When does a real solution exist for this system of trigonometric equations?


by sodemus
Tags: trig equations
sodemus
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#1
Oct10-09, 02:58 PM
P: 32
So, my question is really simple. What are the requirements for a1, a2, C1, C2 for a real solution to exist for the following system of equations?

sin(theta1) + a1*sin(theta2) = C1
cos(theta1) + a2*cos(theta2) = C2

If it would be helpful, I'm not interested in the solution, necessarily (could do that with numerical methods), just interested for what parameter configurations it exists over real numbers.

I have tried assuming a complex solution and then imposing conditions on that but it hasn't really taken me anywhere...
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tiny-tim
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#2
Oct11-09, 03:11 AM
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Hi sodemus!

(have a theta: θ and try using the X2 tag just above the Reply box )

Put the θ1s on the left, then get rid of them by squaring and adding

then use standard trigonometric identities to get a quadratic equation in sin2θ2.
sodemus
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#3
Oct11-09, 04:35 AM
P: 32
Hey Tim, thanks for the reply!
Well, I knew that would be a possibility, I just didn't want to go there since it doesn't really yield a quadratic equation. As far as I can see it does actually give me a intrinsically 4th order equation, doesn't it? And a pretty ugly looking one as well.

I was thinking that through assuming a real solution one could see the parametric conditions and get away with solving a lesser order equation but without actually solving the original equation, but that might not be possible.

Edit: sorry, did I miss an obvious factorizing!?

tiny-tim
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#4
Oct11-09, 05:05 AM
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When does a real solution exist for this system of trigonometric equations?


Yes, you're right, it is a 4th order equation in sinθ2

I can't see any way of getting round that

(if there was one, it would be an easy solution to general 4th order equations!).
Gerenuk
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#5
Oct11-09, 06:02 PM
P: 1,057
I'm not mathematician, but I suspect there is no easy solution. If you multiply the first equation by imaginary i and add it to the second, you realize that this problem is equivalent to the task of finding an intersection between an ellipse and a displaced circle. Maybe that helps? I'll think about that :)
sodemus
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#6
Oct11-09, 06:30 PM
P: 32
I don't think it's really an ellipse, and certainly not for all parameter values (I might be wrong though). Look at the where a1+1>C1 for example. Anyway, I agree with you, I think it takes at least two steps of calculations and decisions (pretty simple calculations, but kind of messy in the original parameter space) to decide whether the solution is real or complex. I hoped for one closed form expression. Oh, well, math can't be beautiful everywhere. :)

Thanks for the input in any case! I appreciate it!

Edit: d-mn, it looked like such an easy problem at first glance!! Just two freakin' sines and cosines!
Gerenuk
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#7
Oct11-09, 09:47 PM
P: 1,057
I think the following (almost) solves the problem.

If you write
[tex]a_1\sin\theta_2=C_1-\sin\theta_1[/tex]
[tex]a_2\cos\theta_2=C_2-\cos\theta_1[/tex]
you see that the problem is equivalent to finding if there is an intersection between an ellipse at the origin with axis a1 and a2 and a circle of unit radius at (C2,C1).

They just start to have intersections, when both of them touch (at only one point). For this case the tangents have an equal gradient
[tex]
\frac{a_1}{a_2}\tan\theta_2=\tan\theta_1=g
[/tex]
Looking at the two equations you can further derive
[tex]
g=\frac{C_1}{C_2}
[/tex]

Now you can rearrange
[tex]
\frac{C_1}{C_2}=\frac{a_1}{a_2}\tan\theta_2=\tan\theta_1
[/tex]
for both theta1 and theta2 and put it back into the second equation (remember [itex]\tan^2\alpha+1=1/\cos^2\alpha[/itex]).

You finally get
[tex]
\pm\frac{1}{\sqrt{C_1^2+C_2^2}}\pm\frac{1}{\sqrt{\frac{C_1^2}{a_1^2}+\f rac{C_2^2}{a_2^2}}}=1
[/tex]
which is the boundary of the 4-dimensional solution space you are looking for. The tricky part is now to think which direction (slightly smaller/bigger variables) is "inside" that region.

Good luck :)
sodemus
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#8
Oct11-09, 10:14 PM
P: 32
Thank you so much Gerenuk! It is exactly what I was looking for! I think solutions exist when the expression lies between 0 and 1. Would that make sense? Anyway, I apologize for my doubt about the circles, I didn't think in the right coordinates.

Again,
Thanks!

