Let A = Cos( pi/5 )*Cos( 2*pi/5 ) A=cos(π5)⋅cos(2⋅π5)
But Sin( 2*X ) = 2Sin( X )*Cos( X ) sin(2⋅X)=2sin(X)⋅cos(X)
=> Cos( X ) = Sin( 2*X ) / [ 2*Sin( X ) ] ------ > (1 ) ⇒cos(X)=sin(2⋅X)2⋅sin(X)−−−−−−>(1)
When X = Pi / 5 Then
Cos( Pi/5 ) = Sin( 2*pi/5 ) / [ 2*Sin( Pi/5 ) ]
And A = Cos( pi/5 )*Cos( 2*pi/5 )
= [ Sin( 2*Pi/5 ) * Cos( 2*Pi/5 ) ] / [ 2*Sin(Pi/5) ]
Following Equation ------ > ( 1 ) Then
Sin(2*Pi/5) * Cos(2*Pi/5) = [ Sin ( 4*Pi/5 ) ] / 2 Then
And A = Cos( pi/5 )*Cos( 2*pi/5 )
= [ Sin( 2*Pi/5 ) * Cos( 2*Pi/5 ) ] / [ 2*Sin(Pi/5) ]
= [ Sin ( 4*Pi/5 ) ] / [ 2 * 2 * Sin( Pi/5 ) ]
But Sin ( 4*Pi/5 ) = Sin ( Pi - Pi/5 )
= Sin ( Pi )*Cos( Pi/5 ) - Cos( Pi )*Sin( Pi/5 ) = Sin ( Pi/5 )
Then A = [ Sin ( 4*Pi/5 ) ] / [ 2 * 2 * Sin( Pi/5 ) ]
= [ Sin (Pi/5 ) ] / [ 2 *2 *Sin (Pi/5 ) ] = 1/4