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How to simplify certain solutions of trigonometric equation?

Started byRainer Rosenthal <r.rosenthal@web.de>
First post2015-03-08 20:25 +0100
Last post2015-03-10 01:12 +0100
Articles 10 — 2 participants

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  How to simplify certain solutions of trigonometric equation? Rainer Rosenthal <r.rosenthal@web.de> - 2015-03-08 20:25 +0100
    Re: How to simplify certain solutions of trigonometric equation? acer <maple@rogers.com> - 2015-03-08 18:48 -0700
      Re: How to simplify certain solutions of trigonometric equation? Rainer Rosenthal <r.rosenthal@web.de> - 2015-03-09 11:05 +0100
        Re: How to simplify certain solutions of trigonometric equation? acer <maple@rogers.com> - 2015-03-09 06:09 -0700
          Re: How to simplify certain solutions of trigonometric equation? Rainer Rosenthal <r.rosenthal@web.de> - 2015-03-09 14:40 +0100
            Re: How to simplify certain solutions of trigonometric equation? acer <maple@rogers.com> - 2015-03-09 07:36 -0700
              Re: How to simplify certain solutions of trigonometric equation? acer <maple@rogers.com> - 2015-03-09 07:48 -0700
              Re: How to simplify certain solutions of trigonometric equation? Rainer Rosenthal <r.rosenthal@web.de> - 2015-03-09 17:56 +0100
                Re: How to simplify certain solutions of trigonometric equation? acer <maple@rogers.com> - 2015-03-09 10:15 -0700
                  Re: How to simplify certain solutions of trigonometric equation? Rainer Rosenthal <r.rosenthal@web.de> - 2015-03-10 01:12 +0100

#1105 — How to simplify certain solutions of trigonometric equation?

FromRainer Rosenthal <r.rosenthal@web.de>
Date2015-03-08 20:25 +0100
SubjectHow to simplify certain solutions of trigonometric equation?
Message-ID<cm3lseFejjsU1@mid.individual.net>
I was searching for solutions of the following equation:

  sin(x) + sin(2x) + sin(3x) = cos(x) + cos(2x) + cos(3x)

So I defined funtion f:
f := x -> sin(x) + sin(2*x) + sin(3*x) - (cos(x) + cos(2*x) + cos(3*x));

and solved as follows:
p := [solve(f(x)=0,x)];

I got six different solutions:

p[1] = 2/3*Pi
p[2] = -2/3*Pi
p[3] = arctan(2*(1/2*(2+sqrt(2))^(3/2)-3/2*sqrt(2+sqrt(2)))/(sqrt(2+sqrt(2))))
p[4] = -arctan(2*(-1/2*(2+sqrt(2))^(3/2)+3/2*sqrt(2+sqrt(2)))/(sqrt(2+sqrt(2))))-Pi
p[5] = arctan(2*(1/2*(2-sqrt(2))^(3/2)-3/2*sqrt(2-sqrt(2)))/(sqrt(2-sqrt(2))))
p[6] = -arctan(2*(-1/2*(2-sqrt(2))^(3/2)+3/2*sqrt(2-sqrt(2)))/(sqrt(2-sqrt(2))))+Pi]

I hoped for a simple zero for each f(p[i]), but most of the time I got
difficult expressions, which were only simplified by "simplify":

I usd these Maple-lines:
for i to nops(p) do   print(Check_Solution_Number_,i);
   simplify(f(p[i]));
od;

and received the following output:


                      Check_Solution_Number_, 1


                                  0


                      Check_Solution_Number_, 2


                                  0


                      Check_Solution_Number_, 3


  (-2 + sqrt(2) + sin(3/8 Pi) sqrt(4 - 2 sqrt(2))

         - cos(3/8 Pi) sqrt(4 - 2 sqrt(2)))/sqrt(4 - 2 sqrt(2))


                      Check_Solution_Number_, 4


  (2 - sqrt(2) - sin(2 %2) %1 + sin(3 %2) %1 - cos(2 %2) %1

         + cos(3 %2) %1)/sqrt(4 - 2 sqrt(2))

  %1 := sqrt(4 - 2 sqrt(2))

  %2 := arctan(-sqrt(2) + 1)


