homework help 657

##Cos (pi/5) =
(sqrt (5)+1)/4##.

Let ##a = cos(pi/5)##, ##b = cos(2*pi/5)##. Thus ##cos(4*pi/5) = -a##. From the double angle formulas:

##b = 2a^2-1##
##-a = 2b^2-1##

Subtracting,

##a+b = 2(a^2-b^2)
= 2(a+b)(a-b)##

##a+b## is not zero, as both terms are positive, so ##a-b## must be ##1/2##. Then

##a-1/2 = 2a^2-1##
##4a^2-2a-1 = 0##

and the only positive root is

##a = cos (pi/5) = (sqrt(5)+1)/4##.

And ##b = cos (2*pi/5)
= a-1/2 = (sqrt(5)-1)/4##.