Mixing
4 equations with 4 unknowns
$begingroup$
I have reduced a problem I'm trying to solve, to the following $4$ equations:
$$cos(theta) = frac{A}{b+c}$$
$$sin(theta) = frac{D-e}{b-c}$$
$$sin(theta) = frac{c}{e}$$
$$cos(theta)^2 + sin(theta)^2 = 1$$
where only $A$ and $D$ are known.
I having a hard time solving this analytically. Is there an analytic solution to these equations or can a solution only be found numerically? Is there a(n) (easy) way to tell whether it is one or the other?
analysis numerical-methods
asked Feb 1 at 22:59
JensJens
4,03721032
$endgroup$
|
show 1 more comment
$begingroup$
I have reduced a problem I'm trying to solve, to the following $4$ equations:
$$cos(theta) = frac{A}{b+c}$$
$$sin(theta) = frac{D-e}{b-c}$$
$$sin(theta) = frac{c}{e}$$
$$cos(theta)^2 + sin(theta)^2 = 1$$
where only $A$ and $D$ are known.
I having a hard time solving this analytically. Is there an analytic solution to these equations or can a solution only be found numerically? Is there a(n) (easy) way to tell whether it is one or the other?
analysis numerical-methods
asked Feb 1 at 22:59
JensJens
4,03721032
$endgroup$
2
$begingroup$
You only have 2 equations... The last equation is always satisfied and $theta$ can be eliminated. In the end you just have $frac ce = frac{D-e}{b-c}$ and $frac{c^2}{e^2} + frac{A^2}{(b+c)^2}=1$.
$endgroup$
– PierreCarre
Feb 1 at 23:11
$begingroup$
Thanks @JohnOmielan, i'll correct the comment.
$endgroup$
– PierreCarre
Feb 1 at 23:16
$begingroup$
@PierreCarre So the last equation is not independent of the others, is that right? I had a suspicion this was the case, but how does one determine this is the case?
$endgroup$
– Jens
Feb 1 at 23:26
$begingroup$
The last equation is an universal condition. You may as well consider $0=0$ as your last equation.
$endgroup$
– PierreCarre
Feb 1 at 23:27
$begingroup$
Would you accept a solution where given known $,A,D,$ then $b$ is the solution of a quadratic equation in $sin(theta),,$ and $,c,$ the solution of a linear equation in $,b,sin(theta),$ and $,e,$ is linear in $,b,c,sin(theta).,$?
$endgroup$
– Somos
Feb 2 at 0:19
|
show 1 more comment
$begingroup$
I have reduced a problem I'm trying to solve, to the following $4$ equations:
$$cos(theta) = frac{A}{b+c}$$
$$sin(theta) = frac{D-e}{b-c}$$
$$sin(theta) = frac{c}{e}$$
$$cos(theta)^2 + sin(theta)^2 = 1$$
where only $A$ and $D$ are known.
I having a hard time solving this analytically. Is there an analytic solution to these equations or can a solution only be found numerically? Is there a(n) (easy) way to tell whether it is one or the other?
analysis numerical-methods
asked Feb 1 at 22:59
JensJens
4,03721032
$endgroup$
I have reduced a problem I'm trying to solve, to the following $4$ equations:
$$cos(theta) = frac{A}{b+c}$$
$$sin(theta) = frac{D-e}{b-c}$$
$$sin(theta) = frac{c}{e}$$
$$cos(theta)^2 + sin(theta)^2 = 1$$
where only $A$ and $D$ are known.
I having a hard time solving this analytically. Is there an analytic solution to these equations or can a solution only be found numerically? Is there a(n) (easy) way to tell whether it is one or the other?
analysis numerical-methods
analysis numerical-methods
asked Feb 1 at 22:59
JensJens
4,03721032
asked Feb 1 at 22:59
JensJens
4,03721032
asked Feb 1 at 22:59
JensJens
4,03721032
asked Feb 1 at 22:59
JensJens
4,03721032
asked Feb 1 at 22:59
JensJens
4,03721032
4,03721032
2
$begingroup$
You only have 2 equations... The last equation is always satisfied and $theta$ can be eliminated. In the end you just have $frac ce = frac{D-e}{b-c}$ and $frac{c^2}{e^2} + frac{A^2}{(b+c)^2}=1$.