                      Check_Solution_Number_, 5


  - (2 + sqrt(2) - sin(2 %2) %1 - sin(3 %2) %1 + cos(2 %2) %1

         + cos(3 %2) %1)/sqrt(4 + 2 sqrt(2))

  %1 := sqrt(4 + 2 sqrt(2))

  %2 := arctan(-1 - sqrt(2))


                      Check_Solution_Number_, 6


  (2 + sqrt(2) - sin(1/8 Pi) sqrt(4 + 2 sqrt(2))

         - cos(1/8 Pi) sqrt(4 + 2 sqrt(2)))/sqrt(4 + 2 sqrt(2))


Which options of "simplify" could do better?
(I still use Maple V).

Cheers,
Rainer

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#1106

Fromacer <maple@rogers.com>
Date2015-03-08 18:48 -0700
Message-ID<fe5f8fd2-5868-4bee-beb3-92335fd0b585@googlegroups.com>
In reply to#1105
Le dimanche 8 mars 2015 15:22:56 UTC-4, Rainer Rosenthal a écrit :
> I was searching for solutions of the following equation:
> 
>   sin(x) + sin(2x) + sin(3x) = cos(x) + cos(2x) + cos(3x)
> 
> So I defined funtion f:
> f := x -> sin(x) + sin(2*x) + sin(3*x) - (cos(x) + cos(2*x) + cos(3*x));
> 
> and solved as follows:
> p := [solve(f(x)=0,x)];
> 
> I got six different solutions:
> 
> p[1] = 2/3*Pi
> p[2] = -2/3*Pi
> p[3] = arctan(2*(1/2*(2+sqrt(2))^(3/2)-3/2*sqrt(2+sqrt(2)))/(sqrt(2+sqrt(2))))
> p[4] = -arctan(2*(-1/2*(2+sqrt(2))^(3/2)+3/2*sqrt(2+sqrt(2)))/(sqrt(2+sqrt(2))))-Pi
> p[5] = arctan(2*(1/2*(2-sqrt(2))^(3/2)-3/2*sqrt(2-sqrt(2)))/(sqrt(2-sqrt(2))))
> p[6] = -arctan(2*(-1/2*(2-sqrt(2))^(3/2)+3/2*sqrt(2-sqrt(2)))/(sqrt(2-sqrt(2))))+Pi]
> 
> I hoped for a simple zero for each f(p[i]), but most of the time I got
> difficult expressions, which were only simplified by "simplify":
> 
> I usd these Maple-lines:
> for i to nops(p) do   print(Check_Solution_Number_,i);
>    simplify(f(p[i]));
> od;
> 
> and received the following output:
> 
> 
>                       Check_Solution_Number_, 1
> 
> 
>                                   0
> 
> 
>                       Check_Solution_Number_, 2
> 
> 
>                                   0
> 
> 
>                       Check_Solution_Number_, 3
> 
> 
>   (-2 + sqrt(2) + sin(3/8 Pi) sqrt(4 - 2 sqrt(2))
> 
>          - cos(3/8 Pi) sqrt(4 - 2 sqrt(2)))/sqrt(4 - 2 sqrt(2))
> 
> 
>                       Check_Solution_Number_, 4
> 
> 
>   (2 - sqrt(2) - sin(2 %2) %1 + sin(3 %2) %1 - cos(2 %2) %1
> 
>          + cos(3 %2) %1)/sqrt(4 - 2 sqrt(2))
> 
>   %1 := sqrt(4 - 2 sqrt(2))
> 
>   %2 := arctan(-sqrt(2) + 1)
> 
> 
>                       Check_Solution_Number_, 5
> 
> 
>   - (2 + sqrt(2) - sin(2 %2) %1 - sin(3 %2) %1 + cos(2 %2) %1
> 
>          + cos(3 %2) %1)/sqrt(4 + 2 sqrt(2))
> 
>   %1 := sqrt(4 + 2 sqrt(2))
> 
>   %2 := arctan(-1 - sqrt(2))
> 
> 
>                       Check_Solution_Number_, 6
> 
> 
>   (2 + sqrt(2) - sin(1/8 Pi) sqrt(4 + 2 sqrt(2))
> 
>          - cos(1/8 Pi) sqrt(4 + 2 sqrt(2)))/sqrt(4 + 2 sqrt(2))
> 
> 
> Which options of "simplify" could do better?
> (I still use Maple V).
> 
> Cheers,
> Rainer


Try something like this,

  seq( radnormal(convert(radnormal(f(t)),radical)), t=p );

or

  seq( simplify(convert(simplify(f(t)),radical)), t=p );

because it looks like your MapleV is missing the step of turning certain trig constants into their radical equivalents.

