@Markiyan Hirnyk 

Thank you!
You are right.  I did not notice that in the third quarter the plots are superimposed on one another. But Maple is also wrong, the solutions  {a=0, b=0}  and  {a=1, b=1}  are lost.

@Markiyan Hirnyk

Try  RealDomain:-solve(sys, symbolic=false);

@Markiyan Hirnyk

Try  RealDomain:-solve(sys, symbolic=false);

@Markiyan Hirnyk 

Take a closer look:

sys := [sqrt(sin(x)^2+1/sin(x)^2)+sqrt(cos(y)^2+1/cos(y)^2) = sqrt(20*y/(x+y)), sqrt(sin(y)^2+1/sin(y)^2)+sqrt(cos(x)^2+1/cos(x)^2) = sqrt(20*x/(x+y))]:

A := lhs(sys[1])+lhs(sys[2]);

@Markiyan Hirnyk 

Take a closer look:

sys := [sqrt(sin(x)^2+1/sin(x)^2)+sqrt(cos(y)^2+1/cos(y)^2) = sqrt(20*y/(x+y)), sqrt(sin(y)^2+1/sin(y)^2)+sqrt(cos(x)^2+1/cos(x)^2) = sqrt(20*x/(x+y))]:

A := lhs(sys[1])+lhs(sys[2]);

st:=time(): Tuples(10, 4): time()-st;
                42.828

st:=time(): CartProdSeq([$0..4] $ 10): time()-st;
                46.593

@Markiyan Hirnyk

You are wrong on two points:

1. You forgot about the condition: acute triangle

2. This inequality requires symbolic, ie, the exact solution, rather than the approximate one.

@ThU 

You are right. Shoelace's formula is a wonderful formula! It is used in the text of the procedure  Area  for calculating the area of any region bounded non-selfintersecting line, if the whole or a part of the boundary is a broken line.

See  http://www.maplesoft.com/applications/view.aspx?SID=146470

@ThU 

You are right. Shoelace's formula is a wonderful formula! It is used in the text of the procedure  Area  for calculating the area of any region bounded non-selfintersecting line, if the whole or a part of the boundary is a broken line.

See  http://www.maplesoft.com/applications/view.aspx?SID=146470

@Carl Love

Work of your procedure:

P(3, 9);

           1

My procedure also works incorrectly if  N2 - N1<=1

@Carl Love

Work of your procedure:

P(3, 9);

           1

My procedure also works incorrectly if  N2 - N1<=1

To M. Hirnyk!

Of course, my answer is wrong. I partially solved by hand, and it seemed to me that  tan(tan^(-1)(4))=1/4

To C. Love!

To complete the solution should be checked, ie substitution of the answer to the original equation. The fact that the two equations  A=B  and  tan(A)=tan(B)  are not equivalent  (the second equation is a consequence of the first). Therefore, you may receive extraneous solutions.

1. 

U := [7, 9, 14]:

A := [-2, 7, 8, 12, 9, -78, 0]:

remove(x->is(x in U), A);

      [-2, 8, 12, -78, 0]

2.

C:=[[-2, 2], [7, 0], [8, 6], [12, 12], [9, 6], [-78, 45], [0, 7]]:

remove(x->is(x[1] in U), C);

         [[-2, 2], [8, 6], [12, 12], [-78, 45], [0, 7]]

3.

E := map(x->`if`(not x[1] in U, x, NULL), C);

         E := [[-2, 2], [8, 6], [12, 12], [-78, 45], [0, 7]]

minimize(3-4*cos(z)+cos(2*z), z = 0 .. infinity);

maximize(3-4*cos(z)+cos(2*z), z = 0 .. infinity);

minimize(3-4*t+2*t^2-1, t = -1 .. 1);  # Change  cos(z)=t

maximize(3-4*t+2*t^2-1, t = -1 .. 1);

PS.  It seems that Maple does not understand that there is a periodic function with period 2*Pi.

minimize(3-4*cos(z)+cos(2*z), z = 0 .. 2*Pi);

maximize(3-4*cos(z)+cos(2*z), z = 0 .. 2*Pi);

                                     0

                                     8

@Markiyan Hirnyk 

1. Of course, you are right in that. I just do not understand why any problem you want to solve exclusively with Maple. It's interesting, but not always effective. Even if the result is obtained by using Maple, it should be tested as sometimes bugs are possible.

2. For  x = -1  it is obvious that the total sum is 0 since each term of the series is 0. It can also be set with Maple, but it requires some effort. The fact is that Maple does not understand that  3^n  is odd for any non-negative integer. For the success the first term of the series should be calculated separately. See

is((3^n)::odd) assuming n::nonnegint;

                           FAIL

eval((x^(3^n)+(x^(3^n))^2)/(1-x^(3^(n+1))), {n = 0, x = -1})+eval(sum((x^(3^n)+(x^(3^n))^2)/(1-x^(3^(n+1))), n = 1 .. infinity), x = -1) ;

                              0