Solve this equation!
Eugene Gath
with 0<x<90 (degrees).
Hauke Reddmann
: Find (with proof) all solutions of tan(2x)*tan(3x)*tan(4x) = tan(x),
Look up the formula tan(a+b)=? (if I didn't screw it up, it's
(tan a+tan b)/(1-tan a tan b), but look it up anyway) and
replace the multiple angle tan's with tan x dependents.
Solve after tan x (factorize if possible).
Take arctan of solutions.
--
Hauke Reddmann fc3...@math.uni-hamburg.de
<:-EX8
John McGowan
> Eugene Gath (ga...@ul.ie) wrote:
> : Find (with proof) all solutions of tan(2x)*tan(3x)*tan(4x) = tan(x),
> : with 0<x<90 (degrees).
>
> Look up the formula tan(a+b)=? (if I didn't screw it up, it's
> (tan a+tan b)/(1-tan a tan b), but look it up anyway) and
> replace the multiple angle tan's with tan x dependents.
> Solve after tan x (factorize if possible).
> Take arctan of solutions.
What an ungodly mess that is.
Try the following:
rewrite as:
[sin(x)cos(3x)][cos(2x)cos(4x)]=[cos(x)sin(3x)][sin(2x)sin(4x)]
(we will have to watch out for extraneous roots we introduced...
namely those where one of the cosines is zero, or one of the tangents
is undefined)
Now use the forumla for sin(a)cos(b)=..., cos(a)cos(b)=..., etc. to
rewrite as:
(1/4)[sin(4x)-sin(2x)][cos(6x)+cos(2x)]=(1/4)[sin(4x)+sin(2x)][cos(2x)-cos(6x)]
use sin(4x)=2sin(2x)cos(2x) and cancel factors of 1/4 and 2 to get:
sin(2x)(2cos(2x)-1)[cos(6x)+cos(2x)]=sin(2x)(2cos(2x)+1)[cos(2x)-cos(6x)]
one possibility is now sin(2x)=0 (or, for 0<=x<=90, only x=0 or 90,
but for x=90 degrees, tan(x) is undefined... an extraneous root) that
is, x=0.
cancel sin(2x) to leave:
[2cos(2x)-1][cos(6x)+cos(2x)]=[2cos(2x)+1][cos(2x)-cos(6x)]
Let me write C=cos(2x) and D=cos(6x) (the equation is:
(2C-1)(C+D)=(2C+1)(C-D))
to get:
2C^2+2CD-C-D=2C^2-2CD+C-D
or 2CD-C=0
Thus... C(2D-1)=cos(2x)(2cos(6x)-1)=0 is the result.
Cos(2x) =0 has tan(2x) undefined (an extraneous root).
For cos(6x)=1/2 we have 6x=n*360+/-60 degrees (n an integer) or
x = n*60+/-10
For 0<=x<=90 we can have n=0, x=+10 or n=1 and x=60+10=70 or 60-10=50
Thus the solution set for 0<=x<90 (no roots when a tangent is undefined)
is given by:
{0, 10, 50, 70} (in degrees)
(now imagine using the procedure suggested originally, viz. obtaining
an equation in tan(x) and solving for that... one would have to
recognize the formula for the tangent of 10, 50 and 70 degrees! Here
the solution is immediate, but we got the result down to 2x and 6x
and stopped there, as it sufficed)
Regards,
--
John McGowan | jmcg...@inch.com [Internet Channel]
| jmcg...@mail.coin.missouri.edu [COIN]
--------------+-----------------------------------------------------
Peter L. Montgomery
>Subject: Re: Solve this equation!
>
>> Eugene Gath (ga...@ul.ie) wrote:
>> : Find (with proof) all solutions of tan(2x)*tan(3x)*tan(4x) = tan(x),
>> : with 0<x<90 (degrees).
>>
>
>rewrite as:
...
>
>Let me write C=cos(2x) and D=cos(6x) (the equation is:
>
> (2C-1)(C+D)=(2C+1)(C-D))
>
>to get:
>
> 2C^2+2CD-C-D=2C^2-2CD+C-D
>
>or 2CD-C=0
After we get the answer, we can go back and derive the identity
cos(9x) sin(x)
tan(x) - tan(2x) tan(3x) tan(4x) = ------------------
cos(3x)^2 cos(4x)
whenever both sides are defined.