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#1107

FromRainer Rosenthal <r.rosenthal@web.de>
Date2015-03-09 11:05 +0100
Message-ID<cm59dtFqqh2U1@mid.individual.net>
In reply to#1106
Am 09.03.2015 um 02:48 schrieb acer:
> 
> Try something like this,
> 
>   seq( radnormal(convert(radnormal(f(t)),radical)), t=p );
> 
> or
> 
>   seq( simplify(convert(simplify(f(t)),radical)), t=p );
> 
> because it looks like your MapleV is missing the step of turning certain trig constants into their radical equivalents.
> 

Thank you. There was partial success as the expressions became
slightly simpler.

Let me tell you what I got for solution #3, i.e. for the expression

S3 := (-2+sqrt(2)+sin(3/8*Pi)*sqrt(4-2*sqrt(2))-cos(3/8*Pi)*sqrt(4-2*sqrt(2)))/(sqrt(4-2*sqrt(2)));

With "radnormal(convert(radnormal(f(t)),radical))" this becomes
S31 := -1/2*sqrt(4-2*sqrt(2))-cos(3/8*Pi)+sin(3/8*Pi);    # quite nice, but still not 0

With "simplify(convert(simplify(f(t)),radical))" this becomes
S32 := 1/4*(-2+sqrt(2+sqrt(2))*sqrt(2)*sqrt(2-sqrt(2)))*sqrt(2)/(sqrt(2-sqrt(2))); # not 0, but no trig

The last one is without trig (nice) and can be seen to be 0 as follows:

S32 is equal to
S32a := 1/4*(-2+T)*sqrt(2)/(sqrt(2-sqrt(2)));
where
T := sqrt(2+sqrt(2))*sqrt(2)*sqrt(2-sqrt(2));

My Maple V refused to show me that this is 0. I had to square it for that purpose:
expand(T^2) gived 4, so T must be 2 as T > 0.
From that we can see S32a = 0, since factor (-2+T) = 0.

Well, so far this helped quite well, but I would be glad to see a handsome
formula F such that F(S3) = 0 will be stated by Maple (V).

Cheers,
Rainer

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#1108

Fromacer <maple@rogers.com>
Date2015-03-09 06:09 -0700
Message-ID<110212e8-6429-46db-83a5-703955762a40@googlegroups.com>
In reply to#1107
On Monday, March 9, 2015 at 6:02:39 AM UTC-4, Rainer Rosenthal wrote:
> Am 09.03.2015 um 02:48 schrieb acer:
> > 
> > Try something like this,
> > 
> >   seq( radnormal(convert(radnormal(f(t)),radical)), t=p );
> > 
> > or
> > 
> >   seq( simplify(convert(simplify(f(t)),radical)), t=p );
> > 
> > because it looks like your MapleV is missing the step of turning certain trig constants into their radical equivalents.
> > 
> 
> Thank you. There was partial success as the expressions became
> slightly simpler.
> 
> Let me tell you what I got for solution #3, i.e. for the expression
> 
> S3 := (-2+sqrt(2)+sin(3/8*Pi)*sqrt(4-2*sqrt(2))-cos(3/8*Pi)*sqrt(4-2*sqrt(2)))/(sqrt(4-2*sqrt(2)));
> 
> With "radnormal(convert(radnormal(f(t)),radical))" this becomes
> S31 := -1/2*sqrt(4-2*sqrt(2))-cos(3/8*Pi)+sin(3/8*Pi);    # quite nice, but still not 0
> 
> With "simplify(convert(simplify(f(t)),radical))" this becomes
> S32 := 1/4*(-2+sqrt(2+sqrt(2))*sqrt(2)*sqrt(2-sqrt(2)))*sqrt(2)/(sqrt(2-sqrt(2))); # not 0, but no trig
> 
> The last one is without trig (nice) and can be seen to be 0 as follows:
> 
> S32 is equal to
> S32a := 1/4*(-2+T)*sqrt(2)/(sqrt(2-sqrt(2)));
> where
> T := sqrt(2+sqrt(2))*sqrt(2)*sqrt(2-sqrt(2));
> 
> My Maple V refused to show me that this is 0. I had to square it for that purpose:
> expand(T^2) gived 4, so T must be 2 as T > 0.
> From that we can see S32a = 0, since factor (-2+T) = 0.
> 
> Well, so far this helped quite well, but I would be glad to see a handsome
> formula F such that F(S3) = 0 will be stated by Maple (V).
> 
> Cheers,
> Rainer


