F.4 m2 ~~20 marks...

http://img134.imageshack.us/img134/3376/triangle1.png

圖片參考：http://img134.imageshack.us/img134/3376/triangle1.png

(a)(1) Since ABC is equilateral, angle BAC = π/3
Draw a line AG parallel to DE, angle BAG = θ
y = π/3 - θ
Angle ACF = y = π/3 - θ
PR = AC cos y = x cos(π/3 - θ)
(2) PQ = AB cos BAGv
= xcosθ
(3) Angle ABQ = π/2 - θ
z = π/3 - (π/2 - θ)
= θ - π/6
QR = BC sinz
= xsin(θ - π/6)
= xcos(π/2 - θ + π/6)
= xcos(2π/3 - θ)
(b) since PR = PQ + QR
x cos(π/3 - θ) = xcosθ + xcos(2π/3 - θ)
cosθ + cos(2π/3 - θ) = cos(π/3 - θ)
(c) (1) when θ = π/4,
cos(π/4) + cos(2π/3 - π/4) = cos(π/3 - π/4)
(√2)/2 + cos(5π/12) = cos(π/12)
(√2)/2 + sin(π/2 - 5π/12) = cos(π/12)
(√2)/2 + sin(π/12) = cos(π/12)
cos(π/12) - sin(π/12) = (√2)/2 ... (1)
(2) [cos(π/12) - sin(π/12)]^2 = 1/2
1 - 2sin(π/12)cos(π/12) = 1/2
2sin(π/12)cos(π/12) = 1/2
1 + 2sin(π/12)cos(π/12) = 3/2
[cos(π/12) + sin(π/12)]^2 = 6/4
cos(π/12) + sin(π/12) = (√6)/2 ...(2)
(d) (1) + (2) => 2cos(π/12) = (√2)/2 + (√6)/2
cos(π/12) = (√2 + √6)/4
sec(π/12) = 4/(√6 + √2)
sec(π/12) = 4(√6 - √2)/(6 - 2)
sec(π/12) = √6 - √2