F.4 Amath題

2009-04-06 3:36 am

E題係PAST PAPER 2003Q10.

Given two acute angle A and B.

a.Show that sinA + sinB / cosA + cosB =tan(A+B /2).

b.If 3sinA - 4cosA = 4cosB - 3sinB , find the value of tan(A+B).

我係唔明B果PART
我做到 tan(A+B /2)=4/3
之後唔識點做
點樣整走入面分母個2=.=??
識既可以詳細解釋比我聽嗎= . =?

更新1:

By Double angle formula , tan 2x = 2 tan x / (1- tan^2 x) we have tan ( A+B) = 2 tan ( A+B)/2 / [ 1-[ tan ( A+B)/2]^2 =2 (4/3) / [ 1- (4/3)^2] = -24 /7// E度開始唔明=.= 無啦啦點解走個tan2x出來既=.= 諗唔通=.=

回答 (1)

So Much

2009-04-06 4:24 am

✔ 最佳答案

(sinA + sinB )/ (cosA + cosB )

=[ sin (A+B)/2 cos (A-B)/2 ] / [ cos (A+B)/2 cos (A-B)/2]

=sin (A+B)/2 / cos (A+B ) /2

= tan (A+B)/2 //

b) 3sinA - 4cosA = 4cosB - 3sinB

=>3sinA+3sinB=4cosA+4cosB

=>( sin A + sin B) / ( cosA + cos B) =4/3

=> tan ( A+B)/2 = 4/3 --by part a

By Double angle formula , tan 2x = 2 tan x / (1- tan^2 x)

we have tan ( A+B)

= 2 tan ( A+B)/2 / [ 1-[ tan ( A+B)/2]^2

=2 (4/3) / [ 1- (4/3)^2]

= -24 /7//

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