I found this using Maple, realizing from McGowan's solution
that 2 cos(6x) - 1 = cos(9x)/cos(3x) is a factor of the left side.
After simplifying the quotient, the above formula follows.
We can prove it by
cos(3x) cos(4x) [tan(x) - tan(2x) tan(3x) tan(4x)]
= tan(x) cos(3x) cos(4x) - tan(2x) sin(3x) sin(4x)
= sin(x) [2 cos(2x) - 1] cos(4x) - 2 sin(2x)^2 sin(3x)
= sin(x) [[2 cos(2x) - 1] cos(4x) - [1 - cos(4x)] [1 + 2 cos(2x)]]
= sin(x) [4 cos(2x) cos(4x) - 2 cos(2x) - 1]
= sin(x) [2 cos(6x) - 1]
= sin(x) cos(9x)/cos(3x)
--
Peter L. Montgomery pmon...@cwi.nl San Rafael, California
Mathematically gifted, unemployed, U.S. citizen. Interested in computer
architecture, program optimization, computer arithmetic, cryptography,
compilers, computational mathematics. 17 years industrial experience.
Etherman
> Find (with proof) all solutions of tan(2x)*tan(3x)*tan(4x) = tan(x),
> with 0<x<90 (degrees).
I assume that since you restrict x so that 0<x<90 that you want
only real values for x. The equation then has no solution.
Pf: Let u=tan(x). Then using double, triple and quadruple angle
formulae we have
2u
tan(2x)= -------
1-u^2
3u-u^3
tan(3x)= --------
1-3u^2
4u-4u^3
tan(4x)= -----------
1-6u^2+u^4
Putting these back into the equation yeilds
2u(3-u^3)(4u-4u^3)
--------------------------- = u
(1-u^2)(1-3u^2)(1-6u^2+u^4)
Which upon simplification yeilds
3u^6 - 22u^4 + 18u^2 - 1 = 0
Let v = u^2 gives
3v^3 - 22v^2 + 18 v - 1 = 0
By inspection we have the three roots
322z^(-1/3) 22
v1 = z + ------------ + ----
81 9
1 161z^(-1/3) 22 1 322z^(-1/3)
v2 = - -z^(1/3) - ----------- + ---- + -isqrt(3)(z^(1/3) - -------------)
2 81 9 2 81
1 161z^(-1/3) 22 1 322z^(-1/3)
v3 = - -z^(1/3) - ----------- + ---- - -isqrt(3)(z^(1/3) - -------------)
2 81 9 2 81
10847 1
where z= ------ + ---isqrt(65381)sqrt(3)
1458 162
We have six values for u
u1 = sqrt(v1)
u2 = sqrt(v2)
u3 = sqrt(v3)
u4 = -sqrt(v1)
u5 = -sqrt(v2)
u6 = -sqrt(v3)
Note that all the roots are complex. When we take the arctan of any of these
we have a complex number. So none of these are solns. There is another
point that was glossed over in the arithmetic. When we had
2u(3-u^3)(4u-4u^3)
--------------------------- = u
(1-u^2)(1-3u^2)(1-6u^2+u^4)
we simplified by dividing by u. This assumes that u=!0. u=0
is clearly a soln. However x=arctan(0)=0 which is not in the
allowed interval.
Etherman
PS: Believe it or not it was quicker to do this than to have
Maple solve it. When I tried Maple ran for about 15 min on
my Pentium. After seeing the solution it offered I can understand why.
I'll try to post it for those with Maple.
George Baloglou
Etherman <rc...@The.Ether> writes:
Let me pretend that I am not aware of the cubic formula(s)--which
could very well be the case, both for myself and others--and try
some more recent stuff, namely differentiation :-)
The derivative of 3v^3 - 22v^2 + 18v - 1 is 9v^2 - 44v + 18,
with irrational roots very near 4/9 and 40/9 [quadratic formula
not forgotten!]; this tells us that the cubic's local maximum
is *positive*, approximately equal to 3. We conclude that the
cubic has *three* real roots, none of which can be negative;
therefore, there must be three solutions to the original
problem.
So far, I *deliberately* avoided the use of technology; from
here on, it gets rougher and rougher: *where* are the roots?
