To prove that cos(pi/5)-cos(2xpi/5)=0.5

From proofs using multiple angle theorem on
http://mathworld.wolfram.com/TrigonometryAnglesPi5.html

cos(pi/5)=1/4 x [1+sqrt5]

cos(2pi/5)=1/4 x [-1+sqrt5]

subtracting gives 1/4 +1/4 = 1/2

I'm gonna try a different tact

cos(Ø)-cos(2Ø)=1/2
using double angle formulas
cos(2Ø)=cos^2(Ø)-sin^2(Ø)=cos^2(Ø)-1+cos^2(Ø)=2cos^2(Ø)-1

so

cos(Ø)-2cos^(Ø)+1=1/2
-2cos^2(Ø)+cos(Ø)+1/2=0
let x=cos(Ø)
-2x^2+x+1/2=0
4x^2-2x-1=0

use quadratic eqn

and

x=(1+√5)/4 or x=(1-√5)/4
cos(Ø)=(1+√5)/4
or
cos(Ø)=(1-√5)/4

see if either oe of these Ø's = ∏/5

for the other question check out both of the quadrtic solutions to confirm answer.

Rap

Without a calculator ?

Using a similar developement

if x=cos(phi), then cos(2phi)=2x^2-1 from double angle developement

so x*[2x^2-1]=1/4
or
2x^3-x=1/4
rearranging you get
8x^3-4x-1=0
This is a cubic polynomial (potential for three real roots)

4x^3-4x-1=(2x+1)(4x^2-2x-1)=(2x+1)

so x=-1/2, 1/4[1+sqrt(5)], 1/4[1-sqrt(5)]

so now find all three phi's where x=cos(phi) and see if one of then is pi/5

2pi/3, pi/5, and 3pi/5 are potential solutions to polynomial.

neato

Rap

If I'm not wrong you took the problems from ''RUDIMENTS OF MATHEMATICS ''; here's a solution:-
pi/5 = 36degrees , 2pi/5=72 deg. let A= 18 degree then 5A=90=3A+2A
sin2A=sin(90-3A) , so 2 sinAcosA= cos3A=4 (cosA)^3 - 3cosA
i.e. cosA (2sinA +3 - 4 (cosA)^2 ) = 0
i.e. 2 sinA - 4 ( 1 - (sinA)^2 ) +3 =0
i.e. 4 (sinA)^2 + 2 sinA +3 =0
i.e. sinA= ( - 1 + (sqrt 5) ) /4 , by solving the quadratic in sinA , the other negative root must be rejected because sin18 can not be negative. So cos36 = c0s(2x18)=1 - 2 (sin18)^2 (after some tedious
= (2 + 2 ( sqrt5) ) / 8 but simple calculations)
=( 1 + (sqrt5) )/4
cos72 = c0s(90-18)=sin18 =( -1 + (sqrt5) )/4
So cos(pi/5) - cos(2pi/5) = c0s36 - cos 72 = ( 1+(sqrt5) - (-1) - (sqrt5) ) /4 = (1 + 1)/4 = 0.5 , similarly the other result holds. ( sqrt means square root)