中四附加數

1) sin θ=t-1/t+1
sin θ = y/r
y = t-1
r = t+1
x = √(t+1)^2 - (t-1)^2)
x = √4t
x = 2√t
由於係第2個象限,所以cos同cot都係-ve.
cos θ = - x/r
= - 2√t / t+1
cot θ = x/y
= - 2√t / t-1

2007-11-19 20:57:02 補充：
2) for θ , 由於係第2個象限 所以 x <= 0 , y >= 0 cos θ = x / r x = -1 , r = 2 y = √3sec θ = r/x = 2/-1cos θ = -1/2 for α ,由於係第3個象限 所以 x <= 0 , y <=0cot α = 1 x = y r = √x^2 + y^2 r = x√2sin α= y / r = y / x√2csc α= r / x = x√2 / x

2007-11-19 20:57:29 補充：
(2sinα-secθ/2cosθ-cscα)= (2y / x√2 - (2/-1))/(-1 - x√2 / x )= 2/√2 + 2 / -1 - √2 ( 由於 x = y)= 2√2 / 2 + 2 / -1 - √2=√2 + 2 / -1 -√2= (√2 + 2 )(-1+√2) / (-1-√2 )(-1+√2)= √2 頭先食飯 , 所以咁遲 sorry

B
|\
|. \
|....\
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|__θ(_\
A ... ... C

1) ABC為直角3角形,
因為sinθ=(t-1)/(t+1)
所以 AB=t-1 , BC=t+1
AB^2= BC^2 + AC^2
所以 AC=2(t)^1/2

所以 cos θ = -[2(t)^1/2] ／ (t+1) (因為π/2< θ< π,所以θ係第2區間) (所以cos θ & tan θ係負)
cotθ = 1/tanθ = cos θ/sin θ = -[2(t)^1/2] ／ (t-1)

2)sec θ=1/cosθ= -2

因為π< α< 3π/2　　，　　cot α=1
　　　　　　　所以　　α= 225 °
所以sinα=sin225°= - 1/(2)^(1/2) & cscα=1/sinα= - 2^(1/2)

(2sinα-secθ/2cosθ-cscα)={ 2[- 1/(2)^(1/2) ] - (-2) } / { 2 (-1/2) - [ - 2^(1/2) ] }
=2^(1/2)