Math Challenge, Experts Only?

Question

Prove that for all integers n, Tan(nπ/7) are roots to the equation:

x^7 - 21x^5 + 35x^3 - 7x = 0

Answer 1

tan(x + y) =
= (sinx cosy + cosx siny) / (cosx cosy - sinx siny) =
= (tanx + tany) / (1 - tanx tany)

tanx = z

tan(2x) =
= tan(x + x) =
= 2z/(1 - z²)

tan(3x) =
= (tan(2x) + tan(x))/(1 - tan(2x)tan(x)) =
= (2z/(1 - z²) + z)/(1 - 2z/(1 - z²) z) =
= (2z + (1 - z²)z)/((1 - z²) - 2z²) =
= z(3 - z²)/(1 - 3z²)

tan(4x) =
= 2tan(2x)/(1 - tan²(2x)) =
= 2[2z/(1 - z²)] /[1 - (2z/(1 - z²)²)] =
= 4z(1 - z²) / [(1 - z²)² - 4z²)] =
= 4z(1 - z²) / (1 - 6z² + z^4)

tan(7x) =
= (tan(3x) + tan(4x)) / (1 - tan(3x)tan(4x)) =
= (z(3 - z²)/(1 - 3z²) + 4z(1 - z²)/(1 - 6z² + z^4))/denominator =
=1/d (z(3 - z²)(1 - 6z² + z^4) + 4z(1 - z²)(1 - 3z²)] =
=1/d z[ 3 - 18z² + 3z^4 - z² + 6z^4 - z^6 + 4 - 12z² - 4z² + 12z^4]=
= -1/d (z^7 - 21z^5 + 35z^3 - 7z)


Therefore if x = nπ/7, then
0 = tan(nπ) = tan(7nπ/7) = tan(x) =
= -1/d (z^7 - 21z^5 + 35z^3 - 7z)

Answer 2

the gap is the hypotenuse of the perfect triangle with one leg alongside the top of the prism and the different leg fashioned by using the hypotenuse of the perfect triangle fashioned by using the width and length of the prism. it truly is a tad difficult, I understand: Take the 3x5 oblong base of the prism. Draw a diagonal there. This diagonal, by using the Pythagorean theorem, could have the size of sqrt(3^2 + 5^2). i visit call this x. x = sqrt(9 + 25) x = sqrt(34) this is going to be between the legs of the only appropriate appropriate triangle. the different leg of the perfect triangle would be the size of the prism no longer yet used - the top. i'm going to call this y. y = 6. x and y are at appropriate angles. which ability the the on the instant line connecting them is the hypotenuse of the triangle. This hypotenuse, which i pass to call z, is nearly a "diagonal" of the forged. So, lower back, use the Pythagorean theorem: z^2 = (sqrt(34))^2 + 6^2 z^2 = 34 + 36 z^2 = 70 z = sqrt(70) it truly is the gap you're in seek of, sqrt(70) contraptions.

Answer 3

just graph them on the claculator

Answer 4

No, make me