Solution to a system of transcendental equations

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The discussion revolves around solving a system of equations involving trigonometric functions, specifically to express variables a and b in terms of c1, c2, and θ. The initial equations are transformed by squaring and adding, leading to a relationship between the magnitudes of the vectors involved. Participants suggest using linear algebra techniques, such as matrix rotation, to simplify the problem, while others propose direct algebraic manipulations. A solution is ultimately provided, showing that a and b can be expressed as combinations of c1 and c2 with trigonometric functions of θ. The conversation emphasizes the importance of recognizing the geometric interpretation of the equations for a clearer solution path.
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I got stuck while solving a problem in statics unable to solve a system of equations of the form obtained as below,

a cosθ - b sinθ = c1
a sinθ + b cosθ = c2

I wish to obtain a and b in terms of c1 and c2.Squaring and adding I obtained,

a2 + b2 = c12 + c22

I need another relation in a,b,c1 and c2 to achieve the result.How do I proceed next?
 
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a and b will depend not only on c_1 and c_2 but on θ.

Think about a rotation of θ about the origin.
 
PeroK said:
a and b will depend not only on c_1 and c_2 but on θ.

Think about a rotation of θ about the origin.

Well I did forget to mention c1 and c2 are functions in θ.How do I solve for a and b in terms of c1, c2 and θ?
 
Soumalya said:
Well I did forget to mention c1 and c2 are functions in θ.How do I solve for a and b in terms of c1, c2 and θ?


If you know a little linear algebra, you can express the system of equations in matrix form, then multiple by the inverse rotation matrix.
 
PeroK said:
If you know a little linear algebra, you can express the system of equations in matrix form, then multiple by the inverse rotation matrix.

Well I am not quite used to linear algebra in practice but I will have a look if things appear easier in comparison to conventional algebra.
 
3
Soumalya said:
Well I am not quite used to linear algebra in practice but I will have a look if things appear easier in comparison to conventional algebra.
If you recognise what you have as vector rotation, you can write down the answer without any calculations!

The equation you derived shows that the magnitude of a vector is invariant under rotation.
 
PeroK said:
3

If you recognise what you have as vector rotation, you can write down the answer without any calculations!

The equation you derived shows that the magnitude of a vector is invariant under rotation.

I apologize for using boldface type alphabets but the equation is a simple transcendental equation I used bold alphabets to highlight the quantities primarily involved in my problem.Noting that it's a simple scalar equation how can I solve for a and b in terms of c1, c2 and θ?
 
Soumalya said:
I got stuck while solving a problem in statics unable to solve a system of equations of the form obtained as below,

a cosθ - b sinθ = c1
a sinθ + b cosθ = c2

I wish to obtain a and b in terms of c1 and c2.Squaring and adding I obtained,

a2 + b2 = c12 + c22

I need another relation in a,b,c1 and c2 to achieve the result.How do I proceed next?


These are not transcedental equations. If just Θ is unknown It is not even system of equations since you have only one indepedent variable in each equation. Such trigonometric equations can be solved by transforming sum of two trigonometric function in one trigonometric function. See method http://www.sinclair.edu/centers/mathlab/pub/findyourcourse/worksheets/DiffEq/ConvertingTheForm_asinx.pdf .
Or one can use theory of complex numbers.
 
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Soumalya said:
I apologize for using boldface type alphabets but the equation is a simple transcendental equation I used bold alphabets to highlight the quantities primarily involved in my problem.Noting that it's a simple scalar equation how can I solve for a and b in terms of c1, c2 and θ?

Why not

##a = c_1 cos(\theta) + c_2 sin(\theta)##
##b = -c_1 sin(\theta) + c_2 cos(\theta)##
 
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  • #10
PeroK said:
Why not

##a = c_1 cos(\theta) + c_2 sin(\theta)##
##b = -c_1 sin(\theta) + c_2 cos(\theta)##

Yes!

Thank You :)
 
  • #11
PeroK said:
Why not

##a = c_1 cos(\theta) + c_2 sin(\theta)##
##b = -c_1 sin(\theta) + c_2 cos(\theta)##

Could you briefly outline the procedure that was used to obtain the solution?
 
  • #12
Soumalya said:
Could you briefly outline the procedure that was used to obtain the solution?

I just wrote it down.

##(c_1, c_2)## is the vector obtained by rotating the vector ##(a, b)## by ##-\theta## about the origin.

So, to obtain ##(a, b)## you rotate ##(c_1, c2)## by ##+\theta##
 
  • #13
PeroK said:
I just wrote it down.

##(c_1, c_2)## is the vector obtained by rotating the vector ##(a, b)## by ##-\theta## about the origin.

So, to obtain ##(a, b)## you rotate ##(c_1, c2)## by ##+\theta##

Is there an easy analytical way to arrive to the solution using ordinary algebra rather than using linear algebra? :confused:
 
  • #14
Soumalya said:
Is there an easy analytical way to arrive to the solution using ordinary algebra rather than using linear algebra? :confused:

Multiply the first equation by ##cos\theta## and the second by ##sin\theta## and add to get ##a##.
 
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  • #15
Thank You once again :)
 
  • #16
Soumalya said:
I got stuck while solving a problem in statics unable to solve a system of equations of the form obtained as below,

a cosθ - b sinθ = c1
a sinθ + b cosθ = c2

I wish to obtain a and b in terms of c1 and c2.Squaring and adding I obtained,

a2 + b2 = c12 + c22

I need another relation in a,b,c1 and c2 to achieve the result.How do I proceed next?

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  • #18
Soumalya said:
I got stuck while solving a problem in statics unable to solve a system of equations of the form obtained as below,

a cosθ - b sinθ = c1
a sinθ + b cosθ = c2

I wish to obtain a and b in terms of c1 and c2.Squaring and adding I obtained,

a2 + b2 = c12 + c22

I need another relation in a,b,c1 and c2 to achieve the result.How do I proceed next?

Multiply the first equation by cos(\theta): a cos^2(\theta)- b cos(\theta)sin(\theta)= c_1 cos(\theta)
Multiply the second equation by sin(\theta): a sin^2(\theta)+ b sin(\theta)cos(\theta)= c_2 sin(\theta)
Add those two equations: a= c_1 cos(\theta)a+ c_2 sin(\theta).

Multiply the first equation by sin(\theta), the second by cos(\theta), and subtract to get b.
 
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