Question 1181682
De Moivre's theorem:  {{{(cos(x) + i*sin(x))^n = cos(n*x) + i*sin(n*x)}}} for any real number {{{x}}} and integer {{{n}}}.

===> {{{(cos(x) + i*sin(x))^3 = cos(3*x) + i*sin(3*x)}}}


Expanding the left side of the preceding equation, we get


{{{(cos^3(x) -3cosx*sin^2(x)) + i*(3cos^2(x)*sinx - sin^3(x)) =cos(3x) + i*sin(3x) }}} after rearranging and combining like terms.


===> {{{cos^3(x) -3cosx*sin^2(x) = cos(3x)}}} and {{{3cos^2(x)*sinx - sin^3(x) = sin(3x)}}}


after equating corresponding real and imaginary parts.


Therefore, {{{highlight(cot(3x)) = cos(3x)/sin(3x) = highlight((cos^3(x) -3cosx*sin^2(x))/ ( 3cos^2(x)*sinx - sin^3(x) ))}}}