a-maths~~~~~~~

1(a)
cos4θ
=cos(2θ+2θ)
=cos2θcos2θ-sin2θsin2θ
=(2cos^2θ-1)^2-(2sinθcosθ)^2
=4cos^4θ-4cos^2θ+1-4(1-cos^2θ)(cos^2θ)
=4cos^4θ-4cos^2θ+1+4cos^4θ-4cos^2θ
=8cos^4θ-8cos^2θ+1
(b)
令x=cosθ
則16x⁴-16x²+1=0化成
16cos⁴θ- 16cos²θ+1=0
16cos⁴θ- 16cos²θ+2=1
2cos4θ=1
cos4θ=1/2
4θ=60,300,420,660
θ=15,75,105,165
x=0.966,0.259,-0.259,-0.966


2007-11-21 22:01:41 補充：
補充:下面個位(a)part方法簡明些﹐值得參考﹐不過我做個時只係諗到cos4θ=cos(2θ+2θ)一定做到所需結果便是了。

a) cos4θ
= 2cos² 2θ - 1
= 2(2cos² θ - 1)² - 1
= 2(4cosθ⁴θ - 4cos² θ + 1) - 1
= 8cosθ⁴θ - 8cos² θ + 1

b) Let x = cos θ
Then, 16x⁴-16x²+1=0
=> 2(8cosθ⁴θ - 8cos² θ + 1) - 1 = 0
=> 2cos 4θ = 1
=> cos 4θ = 1/2
=> 4θ = 60, 300, 420, 660
=> θ = 15, 75, 105, 165
=> x = cos θ = 0.966, 0.259, -0.259, -0.959