"If 11γ=π then prove that sec⁡(γ)sec⁡(2γ)sec⁡(3γ)sec⁡(4γ)sec⁡(5γ)=32" was answered by our community

Naomi Hopkins

Answered question

2022-04-24

11 γ = π

then prove that

\sec (γ) \sec (2 γ) \sec (3 γ) \sec (4 γ) \sec (5 γ) = 32

Ezakhenitne

Beginner2022-04-25Added 10 answers

We can prove for positive integer N:

\cos N x = 2^{N - 1} \cos^{N} x + \dots

N = 2 n + 1

and

\cos (2 n + 1) x = 1, (2 n + 1) x = 2 m π

where m is any integer

x = \frac{2 m π}{2 n + 1}

where

m = 0, 1, 2, \dots, 2 n - 1, 2 n

So, the roots of

2^{2 n} \cos^{2 n + 1} x + \dots - 1 = 0

are

\cos \frac{2 m π}{2 n + 1}

where

m = 0, 1, 2, \dots, 2 n - 1, 2 n

\Rightarrow 2^{2 n} \prod_{m = 0}^{2 n} \cos \frac{2 m π}{2 n + 1} = {(- 1)}^{n}

Now

\cos (π - A) = - \cos A

\Rightarrow 2^{2 n} \prod_{r = 1}^{n} {(- 1)}^{n} \cos^{2} \frac{r π}{2 n + 1} = {(- 1)}^{n}

As for

1 \leq r \leq n, 0 \leq \frac{r π}{2 n + 1} < \frac{π}{2} \Rightarrow \cos \frac{r π}{2 n + 1} > 0

\Rightarrow \prod_{r = 1}^{n} \cos \frac{r π}{2 n + 1} = \frac{1}{2^{n}}

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