Question 1084358
{{{ a + b*i  = r*cos(theta) + i*r*sin(theta) =  r*e^(i*theta) }}}
where  
 {{{ r = sqrt(a^2+b^2) }}} and {{{ theta = tan(-1)(b/a) }}}
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{{{ r = sqrt( (cos(x))^2 + (-sin(x))^2  )   = sqrt( cos^2(x) + sin^2(x) ) = sqrt(1) = 1 }}} 

and  {{{ theta = tan^-1(-sin(x) / cos(x)) = tan^-1(-tan(x)) = -x }}}

so:  {{{ (cos(x)+i*sin(x))^5 = (e^(-i*x))^5 = e^(i*-5x) = cos(-5x) + i*sin(-5x) }}}

Using   {{{ cos(-theta) = cos(theta) }}} and {{{ sin(-theta) = -sin(theta) }}}   for {{{ 0<=theta<=pi }}}
lets us simplify to:
  {{{ highlight(cos(5x) - i*sin(5x)) }}}