It could be that in MapleV evala/Normal might get it when radnormal or simplify(...,radical) fail. That is, something like,

 map(u->evala(Normal(simplify(convert(simplify(f(u)),radical)))),p);

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#1109

FromRainer Rosenthal <r.rosenthal@web.de>
Date2015-03-09 14:40 +0100
Message-ID<cm5m0pFu4h2U1@mid.individual.net>
In reply to#1108
Am 09.03.2015 um 14:09 schrieb acer:
> It could be that in MapleV evala/Normal might get it when radnormal or simplify(...,radical) fail.
> That is, something like,
> 
>  map(u->evala(Normal(simplify(convert(simplify(f(u)),radical)))),p);
> 
Wow, solutions #3 and #6 are verified by a clear "0" now, nice!

map(u->evala(Normal(simplify(convert(simplify(f(u)),radical)))),p);
gives the following list:

  [0,                      # Check solution #1
   0,                      # Check solution #2
   0,                      # Check solution #3
   not so easy,            # Check solution #4
   not so easy,            # Check solution #5
   0]                      # Check solution #6

Many thanks!
I'd love to find something for the last two checks for #4 and #5.

For example, check #4 results in this difficult expression:
1/2*sqrt(4-2*sqrt(2))-cos(2*arctan(-sqrt(2)+1))+sin(3*arctan(-sqrt(2)+1))-sin(2*arctan(-sqrt(2)+1))+cos(3*arctan(-sqrt(2)+1))

Cheers,
Rainer

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#1110

Fromacer <maple@rogers.com>
Date2015-03-09 07:36 -0700
Message-ID<58932a2e-122c-4dcd-8304-3bdcb5913803@googlegroups.com>
In reply to#1109
On Monday, March 9, 2015 at 9:37:30 AM UTC-4, Rainer Rosenthal wrote:
> Am 09.03.2015 um 14:09 schrieb acer:
> > It could be that in MapleV evala/Normal might get it when radnormal or simplify(...,radical) fail.
> > That is, something like,
> > 
> >  map(u->evala(Normal(simplify(convert(simplify(f(u)),radical)))),p);
> > 
> Wow, solutions #3 and #6 are verified by a clear "0" now, nice!
> 
> map(u->evala(Normal(simplify(convert(simplify(f(u)),radical)))),p);
> gives the following list:
> 
>   [0,                      # Check solution #1
>    0,                      # Check solution #2
>    0,                      # Check solution #3
>    not so easy,            # Check solution #4
>    not so easy,            # Check solution #5
>    0]                      # Check solution #6
> 
> Many thanks!
> I'd love to find something for the last two checks for #4 and #5.
> 
> For example, check #4 results in this difficult expression:
> 1/2*sqrt(4-2*sqrt(2))-cos(2*arctan(-sqrt(2)+1))+sin(3*arctan(-sqrt(2)+1))-sin(2*arctan(-sqrt(2)+1))+cos(3*arctan(-sqrt(2)+1))
> 
> Cheers,
> Rainer

Another possibility, is to convert to expln. (Again, for those coming late to the party, the problem here is to use MapleV. The earliest I have access to right now is MapleV R5. I'm not sure which Rainer is using.)

