We all know that cos⁡(π6)=32, cos⁡(π4)=22 and cos⁡(π3)=12. One can also prove that cos⁡(π5)=5+14. But it seems that cos⁡(π7) cannot be put in a closed form algebraic expression. Is there a proof of such a claim? I feel like this has to do with the Abel–Ruffini theorem.The same for cos⁡(π8) and cos⁡(π9). But cos⁡(π10)=58+58

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We all know that cos⁡(π6)=32, cos⁡(π4)=22 and cos⁡(π3)=12. One can also prove that cos⁡(π5)=5+14. But it seems that cos⁡(π7) cannot be put in a closed form algebraic expression. Is there a proof of such a claim? I feel like this has to do with the Abel–Ruffini theorem.The same for cos⁡(π8) and cos⁡(π9). But cos⁡(π10)=58+58

Friegordigh7r7 · Accepted Answer

For any n≥3,  cos⁡(2πn) is al algebraic number over Q with degree φ(n)2, since its minimal polynomial can be computed from Φn(x) in a straightforward way. Our case is given by n=14, where the minimal polynomial of cos⁡(π7) is given by q(x)=8x3−4x2−4x+1. By letting ω=−1+−272, through the cubic formula we getcos⁡(π7)=16(1+723ω13+713ω13)It might be interesting to notice that the LHS is very close to 910

We all know that \cos(\frac{\pi}{6})=\frac{\sqrt 3}{2}, \ \cos(\frac{\pi}{4})=\frac{\sqrt 2}{2}ZS

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