-4

Answer 3

3cos(2A+B)+5cosB
=3cos[(A+B)+A]+5cos[(A+B)-A]
=3[cos(A+B)cosA-sin(A+B)sinA]+5[cos(A+B)cosA+sin(A+B)sinA]
=8cos(A+B)cosA+2sin(A+B)sinA
=0
sin(A+B)sinA=-4cos(A+B)cosA
tan(A+B)*tanA=-4