Also, it might help to simplify the values of list p before applying f.

  map(z->radnormal(expand(convert(f(simplify(z)),expln))),p);

In my MapleV R5 just this is enough,

  map(z->radnormal(expand(f(simplify(z)))),p);

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#1111

Fromacer <maple@rogers.com>
Date2015-03-09 07:48 -0700
Message-ID<2ae20fbe-d745-474d-b049-7444706395ab@googlegroups.com>
In reply to#1110
On Monday, March 9, 2015 at 10:36:12 AM UTC-4, acer wrote:
> On Monday, March 9, 2015 at 9:37:30 AM UTC-4, Rainer Rosenthal wrote:
> > Am 09.03.2015 um 14:09 schrieb acer:
> > > It could be that in MapleV evala/Normal might get it when radnormal or simplify(...,radical) fail.
> > > That is, something like,
> > > 
> > >  map(u->evala(Normal(simplify(convert(simplify(f(u)),radical)))),p);
> > > 
> > Wow, solutions #3 and #6 are verified by a clear "0" now, nice!
> > 
> > map(u->evala(Normal(simplify(convert(simplify(f(u)),radical)))),p);
> > gives the following list:
> > 
> >   [0,                      # Check solution #1
> >    0,                      # Check solution #2
> >    0,                      # Check solution #3
> >    not so easy,            # Check solution #4
> >    not so easy,            # Check solution #5
> >    0]                      # Check solution #6
> > 
> > Many thanks!
> > I'd love to find something for the last two checks for #4 and #5.
> > 
> > For example, check #4 results in this difficult expression:
> > 1/2*sqrt(4-2*sqrt(2))-cos(2*arctan(-sqrt(2)+1))+sin(3*arctan(-sqrt(2)+1))-sin(2*arctan(-sqrt(2)+1))+cos(3*arctan(-sqrt(2)+1))
> > 
> > Cheers,
> > Rainer
> 
> Another possibility, is to convert to expln. (Again, for those coming late to the party, the problem here is to use MapleV. The earliest I have access to right now is MapleV R5. I'm not sure which Rainer is using.)
> 
> Also, it might help to simplify the values of list p before applying f.
> 
>   map(z->radnormal(expand(convert(f(simplify(z)),expln))),p);
> 
> In my MapleV R5 just this is enough,
> 
>   map(z->radnormal(expand(f(simplify(z)))),p);

Actually, in my MapleV R5 it is enough to just do,

  map(z->simplify(expand(f(z))),p);

Without access to the particular MapleV version Rainer is using it's tricky to figure out the best way to get cos(3*arctan(-2^(1/2)+1)) converted to just radicals.

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#1112

FromRainer Rosenthal <r.rosenthal@web.de>
Date2015-03-09 17:56 +0100
Message-ID<cm61fsF2o63U1@mid.individual.net>
In reply to#1110
Am 09.03.2015 um 15:36 schrieb acer:
> In my MapleV R5 just this is enough,
> 
>   map(z->radnormal(expand(f(simplify(z)))),p);
> 

Well, that wasn't enough for my Maple Version

                     Maple 6
                   Jan 31 2000

(Hmm, I thought it was Maple V. And maybe it's just Maple V, R6?).

It gave a clear 0 for check #4 and #5 (great! thanks!), but
it failed for #3 and #6. Following the advice of my nephew
Frithjof I combined all the tricks:

convrad := x -> convert(x,radical);
squeeze := radnormal@expand@evala@Normal@simplify@convrad@simplify@f@simplify;

And, alas! There we are:

map(squeeze,p);
                          [0, 0, 0, 0, 0, 0]

What a neat result! Many thanks for your help. That was fun!

Cheers,
Rainer

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#1113

Fromacer <maple@rogers.com>
Date2015-03-09 10:15 -0700
Message-ID<ba41d2e8-fc39-4440-8906-4b403d748c4d@googlegroups.com>
In reply to#1112
On Monday, March 9, 2015 at 12:53:19 PM UTC-4, Rainer Rosenthal wrote:
> Am 09.03.2015 um 15:36 schrieb acer:
> > In my MapleV R5 just this is enough,
> > 
> >   map(z->radnormal(expand(f(simplify(z)))),p);
> > 
> 
> Well, that wasn't enough for my Maple Version
> 
>                      Maple 6
>                    Jan 31 2000
> 
> (Hmm, I thought it was Maple V. And maybe it's just Maple V, R6?).
> 
> It gave a clear 0 for check #4 and #5 (great! thanks!), but
> it failed for #3 and #6. Following the advice of my nephew
> Frithjof I combined all the tricks:
> 
> convrad := x -> convert(x,radical);
> squeeze := radnormal@expand@evala@Normal@simplify@convrad@simplify@f@simplify;
> 
> And, alas! There we are:
> 
> map(squeeze,p);
>                           [0, 0, 0, 0, 0, 0]
> 
> What a neat result! Many thanks for your help. That was fun!
> 
> Cheers,
> Rainer

