maths~~~

a)
i)
AC 與 A'C' 所成銳角 = π - θ - ㄥDAC = π - θ - 2π/3 = (π/3 - θ)
所以A'C' = AC cos (π/3 - θ)
A'C' = x cos (π/3 - θ)
ii)
AB 與 A'B' 所成銳角 = θ
所以A'B' = AB cosθ
A'B' = x cosθ
iii)
BC 與 B'C' 所成銳角 = π - θ - ㄥABC = π - θ - π/3 = ( 2π/3 - θ )
所以 B'C' = BC cos (2π/3 - θ)
B'C' = x cos (2π/3 - θ)
b)
如圖 A'B' + B'C' = A'C' , 由 a) 結果 ,
x cosθ + x cos (2π/3 - θ) = x cos (π/3 - θ)
因 x<>0 , 約去 x 得
cosθ + cos (2π/3 - θ) = cos (π/3 - θ) ......*
c)i) 令 * 中 θ = π/4 ,
cos π/4 + cos (2π/3 - π/4) = cos (π/3 - π/4)
√2 / 2 + cos (5π/12) = cos (π/12)
cos (π/12) - cos (5π/12) = √2 / 2
cos (π/12) - sin (π/2 - 5π/12) = √2 / 2
cos (π/12) - sin (π/12) = √2 / 2 ......☆
ii)
☆ 式 兩邊平方 :
cos^2 (π/12) + sin^2 (π/12) - 2 cos (π/12) sin (π/12) = 1/2
1 - 2 cos (π/12) sin (π/12) = 1/2
2 cos (π/12) sin (π/12) = 1/2
又
[cos (π/12) + sin (π/12)]^2
= cos^2 (π/12) + sin^2 (π/12) + 2 cos (π/12) sin (π/12)
= 1 + 1/2 = 3/2
cos (π/12) + sin (π/12) = +/- √(3/2) = √6 / 2
或 - √6 / 2 (捨去因 π/12 於第一象限)
cos (π/12) + sin (π/12) = √6 / 2 ........¢
d)
☆ + ¢ :
2 cos (π/12) = √2 / 2 + √6 / 2
cos (π/12) = (√2 + √6) / 4
sec (π/12) = 4 / (√2 + √6)
sec (π/12) = 4(√6 - √2) / (6 - 2)
sec (π/12) = √6 - √2