why is this nontrivial? find cubic poly w/ roots cos{2pi/7}, cos{4pi/7}, cos{6pi/7}

caffeineinmyveins

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I recently came across a problem that was marked as a hard one (and the book does a good job at picking hard problems).

Find a cubic polynomial whose roots are cos{2pi/7}, cos{4pi/7}, and cos{6pi/7}.

The solution given in the solutions manual painstakingly constructs the polynomial 8x^3+4x^2-4x-1, but without any other restrictions on the problem, why can't you just make something like (x-cos{2pi/7})(x-cos{4pi/7})(x-cos{6pi/7})? I don't understand. Do you think the authors were missing some extra conditions, like the condition that the coefficients had to be integers? Or is there something wrong with the trivial solution?
 
I recently came across a problem that was marked as a hard one (and the book does a good job at picking hard problems).

Find a cubic polynomial whose roots are cos{2pi/7}, cos{4pi/7}, and cos{6pi/7}.

The solution given in the solutions manual painstakingly constructs the polynomial 8x^3+4x^2-4x-1, but without any other restrictions on the problem, why can't you just make something like (x-cos{2pi/7})(x-cos{4pi/7})(x-cos{6pi/7})? I don't understand. Do you think the authors were missing some extra conditions, like the condition that the coefficients had to be integers? Or is there something wrong with the trivial solution?

I imagine they are taking "polynomial" to mean explicitly showing the coefficients as numbers, not implicitly in terms of trig functions.

If they had said "polynomial with integer coefficients", it would have been clearer what the goal is, and would also motivate you to expect a nice answer. But that condition is not necessary, as long as it is clear that the polynomial is to be exhibited in standard form.
 
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