Question 868206
<pre>
{{{y = cos(4pi/5)}}}

{{{cos(5*(4pi/5))=cos(4pi) = 1}}}

Let A = {{{4pi/5}}}

{{{cos(5A) = 1}}}

Then {{{4pi/5}}} must be one of the solutions to:

{{{cos(4A+A)=1}}}

Use identity {{{cos(alpha+beta)=cos^2(alpha)-sin^2(beta)}}}

{{{cos(4A)cos(A)-sin(4A)sin(A)=1}}}

Use identity {{{cos(2theta)=2cos^2(theta)-1}}} on {{{cos(4A)}}}
Use identity {{{sin(2theta)=2sin(theta)cos(theta)}}} on {{{sin(4A)}}}

{{{(2cos^2(2A)-1)cos(A)-2sin(2A)cos(2A)sin(A)=1}}}

Use identity {{{cos(2theta)=2cos^2(theta)-1}}} on {{{cos(2A)}}}
Use identity {{{sin(2theta)=2sin(theta)cos(theta)}}} on {{{sin(2A)}}}

{{{(2(2cos^2(A)-1)^2-1)cos(A)-2(2sin(A)cos(A))(2cos^2(A)-1)sin(A)=1}}}

Simplify:

{{{(2(4cos^4(A)-4cos^2(A)+1)-1)cos(A)-4sin^2(A)cos(A)(2cos^2(A)-1)=1}}}

{{{(8cos^4(A)-8cos^2(A)+2-1)cos(A)-8sin^2(A)cos^3(A)+4sin^2(A)cos(A)=1}}}

{{{(8cos^4(A)-8cos^2(A)+1)cos(A)-8sin^2(A)cos^3(A)+4sin^2(A)cos(A)=1}}}

{{{8cos^5(A)-8cos^3(A)+cos(A)-8sin^2(A)cos^3(A)+4sin^2(A)cos(A)=1}}}

Factor sin²(A) out of the last two terms on the left:

{{{8cos^5(A)-8cos^3(A)+cos(A)-sin^2(A)(8cos^3(A)-4cos(A))=1}}}

Use the identity {{{sin^2(theta)=1-cos^2(theta)}}} on sin²(A)

{{{8cos^5(A)-8cos^3(A)+cos(A)-(1-cos^2(A))(8cos^3(A)-4cos(A))=1}}}

FOIL out the last term on the left:

{{{8cos^5(A)-8cos^3(A)+cos(A)-(8cos^3(A)-4cos(A)-8cos^5(A)+4cos^3(A))=1}}}

{{{8cos^5(A)-8cos^3(A)+cos(A)-(12cos^3(A)-4cos(A)-8cos^5(A))=1}}}

{{{8cos^5(A)-8cos^3(A)+cos(A)-12cos^3(A)+4cos(A)+8cos^5(A)=1}}}

{{{16cos^5(A)-20cos^3(A)+5cos(A)=1}}}

Let cos(A)=x

{{{16x^5-20x^3+5x=1}}}

{{{16x^5-20x^3+5x-1=0}}}

Try a root of 1

1 | 16   0  -20   0   5  -1
  |<u>     16   16  -4  -4   1</u>
    16  16   -4  -4   1   0

So it factors as

{{{(x-1)(16x^4+16x^3-4x^2-4x+1)=0}}}

We try factoring {{{16x^4+16x^3-4x^2-4x+1}}} as
the product of two quadratics.  In fact we might
expect that it is the square of a quadratic because
the leading term {{{16x^4}}} is a square and
also the constant term 1 is a square.

So we try to see if there is a quadratic (4x²+kx-1)
which when squared gives  {{{16x^4+16x^3-4x^2-4x+1}}}

{{{(4x^2+kx-1)^2}}}{{{""=""}}}{{{16x^4+16x^3-4x^2-4x+1}}} 

{{{16x^4+k^2x^2+1+8kx^3-8x^2-2kx}}}{{{""=""}}}{{{16x^4+16x^3-4x^2-4x+1}}}

Equating coefficients of x³:

8k = 16
  k = 2

Equating coefficients of x²:

k²-8 = -4
   k² = 4
    k = ±2

So far, k=2 works

Equating  coefficients of x

-2k=-4
   k=2

So indeed 

{{{(x-1)(16x^4+16x^3-4x^2-4x+1)=0}}}

factors as

{{{(x-1)(4x^2+2x-1)^2=0}}}

So we have solutions

x-1 = 0    and   {{{4x^2+2x-1=0}}}
  x = 1          {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
                 {{{x = (-(2) +- sqrt( (2)^2-4*(4)*(-1) ))/(2*(4)) }}} 
                 {{{x = (-2 +- sqrt(4+16))/8 }}} 
                 {{{x = (-2 +- sqrt(20))/8 }}}
                 {{{x = (-2 +- sqrt(4*5))/8 }}}
                 {{{x = (-2 +- 2sqrt(5))/8 }}}
                 {{{x = (2(-1 +- sqrt(5)))/8 }}}
                 {{{x = (-1 +- sqrt(5))/4 }}}

So the solutions are

{{{cos(A)=1}}},  {{{cos(A)= (-1 + sqrt(5))/4 }}}, {{{cos(A)= (-1 - sqrt(5))/4 }}}.

Since {{{4pi/5}}} is in Q2, its cosine must be negative.  The
only solution which is negative is

{{{cos(A)= (-1 - sqrt(5))/4 }}}.

thus

{{{cos(4pi/5)= (-1 - sqrt(5))/4 }}}.

Edwin</pre>