# trigo q

Given two acute angles α and β.

a) Show that (sin α + sin β) / (cos α + cos β) = tan [(α+β)/2]

b) If 3 sin α - 4 cos α = 4 cos β - 3 sin β, find the value of tan(α+β)

Please do (b).

Thanks.

Why is it 2x/(1 - x^2)?

### 2 Answers

- 8 years agoFavorite Answer
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.

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- myisland8132Lv 78 years ago
(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

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