Question 1062978
{{{ arc cos( -sqrt(2)/2 ) }}} in quadrant 2
( I like to write it this way )
This is the same as:
{{{ arc cos( -1/sqrt(2)) ) }}}
Since 
{{{ 1^2 + 1^2 = sqrt(2)^2 }}}
{{{ 2 = 2 }}}
The terminal point is ( -1,1 )
This gives me an angle of {{{ 3*pi/4 }}}
{{{ arc cos( -sqrt(2)/2 ) = 3*pi/4 }}}