Question 695901
Consider the following right angled triangle
{{{ drawing( 300, 200, 0, 10, 0, 10,
  triangle( 1, 1, 1, 9, 9, 1 ),
  locate( 1, 1, a ), locate( 1, 10, b ), locate( 9, 1, c ),
  locate( 7.5, 2, A ),
  locate( 4, 0.8, 1 ),
  locate( 1.2, 5, sqrt(2)-1 )
) }}}

Then for this triangle {{{ tan A = opp/adj = (sqrt(2)-1)/1 = sqrt(2)-1 }}},

By Pythagoras {{{ hyp^2 = opp^2 + adj^2 }}} so 
{{{ hyp^2=(sqrt(2)-1)^2 + (1)^2 }}}
{{{ hyp^2=(2 -2sqrt(2) + 1 ) + (1) }}}
{{{ hyp^2=4 -2sqrt(2) }}}


so then {{{ sin A = opp/hyp }}} and  {{{ cos A = adj/hyp }}} giving
{{{ sin A*cosA = (opp/hyp)(adj/hyp) }}}
{{{ sin A*cosA = (opp)(adj)/hyp^2 }}}
{{{ sin A*cosA = (sqrt(2)-1)(1)/(4 -2sqrt(2)) }}}
{{{ sin A*cosA = (sqrt(2)-1)/(4 -2sqrt(2)) }}}
{{{ sin A*cosA = (sqrt(2)-1)(4+2sqrt(2))/(4 -2sqrt(2))(4+2sqrt(2)) }}}
{{{ sin A*cosA = (sqrt(2)-1)(4+2sqrt(2))/(4 -2sqrt(2))(4+2sqrt(2)) }}}
{{{ sin A*cosA = (2sqrt(2)sqrt(2)+4sqrt(2)-2sqrt(2)-4)/(16-2sqrt(2)2sqrt(2)) }}}
{{{ sin A*cosA = 2sqrt(2)/(16-8) }}}
{{{ sin A*cosA = 2sqrt(2)/8 }}}
{{{ sin A*cosA = sqrt(2)/4 }}} QED