"How to computeS=csc2π14+csc23π14+csc25π14" was answered by our experts

estroishave4

Answered question

2022-04-23

How to compute

S = \csc^{2} \frac{π}{14} + \csc^{2} \frac{3 π}{14} + \csc^{2} \frac{5 π}{14}

impire7vw

Beginner2022-04-24Added 19 answers

Let

s = \sin, c = \cos

and the subscript as in

s_{k}

denote

\sin (\frac{k π}{14})

and similiarly for other trigonometric ratios.

\csc^{2} \frac{π}{14} + \csc^{2} \frac{3 π}{14} + \csc^{2} \frac{5 π}{14}

= \frac{{(s_{3} s_{5})}^{2} + {(s_{5} s_{1})}^{2} + {(s_{1} s_{3})}^{2}}{{(s_{1} s_{3} s_{5})}^{2}}

= \frac{{(c_{2} - c_{8})}^{2} + {(c_{4} - c_{6})}^{2} + {(c_{2} - c_{4})}^{2}}{4 s_{1}^{2} s_{3}^{2} s_{5}^{2}}

Now numerator is:

{(c_{2} - c_{8})}^{2} + {(c_{4} - c_{6})}^{2} + {(c_{2} - c_{4})}^{2} = 2 c_{2}^{2} + 2 c_{4}^{2} + c_{6}^{2} + c_{8}^{2} - 2 (c_{2} c_{4} + c_{4} c_{6} + c_{2} c_{8})

= 1 + c_{4} + 1 + c_{8} + 1 + \frac{1}{2} (c_{12} + c_{16}) - 2 (c_{6} + c_{2} + c_{10})

= 3 + c_{4} + c_{8} - c_{2} - 2 (c_{2} + c_{6} + c_{10})

= 3 (1 - c_{2} + c_{4} - c_{6})

= \frac{3}{2}

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