Evaluate: csc 2 ⁡ ( π 9 ) + csc 2 ⁡ ( 2 π 9 ) + csc 2 ⁡ ( 4 π 9 ) = ???

Question

Evaluate:      csc    2    ⁡      (          π      9        )    +      csc    2    ⁡      (                  2        π            9        )    +      csc    2    ⁡      (                  4        π            9        )      =    ???

Turynka2f · Accepted Answer

Like my answer here in Sum of tangent functions where arguments are in specific arithmetic series

\tan 9 x = \frac{(\binom{9}{1}) t - (\binom{9}{3}) t^{3} + (\binom{9}{5}) t^{5} - (\binom{9}{7}) t^{7} + t^{9}}{\dots}

where

t = \tan x

If

\tan 9 x = 0, 9 x = n π

where n is any integer
So, the roots of

(\binom{9}{1}) t - (\binom{9}{3}) t^{3} + (\binom{9}{5}) t^{5} - (\binom{9}{7}) t^{7} + t^{9} = 0

are

\tan \frac{r π}{9}

where

n \equiv 0, \pm 1, \pm 2, \pm 3, \pm 4 (\mod 9)

As

\tan 0 = 0,

the roots of

(\binom{9}{1}) - (\binom{9}{3}) t^{2} + (\binom{9}{5}) t^{4} - (\binom{9}{7}) t^{6} + t^{8} = 0

are

\tan \frac{r π}{9}

where

r \equiv \pm 1, \pm 2, \pm 3, \pm 4 (\mod 9)

Putting

c = 1 / t,

the roots of

(\binom{9}{1}) c^{8} - (\binom{9}{3}) c^{6} + (\binom{9}{5}) c^{4} - (\binom{9}{7}) c^{2} + 1 = 0

are

\cot \frac{r π}{9}

where

r \equiv \pm 1, \pm 2, \pm 3, \pm 4 (\mod 9)

\sum_{r = 1}^{4} \cot^{2} \frac{r π}{9} = \frac{(\binom{9}{3})}{(\binom{9}{1})} = ?

But

r = 3 ⟹ \cot^{2} \frac{3 π}{9} = 3

Use

\csc^{2} u = 1 + \cot^{2} u

Evaluate: csc 2 </msup> ⁡ ( π

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