Question 1088789
<pre>
{{{cos(2x)}}}{{{""=""}}}{{{2cos^2(x)-1}}}
{{{1+cos(2x)}}}{{{""=""}}}{{{2cos^2(x)}}}, solving for cos<sup>2</sup>(x),
{{{1/2+expr(1/2)cos(2x)}}}{{{""=""}}}{{{cos^2(x)}}}

{{{cos(3x)}}}{{{""=""}}}{{{cos(2x+x)}}}{{{""=""}}}
{{{cos(2x)cos(x)-sin(2x)sin(x)}}}{{{""=""}}}{{{(2cos^2(x)-1)cos(x)-2sin(x)cos(x)sin(x)}}}{{{""=""}}}
{{{2cos^3(x)-cos(x)-2sin^2(x)cos(x)}}}{{{""=""}}}
{{{2cos^3(x)-cos(x)-2(1-cos^2(x))cos(x)}}}

So we solve this for  cos<sup>3</sup>(x),
{{{cos(3x)}}}{{{""=""}}}{{{2cos^3(x)-cos(x)-2(1-cos^2(x))cos(x)}}}
{{{cos(3x)}}}{{{""=""}}}{{{2cos^3(x)-cos(x)-2cos(x)+2cos^3(x)}}}
{{{cos(3x)}}}{{{""=""}}}{{{4cos^3(x)-3cos(x)}}}
Solving for cos<sup>3</sup>(x)
{{{cos(3x)+3cos(x)}}}{{{""=""}}}{{{4cos^3(x)}}}
{{{expr(1/4)cos(3x)+expr(3/4)cos(x)}}}{{{""=""}}}{{{cos^3(x)}}}

So {{{cos(pi/7)+cos^2(pi/7)-2cos^3(pi/7)}}}{{{""=""}}}

{{{cos(pi/7) + 1/2+expr(1/2)cos(2pi/7) -2(expr(1/4)cos(3pi/7)+expr(3/4)cos(pi/7))}}}{{{""=""}}}

{{{cos(pi/7) + 1/2+expr(1/2)cos(2pi/7) -expr(1/2)cos(3pi/7)-expr(3/2)cos(pi/7)}}}{{{""=""}}}

{{{expr(-1/2)cos(pi/7) + expr(1/2)cos(2pi/7)-expr(1/2)cos(3pi/7)+1/2}}}{{{""=""}}}
 
(expression 1):   {{{expr(-1/2)(cos(pi/7) - cos(2pi/7)+cos(3pi/7)-1)}}}

Expression 1 is what we must evaluate.  

{{{matrix(1,4,Let,z,""="",cos(pi/7)+i*sin(pi/7) )}}}  

{{{matrix(1,10,Then, z^7, ""="",(cos(pi/7)+i*sin(pi/7))^7, ""="", cos(pi)+i*sin(pi),""="",-1+i*0,""="",-1 )}}}

{{{matrix(1,4,So,z^7+1,""="",0) }}}

{{{matrix(1,4,"Factoring,",(z+1)(z^6-z^5+z^4-z^3+z^2-z+1),""="",0) }}}


{{{matrix(1,6,Since,z<>-1,",",z^6-z^5+z^4-z^3+z^2-z+1,""="",0) }}}

(equation 2)  {{{matrix(1,3, 1+z^2+z^4+z^6,""="",z+z^3+z^5) }}}

{{{matrix(1,12,Since,z^n,""="",cos(n*pi/7)+i*sin(n*pi/7),",",the,real,part,of,z^n,is,cos(n*pi/7) )}}}

We set the real part of the left side of equation 2 
equal to the real part of the right side of equation 2.

 {{{matrix(1,3, 1+cos(2pi/7)+cos(4pi/7)+cos(6pi/7),""="",cos(pi/7)+cos(3pi/7)+cos(5pi/7)) }}}

Now we use identities of angles subtracted from pi to reduce
the angles:

{{{cos(4pi/7) = cos(7pi/7-3pi/7)=cos(pi-3pi/7) = -cos(3pi/7)}}}
{{{cos(6pi/7) = cos(7pi/7-pi/7)=cos(pi-pi/7) = -cos(pi/7)}}}
{{{cos(5pi/7) = cos(7pi/7-2pi/7)=cos(pi-2pi/7) = -cos(2pi/7)}}}

Substituting

 {{{matrix(1,3, 1+cos(2pi/7)-cos(3pi/7)-cos(pi/7),""="",cos(pi/7)+cos(3pi/7)-cos(2pi/7)) }}} 

 {{{matrix(1,3, 1,""="",2cos(pi/7)-2cos(2pi/7)+2cos(3pi/7)) }}}

{{{matrix(1,3, 1,""="",2(cos(pi/7)-cos(2pi/7)+cos(3pi/7))) }}}

(equation 3):   {{{matrix(1,3, 1/2,""="",cos(pi/7)-cos(2pi/7)+cos(3pi/7)) }}}

Now we go back to expression (1)

(expression 1):   {{{expr(-1/2)(cos(pi/7) - cos(2pi/7)+cos(3pi/7)-1)}}}

and use equation 3 to substitute 1/2 for the first three terms in 
the parentheses of expression (1):

{{{expr(-1/2)(1/2-1)}}}{{{""=""}}}{{{expr(-1/2)(-1/2)}}}{{{""=""}}}{{{1/4}}}

So {{{p/q}}}{{{""=""}}}{{{1/4}}}

and {{{p+q}}}{{{""=""}}}{{{1+4}}}{{{""=""}}}{{{5}}}

Edwin</pre>