Find the (monic) minimal polynomial of with rational coefficients. Submit the coefficient of .
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If t = tan 1 5 π , then 5 t 4 − 1 0 t 2 + 1 t 5 − 1 0 t 3 + 5 t = tan ( 5 1 5 π ) = tan 3 1 π = 3 1 and hence 0 = 3 ( t 5 − 1 0 t 3 + 5 t ) 2 − ( 5 t 4 − 1 0 t 2 + 1 ) 2 = ( 3 t 2 − 1 ) ( t 8 − 2 8 t 6 + 1 3 4 t 4 − 9 2 t 2 + 1 ) which means that the minimum polynomial of t certainly divides X 8 − 2 8 X 6 + 1 3 4 X 4 − 9 2 X 2 + 1 . It turns out that this last polynomial is in fact irreducible, and hence is the minimum polynomial of t . Thus the answer is 1 3 4 .