Proving that sec⁡π30=2−5+15−65 which I instantly wanted to prove. I know that I can "reduce" the problem to the evaluation of cos⁡π15, as the rest is easy with the use of the half-angle formula. I know that cos obeys the nice relation cos⁡nx=Tn(cos⁡x) where Tn(x)=n2∑k=0|n2|(−1)kn−k(n−kk)(2x)n−2k Thus, setting t=cos⁡π15, T15(t)=−1 The only thing left to do is solve for t. We can narrow down our search to the values 0&lt;t&lt;1. I have never dealt with degree-15 polynomials before, so I was hoping one of you could help me out.

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Proving that sec⁡π30=2−5+15−65  which I instantly wanted to prove.  I know that I can &quot;reduce&quot; the problem to the evaluation of cos⁡π15, as the rest is easy with the use of the half-angle formula.  I know that cos obeys the nice relation  cos⁡nx=Tn(cos⁡x)  where  Tn(x)=n2∑k=0|n2|(−1)kn−k(n−kk)(2x)n−2k  Thus, setting t=cos⁡π15,  T15(t)=−1  The only thing left to do is solve for t. We can narrow down our search to the values 0&amp;lt;t&amp;lt;1.  I have never dealt with degree-15 polynomials before, so I was hoping one of you could help me out.

Madelyn Townsend · Accepted Answer

Another way:  cos⁡6∘=+1+cos⁡12∘2  cos⁡(30∘−18∘)=?  Use Exact Value to find cos⁡18∘,sin⁡18∘

Proving that \sec \frac{\pi}{30}=\sqrt{2-\sqrt5+\sqrt{15-6\sqrt5}} which I instantly wante

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