amath.....trigo [複製鏈接]

1)find the general solution of 3sec^2(x) - 5 tan(x) -5=0

2)In triangle ABC
   if  
        sin C = (sinA + sinB)\(sinA+cosB)
   prove triangle ABC is an right-angle triangle.....

2題都寫錯- -
第一條你唔比條EQUATION係幾多點搵
3sec^2(x) - 5 tan(x) -5 = ?
第二題係邊個over邊個呀- -
用 '/' 啦
第2題有條類似
不過係 sinC = (sin A + sinB) / (cos A + cos B)

1 3 sec²x-5 tan x-5=0
  3 tan²x+3-5 tan x-5=0
   3 tan²x-5 tanx-2=0
(3 tanx+1)(tan x-2)=0
x=180^on+(tan^-1 -1/3) or 180^on+tan^-1 2  where n is an integer---其實我習慣用radian,不過也是用degree算
2. 我覺得可能是寫錯吧---所以用了上面的計
sinC = (sin A + sinB) / (cos A + cos B)
sin C = 2sin[(A+B)/2]cos[(A-B)/2]/2cos[(A+B)/2]cos[(A-B)/2]
sin C = tan[(A+B)/2]
sinC = tan(90^o-C/2) since ABC is a triangle (0^o<C/2<90^o)
2sin²(C/2)=1
sin(C/2)=2^-0.5 since ABC is a triangle,the negative value is then rejected
C/2=45^o or 135^o (rejected) since ABC is a triangle
C=90^o
ABC is a right-angled triangle with the right angle C