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Prove sec2π7+sec22π7+sec23π7=24 using the roots of a polynomial

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Priya Roy


( 5 months ago )

I have to

prove sec2π7+sec22π7+sec23π7=24sec2⁡π7+sec2⁡2π7+sec2⁡3π7=24 by using the roots of the polynomial x321x2+35x7=0x3−21x2+35x−7=0

I tried to factor the polynomial but it didn't work and later found it cannot factorize with rationals. and I saw some similar problems in StackExchange. But the answers are very complex to me. I cannot use Euler's complex number formula since it's not in the syllabus. I do not want the exact answer but guidance to the answer.

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