Prove that: cos(?/5)= (1+?5)/4 where a= ?/5 ==> cos3a = -cos2a and cos3x = 4cos^

Question

Prove that: cos(?/5)= (1+?5)/4

where a= ?/5 ==> cos3a = -cos2a

and cos3x = 4cos^3x-3cosx

and cos2x = 2cos^2x-1

Explanation / Answer

cos(5x) = 16cos^5(x) - 20cos^3(x) + 5cos(x). If x = p/5: cos[5(p/5)] = 16cos^5(p/5) - 20cos^3(p/5) + 5cos(p/5) ==> -1 = 16cos^5(p/5) - 20cos^3(p/5) + 5cos(p/5). If we let u = cos(p/5), we obtain: 16u^5 - 20u^3 + 5u = -1 ==> 16u^5 - 20u^3 + 5u + 1 = 0 ==> (u + 1) * (4u^2 - 2u - 1)^2 = 0. ==> u = -1 and 4u^2 - 2u - 1 = 0. However, cos(p/5) = -1 is absurd. So we drop it and we consider: 4u^2 - 2u - 1 = 0 ==> u = (1 + v5)/4 and u = (1 - v5)/4. Since cos(p/5) > 0 as p/5 is in the first Quadrant, we pick u = (1 + v5)/4. Thus: cos(p/5) = (1 + v5)/4.

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