Edit: Hah! Quite an interesting equation by the way...
Gerenuk
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#9
Oct11-09, 10:21 PM
P: 1,057
Thanks for question btw.

That makes me look up inversion in geometry. The final equation suspiciously looks like a triangle inequality in a transformed space.

Let me know if you somehow check this equation visually/numerically.
Integral
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#10
Oct11-09, 11:28 PM
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Quote Quote by Gerenuk View Post
...
They just start to have intersections, when both of them touch (at only one point). For this case the tangents have an equal gradient

...
Unfortunatly at the point of intersection the graphs cross, therefore the slopes CANNOT be equal.
Gerenuk
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#11
Oct12-09, 12:38 AM
P: 1,057
Quote Quote by Integral View Post
Unfortunatly at the point of intersection the graphs cross, therefore the slopes CANNOT be equal.
Please read all of the post. They *touch* where they *start* to have intersections. The equation determine the parameters for which they touch.
tiny-tim
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#12
Oct12-09, 02:20 AM
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Yes, Gerenuk is right

the original question (which I'd completely forgotten! ) was only to determine whether there were real solutions!
sodemus
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#13
Oct12-09, 08:38 AM
P: 32
Quote Quote by Gerenuk View Post
I think the following (almost) solves the problem.

If you write
[tex]a_1\sin\theta_2=C_1-\sin\theta_1[/tex]
[tex]a_2\cos\theta_2=C_2-\cos\theta_1[/tex]
you see that the problem is equivalent to finding if there is an intersection between an ellipse at the origin with axis a1 and a2 and a circle of unit radius at (C2,C1).

They just start to have intersections, when both of them touch (at only one point). For this case the tangents have an equal gradient
[tex]
\frac{a_1}{a_2}\tan\theta_2=\tan\theta_1=g
[/tex]
Looking at the two equations you can further derive
[tex]
g=\frac{C_1}{C_2}
[/tex]

Now you can rearrange
[tex]
\frac{C_1}{C_2}=\frac{a_1}{a_2}\tan\theta_2=\tan\theta_1
[/tex]
for both theta1 and theta2 and put it back into the second equation (remember [itex]\tan^2\alpha+1=1/\cos^2\alpha[/itex]).

You finally get
[tex]
\pm\frac{1}{\sqrt{C_1^2+C_2^2}}\pm\frac{1}{\sqrt{\frac{C_1^2}{a_1^2}+\f rac{C_2^2}{a_2^2}}}=1
[/tex]
which is the boundary of the 4-dimensional solution space you are looking for. The tricky part is now to think which direction (slightly smaller/bigger variables) is "inside" that region.

Good luck :)
I think you were right the first time around. I can't come up with a single case where the signs matter. They change the situation slightly but not in the when it comes to the binary decision making.
Gerenuk
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#14
Oct12-09, 10:29 AM
P: 1,057
To determine the signs I guess one can consider the a1=a2 simplification and look at the circles.

Then the result is the sign combination + + > and + - < for the equation and thus
[tex]
\left|\frac{1}{\sqrt{C_1^2+C_2^2}}-1\right|<\frac{1}{\sqrt{\frac{C_1^2}{a_1^2}+\frac{C_2^2}{a_2^2}}}
[/tex]
for intersections.
sodemus
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#15
Oct12-09, 12:31 PM
P: 32
Quote Quote by Gerenuk View Post
To determine the signs I guess one can consider the a1=a2 simplification and look at the circles.

Then the result is the sign combination + + > and + - < for the equation and thus
[tex]
\left|\frac{1}{\sqrt{C_1^2+C_2^2}}-1\right|<\frac{1}{\sqrt{\frac{C_1^2}{a_1^2}+\frac{C_2^2}{a_2^2}}}
[/tex]
for intersections.
Agreed. :)
Gerenuk
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#16
Oct12-09, 12:55 PM
P: 1,057
Actually on the bus I just figured that I didn't consider all cases circle sizes and all signs matter. The three quantities indeed have to fulfil the "triangle inequality".
[tex]
\left|\frac{1}{\sqrt{C_1^2+C_2^2}}-\frac{1}{\sqrt{\frac{C_1^2}{a_1^2}+\frac{C_2^2}{a_2^2}}}\right|<1<\frac {1}{\sqrt{C_1^2+C_2^2}}+\frac{1}{\sqrt{\frac{C_1^2}{a_1^2}+\frac{C_2^2} {a_2^2}}}
[/tex]
That's my final result I hope


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