There is no version MapleV R6. After MapleV R5 came Maple 6.

And in Maple 6 I can get all size to become zero without having to use evala/Normal.

kernelopts(version);                                                    
           Maple 6.01, SUN SPARC SOLARIS, June 9 2000 Build ID 79514

f := x -> sin(x) + sin(2*x) + sin(3*x) - (cos(x) + cos(2*x) + cos(3*x)):
p := [solve(f(x)=0,x)]:                                         

map(z->simplify(expand(convert(f(z),radical))),p); # quick                   

                              [0, 0, 0, 0, 0, 0]

map(z->radnormal(expand(convert(f(z),expln))),p); # slower

                              [0, 0, 0, 0, 0, 0]

Applying simplify to z in p before applying f is not necessary. But it seems nice to notice that p's entries can themselves be simplified.

map(simplify,p);
                               
                                         1/2                         1/2
    [2/3 Pi, - 2/3 Pi, 1/8 Pi, -arctan(-2    + 1) - Pi, arctan(-1 - 2   ), 5/8 Pi]

So,

P := map(simplify,p):

map(z->simplify(expand(convert(f(z),radical))),P); # quick   

                              [0, 0, 0, 0, 0, 0]

map(z->radnormal(expand(convert(f(z),expln))),P); # slower

                              [0, 0, 0, 0, 0, 0]

Have a great day.

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#1114

FromRainer Rosenthal <r.rosenthal@web.de>
Date2015-03-10 01:12 +0100
Message-ID<cm6r04F9hnlU1@mid.individual.net>
In reply to#1113
Am 09.03.2015 um 18:15 schrieb acer:
> kernelopts(version);                                                    
>            Maple 6.01, SUN SPARC SOLARIS, June 9 2000 Build ID 79514
> 
> f := x -> sin(x) + sin(2*x) + sin(3*x) - (cos(x) + cos(2*x) + cos(3*x)):
> p := [solve(f(x)=0,x)]:                                         
> 

#### Me:
kernelopts(version);
          Maple 6, IBM INTEL NT, Jan 31 2000 Build ID 16401



> map(z->simplify(expand(convert(f(z),radical))),p); # quick                   
> 
>                               [0, 0, 0, 0, 0, 0]

#### Me (oh, same, great!):
                                [0, 0, 0, 0, 0, 0]

> map(z->radnormal(expand(convert(f(z),expln))),p); # slower
> 
>                               [0, 0, 0, 0, 0, 0]

#### Me (oh, same, great!):
                                [0, 0, 0, 0, 0, 0]

> Applying simplify to z in p before applying f is not necessary. But it seems nice to notice that p's entries can themselves be simplified.
> 
> map(simplify,p);
>                                
>                                          1/2                         1/2
>     [2/3 Pi, - 2/3 Pi, 1/8 Pi, -arctan(-2    + 1) - Pi, arctan(-1 - 2   ), 5/8 Pi]
> 

#### Me (nearly  same):
      [2/3 Pi, - 2/3 Pi, 1/8 Pi, -arctan(-sqrt(2) + 1) - Pi, arctan(-1 - sqrt(2)), 5/8 Pi]

> So,
> 
> P := map(simplify,p):
> 
> map(z->simplify(expand(convert(f(z),radical))),P); # quick   
> 
>                               [0, 0, 0, 0, 0, 0]
> 
> map(z->radnormal(expand(convert(f(z),expln))),P); # slower
> 
>                               [0, 0, 0, 0, 0, 0]
> 

Again, same for me.

> Have a great day.

Oops, already had one today :-)

Have a nice week!
Cheers,
Rainer

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