interesting (sin, cos, tan)

i)a2 / a1 = a3 / a2
cos x / sin x = tan x / cos x
cos x / sin x = (sin x / cos x) / cos x
sin² x = cos³ x ..................................... (1)
1 - cos² x = cos³ x
cos² x (cos x + 1) = 1
cos x + 1 = 1 / cos² x ...........................(2)Common ratio = cos x / sin x , then ap = a1 * (cosx / sinx)ᵖ⁻¹ ,
i.e.
sin x * (cos x / sin x)ᵖ⁻¹ = cos x + 1
sin x * (cos x / sin x)ᵖ⁻¹ = 1 / cos² x .... [By (2)]
(sin x)²⁻ᵖ = (cos x)-ᵖ⁻¹
(2 - p) : (- p - 1) = 2 : 3 ..................... [By (1)]
p = 8


ii)By i) , cos³ x + cos² x - 1 = 0 Let (y - 1/3) be cos x , then
(y - 1/3)³ + (y - 1/3)² - 1 = 0
y³ - 3(1/3)y² + 3(1/9)y - 1/27 + y² - (2/3)y + 1/9 - 1 = 0
y³ - (1/3)y - 25/27 = 0(a + b)³ = a³ + b³ + 3ab(a + b) , let a + b be y , then y³ - (3ab)y - (a³ + b³) = 0Comparing coefficients :3ab = 1/3 ..............(3)
{
a³ + b³ = 25/27 .....(4)By (3) , b = 1/(9a) , sub. into (4) ,a³ + 1/(729a³) = 25/27
729(a³)² - 675a³ + 1 = 0
a³ = (25 ± 3√69) / 54
then
b³ = (25/27 - (25 ± 3√69) / 54) = (25 干 3√69) / 54Therefore y = a + b = ∛((25 ± 3√69) / 54) + ∛((25 干 3√69) / 54)
i.e.
y = ∛((25 + 3√69) / 54) + ∛((25 - 3√69) / 54) = 1.088211cos x = 1.088211 - 1/3 = 0.754877667arccos x = 0.71533 x = 2kπ ± arccos x = 2kπ ± 0.71533 , (k ∈ Z)

(i)
r=cosx/sinx=tanx/cosx=sinx/cos²x
cos³x=sin²x=1-cos²x
a4=rtanx=1,Tn=r^(n-4)
(1-cos²x)(1+cosx)=1+cosx-cos²x-cos³x=cosx
1+cosx=cosx/sin²x=(cosx/sin²x)(cos²x/sinx)r=rcos³x/sin³x
=r^4
ap=r^(p-4)=r^4
p=8

(ii)
cos³x+cos²x-1=0
cosx=(-1/3)+[(25+√621)^(1/3)+(25-√621)^(1/3)]/[3(2)^(1/3)]
x=40.985

2012-04-14 16:49:17 補充：
...002 is right,it is the general solution
x = 6.28319n ± 0.71533
By the way,I skip the steps of solving cos³x+cos²x-1=0 because there has a formular:
http://zh.wikipedia.org/zh-hk/%E4%B8%89%E6%AC%A1%E6%96%B9%E7%A8%8B