Dr. E: Mathematica turns the a=1/15 ; b= 1/30 ; theta = Pi case of your first identity into \[Pi] == 30 (1/4 (-1 + Sqrt[5]) Csc[\[Pi]/15] Hypergeometric2F1[1/20, 1, 21/20, Csc[\[Pi]/15]^20 Sin[\[Pi]/30]^20] Sin[\[Pi]/30] - . . . + (two pages) then In[113]:= FullSimplify@FunctionExpand@% // tim (35 minutes) Out[113]= 5 (5 ArcTan[ Root[25 - 1900 #1^2 + 17510 #1^4 - 41420 #1^6 + 841 #1^8 &, 5]] + 6 ArcTan[ Root[1 - 52 #1^2 + 734 #1^4 - 3428 #1^6 + 3721 #1^8 &, 3]]) == \[Pi] + 20 ArcTan[ Root[1 - 100 #1^2 + 2510 #1^4 - 5300 #1^6 + 25 #1^8 &, 4]] + 25 ArcTan[ Root[625 - 71500 #1^2 + 1905350 #1^4 - 17256140 #1^6 + 40056241 #1^8 &, 3]] Although this checks numerically, testing symbolically mod π, TrigExpand@Expand@Tan[Subtract @@ %] // FullSimplify fatally hung the kernel. --rwg Note to Mathematica developers: Macsyma had a useful ATANCONTRACT: ratsimp(atancontract(2*atan(a)+3*atan(b))) atan((((a^2 - 1) * b^3 - 6 * a * b^2 + (3 - 3 * a^2) * b + 2 * a)/(2 * a * b^3 + (3 * a^2 - 3) * b^2 - 6 * a * b - a^2 + 1))) On Sat, Mar 31, 2018 at 12:31 AM, françois mendzina essomba2 < m_essob@yahoo.fr> wrote:
hello
I noticed that: for some values of a and b...
theta=(1/b)*sum(((-1)^(n+1)/n)*(sin(b*theta)/sin(a*theta))^n*sin(n*theta*(b+a)),n=1..inf) ;
I have deducted:
theta=(sum(((-1)^(n+1)*(sinh(b*theta))^n*sinh((b+a)*n*theta) )/(n*(sinh(a*theta))^n),n=1..inf))/b;
an example :
a=1/15 ; b= 1/30 ; theta = Pi
FME...