Good to hear from you, Dr. E. Mathematica: In[225]:= Sin@(k*x)/(1 - Cos@(k*x)) == (Sum[Cos@(n*x), {n, 0, k}])*(Sum[Sin@(n*x), {n, 0, k}])^(-1) Out[225]= Sin[k x]/(1 - Cos[k x]) == Cot[(k x)/2] —Bill On Wed, Aug 14, 2019 at 10:40 AM françois mendzina essomba2 < m_essob@yahoo.fr> wrote:
Hello
I still did not notice this equality between sinus functions,
I find it harmonious
sin(3*x)/(1-cos(3*x))=(cos(3*x)+cos(2*x)+cos(x)+1)/(sin(3*x)+sin(2*x)+sin(x));
sin(5*x)/(1-cos(5*x))=(cos(5*x)+cos(4*x)+cos(3*x)+cos(2*x)+cos(x)+1)/(sin(5*x)+sin(4*x)+sin(3*x)+sin(2*x)+sin(x));
sin(11*x)/(1-cos(11*x))=
(cos(11*x)+cos(10*x)+cos(9*x)+cos(8*x)+cos(7*x)+cos(6*x)+cos(5*x)+cos(4*x)+cos(3*x)+cos(2*x)+cos(x)+1)/(sin(11*x)+sin(10*x)+sin(9*x)+sin(8*x)+sin(7*x)+sin(6*x)+sin(5*x)+sin(4*x)+sin(3*x)+sin(2*x)+sin(x));
general
sin(k*x)/(1-cos(k*x))=(sum(cos(n*x),n=0..k))*(sum(sin(n*x),n=0..k))^(-1);