MQ55 --- Trigonometry

csc²(π/14) + csc²(3π/14) + csc²(5π/14)
= 1 + ctg² π/14 + 1 + ctg² 3π/14 + 1 + ctg² 5π/14
= 3 + tan² 3π/7 + tan² 2π/7 + tan² π/7
Let θ = kπ/7 (k = 1 , 2 , 3) , then
tan 3θ + tan 4θ = 0
(tan θ + tan 2θ ) / (1 - tan θ tan 2θ) + 2tan 2θ / (1 - tan² 2θ)
tan θ + 3tan 2θ - 3tan θ tan² 2θ - tan³ 2θ = 0 ...... (*)
Let tan θ = x , by (*) :
x + 6x / (1 - x²) - 12x³ / (1 - x²)² - 8x³ / (1 - x²)³ = 0
(1 - x²)³ + 6(1 - x²)² - 12x²(1 - x²) - 8x² = 0
x⁶ - 21x⁴+ 35x² - 7 = 0
Therefore tan kπ/7 (k = 1 , 2 , 3) are 3 roots of x⁶ - 21x⁴+ 35x² - 7 = 0 ,

i.e. tan² kπ/7 (k = 1 , 2 , 3) are 3 roots of x³ - 21x² + 35x - 7 = 0.
∴ tan² π/7 + tan² 2π/7 + tan² 3π/7 = 21 ,
3 + tan² π/7 + tan² 2π/7 + tan² 3π/7 = 21 + 3
csc²(π/14) + csc²(3π/14) + csc²(5π/14) = 24

2012-11-24 18:12:29 補充：
Line 6 :
(tan θ + tan 2θ ) / (1 - tan θ tan 2θ) + 2tan 2θ / (1 - tan² 2θ) = 0

2012-11-29 20:03:12 補充：
Because
tan3θ + tan4θ
= (1 - tan3θ tan4θ) tan7θ
= (1 - tan3θ tan4θ) tan kπ
= 0
for any value of k.

And we can easy to rewrite tan3θ and tan4θ in terms of tan θ and tan2θ.

Using complex number

2012-11-28 15:09:36 補充：
Using
[(1 + i tan x) / (1 - i tan x)]^7 = (1 + I tan 7x) / (1 - i tan 7x)