A maths Trigonometry

1 In triangle ABC, if sin C= (sin A+sin B)/(cos A+cos B), prove that triangle ABC is a right triangle.
SOLUTION
　　　　sin A + sin B
sin C = -------------------- ---
　　　　cos A + cos B

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

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

sin C=(sin (90-C/2) cos(A- B)/2)/(cos (90-C/2) cos (A-B)/2)

sin C=cos (C/2) /sin (C/2)

2(sinC/2)(cosC/2)=co s (C/2) /sin (C/2)

sin^2(C/2)=1/2

So C/2=45

C=90

∆ABC is a right-angled triangle and C is the right-angled
Note: (A+B)/2=(180-C)/2=90-C/2
2
Find the maximum value and minimum value of 3/(4(cos x)^2+5)-1
SOLUTION
Because -1<=cos x<=1
0<=(cos x)^2<=1
0<=4(cos x)^2<=4
5<=4(cos x)^2+5<=9
1/9<=1/(4(cos x)^2+5)<=1/5
3/9<=3/(4(cos x)^2+5)<=3/5
1/3<=3/(4(cos x)^2+5)<=3/5
-2/3<=3/(4(cos x)^2+5)-1<=-2/5
So the maximum value is -2/5, the minimum value is -2/3

2007-04-08 15:27:12 補充：
http://hk.knowledge.yahoo.com/question/?qid=7007022600167佢同我既答案都一樣bor所以應該係書既答案錯