As the cubic happens to be negative at both 0 and 1, we must
have one root in each of the intervals (0, 4/9) and (4/9, 1);
and, as the cubic changes sign between 6 and 7 (-37 to +76),
we know there is a third root in (6, 7). At this time, had I
both the time and the inclination, I would apply Newton's
algorithm starting at 0, 1 and 7, respectively, to get the
three roots with, say, two correct decimals for each; next,
I would use the good old way for finding square roots by
hand to get u and then, exhausted, I might be tempted to
push the inverse tan button on my pocket calculator :-)
[After all, the series for arctan converges a bit too
slowly ...]
The excursion into the shipwrecked man's Mathematics is over!
So, you concluded that there are no solutions, and I believe that
there are three; I tend to think that I am right, hence there must
be something wrong in the way you used the cubic formula. Indeed,
MAPLE's graphic component confirms that there exists a root near
0.175, which is consistent with my analysis (the first step in
Newton's algorithm (for v) starting at 0 yields 1/18, etc); I did
not bother to look for the other roots.
>Etherman
>
>PS: Believe it or not it was quicker to do this than to have
>Maple solve it. When I tried Maple ran for about 15 min on
>my Pentium. After seeing the solution it offered I can understand why.
>I'll try to post it for those with Maple.
>
MAPLE should be able to provide at least one solution, see above;
by the way, I have not checked the details of the reduction of
the original equation to the cubic.
George Baloglou -- (Tenured) Assistant Professor of Mathematics,
State University of New York, College at Oswego, NY 13126, USA
** sometimes my opinions contradict those of my employer **
Etherman
> So, you concluded that there are no solutions, and I believe that
> there are three; I tend to think that I am right, hence there must
> be something wrong in the way you used the cubic formula. Indeed,
> MAPLE's graphic component confirms that there exists a root near
> 0.175, which is consistent with my analysis (the first step in
> Newton's algorithm (for v) starting at 0 yields 1/18, etc); I did
> not bother to look for the other roots.
I may have made an algebra mistake.
> MAPLE should be able to provide at least one solution, see above;
> by the way, I have not checked the details of the reduction of
> the original equation to the cubic.
Maple gave six solutions (or maybe it was seven) but used 99
placeholders! I'll recheck those solutions, but I thought they
were all complex.
Etherman
Eugene Gath
>On Thu, 31 Aug 1995, Eugene Gath wrote:
>> Find (with proof) all solutions of tan(2x)*tan(3x)*tan(4x) = tan(x),
>> with 0<x<90 (degrees).
>I assume that since you restrict x so that 0<x<90 that you want
>only real values for x. The equation then has no solution.
No --the restriction was to avoid having to specify all solutions generated
by the oddness and periodicity of the functions.
>Note that all the roots are complex.
You might recall that a cubic always has at least one real root.
Be wary of Maple's exact solutions of cubics. It applies Cardano's formula,
but makes no attempt to simplify complex roots etc. In earlier realeases of
Maple it would actually give the exact solurion to quartics too, but these
days (MapleV3) that feature seems to have been removed.
Or does someone know how to do this?
As regards the problem, the answer is x=10, 50, 70 degrees.
I received correct solutions from John Mc Gowan (as posted on sci.math) and
also from Ron Winther, a variant of whose solution I give below:
Denote tan(x) by T.
Using the multiple angle formulae tan(2x) = 2T/(1 - T^2),
tan(3x) = T(3 - T^2)/(1 - 3*T^2),and tan(4x) = 4T(1 - T^2)/(1 - 6*T^2 + T^4)
and substituting gives a ratio whose numerator
is T*(3*T^6 - 27*T^4 + 33*T^2 - 1).
Noting that T is not zero for 0<x<90, then 3*T^6 - 27*T^4 + 33*T^2 - 1 = 0.
Now write this sextic as 3T^2(T^2-3)^2=(3T^2-1)^2
This gives T(3-T^2)/(1-3T^2)=+- 1/sqrt(3).
This is just tan(3x)=+- 1/sqrt(3), which gives the three solutions above.
John McGowan
|
v
> 2u(3-u^3)(4u-4u^3)
> --------------------------- = u
> (1-u^2)(1-3u^2)(1-6u^2+u^4)
>
> Which upon simplification yeilds
>
> 3u^6 - 22u^4 + 18u^2 - 1 = 0
Something has to be wrong... consider mutliplying the u*the
denominator. This yields a polynomial of degree 9 (from the u, u^2,
3u^2 and u^4 terms) with a simple zero at u=0. The numerator is of
degree 7 (the u, u^3 and 4u^3 terms) with a triple zero at u=0 (when
you correct the 3-u^3 term to 3u-u^3).
