trigo q

3 sin α - 4 cos α = 4 cos β - 3 sin β
3 sin α + 3 sin β= 4 cos β+ 4 cos α
3 ( sin α + sin β)/ 4( cos α+ cos β) = 1
tan [(α+β)/2] = 4/3
tan [2 * (α+β)/2] = 2tan [(α+β)/2] / 1- tan ^2 [(α+β)/2]
(Identity of tan 2θ)
tan (α+β) = (2 * 4/3 ) / 1 - (4/3)^2
tan (α+β) = (8/3) / (-7/9)
tan (α+β) = -24/7

2012-05-12 17:37:59 補充：
Here I treat (α+β)/2 as θ
And so when it multiplies by 2, it becomes tan 2θ and it is tan (α+β) the term we want.

(a) sin α + sin β

= 2sin[(α + β)/2]cos[(α - β)/2]

cos α + cos β

= 2cos[(α + β)/2]cos[(α - β)/2]

So, (sin α + sin β) / (cos α + cos β) = tan [(α + β)/2]

(b) 3 sin α - 4 cos α = 4 cos β - 3 sin β

3 sin α + 3 sin β = 4 cos α + 4 cos β

3(sin α + sin β) = 4(cos α + cos β)

x = tan[(α+β)/2] = 4/3

tan(α+β) = 2x/(1 - x^2) = (8/3)/(7/9) = 24/7