Taking the numerator -u*denominator=0 then gives u*P(u)=0 where P(u)
is a polynomial of degree 8 (even if you cancelled out the u term, the
result should be of degree 8). So the simplification is wrong.
In fact it simplifies to:
u(u-1)(u+1)(3u^6-27u^4+33u^2-1)=0 or
u(u-1)(u+1)(u[3-u^2]sqr(3)-[1-3u^2])(u[3-u^2]sqr(3)+[1-3u^2])=0
u=0 (x=0) is a root but u=1 has tan(2x) undefined (extraneous root
from clearing the denominator) while:
u(3u-u^2)/(1-3u^2)=+/-1/sqr(3)
is nothing but tan(3x)=+/-1/sqr(3) giving the roots 10,50,70 degrees.
--
John McGowan | jmcg...@inch.com [Internet Channel]
| jmcg...@mail.coin.missouri.edu [COIN]
--------------+-----------------------------------------------------
(I followed a hint I was sent, that knowing the roots --obtained as I did
from the sum and product formulae for sines and cosines-- one can realize
that the equation in tan(x)=u must have roots at tan(3x)=+/-1/sqr(3)
enabling one to factor the sixth order equation... in short, the
simplification that had been suggested by the author was wrong, and
correcting the calculations yields the answer in terms of the equation
for the tangent, after one has first noticed what the solutions are...
which is why I prefer the first method I suggested... the sum, product
formulae for sines and cosines)
la...@inland.com
> On Thu, 31 Aug 1995, Eugene Gath wrote:
>
>> Find (with proof) all solutions of tan(2x)*tan(3x)*tan(4x) = tan(x),
>> with 0<x<90 (degrees).
>
> I assume that since you restrict x so that 0<x<90 that you want
> only real values for x. The equation then has no solution.
>
> formulae we have
>
> 2u
> tan(2x)= -------
> 1-u^2
>
> 3u-u^3
> tan(3x)= --------
> 1-3u^2
>
> 4u-4u^3
> tan(4x)= -----------
> 1-6u^2+u^4
>
> Putting these back into the equation yeilds
>
> --------------------------- = u
> (1-u^2)(1-3u^2)(1-6u^2+u^4)
>
> Which upon simplification yeilds
>
> 3u^6 - 22u^4 + 18u^2 - 1 = 0
>
>
> 3v^3 - 22v^2 + 18 v - 1 = 0
>
>
> 322z^(-1/3) 22
> v1 = z + ------------ + ----
> 81 9
The first "z" on the right should be raised to the 1/3 power. Let us press on.
>
> 1 161z^(-1/3) 22 1 322z^(-1/3)
> v2 = - -z^(1/3) - ----------- + ---- + -isqrt(3)(z^(1/3) - -------------)
> 2 81 9 2 81
>
> 1 161z^(-1/3) 22 1 322z^(-1/3)
> v3 = - -z^(1/3) - ----------- + ---- - -isqrt(3)(z^(1/3) - -------------)
> 2 81 9 2 81
>
> 10847 1
> where z= ------ + ---isqrt(65381)sqrt(3)
> 1458 162
>
> We have six values for u
>
> u1 = sqrt(v1)
> u2 = sqrt(v2)
> u3 = sqrt(v3)
> u4 = -sqrt(v1)
> u5 = -sqrt(v2)
> u6 = -sqrt(v3)
>
> Note that all the roots are complex.
Hold the phone a minute. The formula generates complex numbers, _but_ these
combine in such a way as to generate real roots. Let's apply some fundamental
reasoning to the cubic equation derived above:
3v^2 - 22v^2 + 18v - 1 = 0
We know that any polynomial having odd degree and real coefficients has to have
at least one real root, and Descartes' Rule of Signs guarantees that there is
at least one positive root which implies a positive root for u. Therefore,
the problem has at least one solution in the required range. Of course, there
are really three real roots for v (if we study the algorithm for solving cubic
equations, we'll see that this is why we get all those complex intermediates),
in which case the rule of signs guarantees that they are all positive and we'll
end up with three solutions to the original problem.
Good luck!
--OL
la...@inland.com