$endgroup$
– PierreCarre
Feb 1 at 23:11
$begingroup$
Thanks @JohnOmielan, i'll correct the comment.
$endgroup$
– PierreCarre
Feb 1 at 23:16
$begingroup$
@PierreCarre So the last equation is not independent of the others, is that right? I had a suspicion this was the case, but how does one determine this is the case?
$endgroup$
– Jens
Feb 1 at 23:26
$begingroup$
The last equation is an universal condition. You may as well consider $0=0$ as your last equation.
$endgroup$
– PierreCarre
Feb 1 at 23:27
$begingroup$
Would you accept a solution where given known $,A,D,$ then $b$ is the solution of a quadratic equation in $sin(theta),,$ and $,c,$ the solution of a linear equation in $,b,sin(theta),$ and $,e,$ is linear in $,b,c,sin(theta).,$?
$endgroup$
– Somos
Feb 2 at 0:19
|
show 1 more comment
2
$begingroup$
You only have 2 equations... The last equation is always satisfied and $theta$ can be eliminated. In the end you just have $frac ce = frac{D-e}{b-c}$ and $frac{c^2}{e^2} + frac{A^2}{(b+c)^2}=1$.
$endgroup$
– PierreCarre
Feb 1 at 23:11
$begingroup$
Thanks @JohnOmielan, i'll correct the comment.
$endgroup$
– PierreCarre
Feb 1 at 23:16
$begingroup$
@PierreCarre So the last equation is not independent of the others, is that right? I had a suspicion this was the case, but how does one determine this is the case?
$endgroup$
– Jens
Feb 1 at 23:26
$begingroup$
The last equation is an universal condition. You may as well consider $0=0$ as your last equation.
$endgroup$
– PierreCarre
Feb 1 at 23:27
$begingroup$
Would you accept a solution where given known $,A,D,$ then $b$ is the solution of a quadratic equation in $sin(theta),,$ and $,c,$ the solution of a linear equation in $,b,sin(theta),$ and $,e,$ is linear in $,b,c,sin(theta).,$?
$endgroup$
– Somos
Feb 2 at 0:19
2
2
$begingroup$
You only have 2 equations... The last equation is always satisfied and $theta$ can be eliminated. In the end you just have $frac ce = frac{D-e}{b-c}$ and $frac{c^2}{e^2} + frac{A^2}{(b+c)^2}=1$.
$endgroup$
– PierreCarre
Feb 1 at 23:11
$begingroup$
You only have 2 equations... The last equation is always satisfied and $theta$ can be eliminated. In the end you just have $frac ce = frac{D-e}{b-c}$ and $frac{c^2}{e^2} + frac{A^2}{(b+c)^2}=1$.
$endgroup$
– PierreCarre
Feb 1 at 23:11
$begingroup$
Thanks @JohnOmielan, i'll correct the comment.
$endgroup$
– PierreCarre
Feb 1 at 23:16
$begingroup$
Thanks @JohnOmielan, i'll correct the comment.
$endgroup$
– PierreCarre
Feb 1 at 23:16
$begingroup$
@PierreCarre So the last equation is not independent of the others, is that right? I had a suspicion this was the case, but how does one determine this is the case?
$endgroup$
– Jens
Feb 1 at 23:26
$begingroup$
@PierreCarre So the last equation is not independent of the others, is that right? I had a suspicion this was the case, but how does one determine this is the case?
$endgroup$
– Jens
Feb 1 at 23:26
$begingroup$
The last equation is an universal condition. You may as well consider $0=0$ as your last equation.
$endgroup$
– PierreCarre
Feb 1 at 23:27
$begingroup$
The last equation is an universal condition. You may as well consider $0=0$ as your last equation.
$endgroup$
– PierreCarre
Feb 1 at 23:27
$begingroup$
Would you accept a solution where given known $,A,D,$ then $b$ is the solution of a quadratic equation in $sin(theta),,$ and $,c,$ the solution of a linear equation in $,b,sin(theta),$ and $,e,$ is linear in $,b,c,sin(theta).,$?
$endgroup$
– Somos
Feb 2 at 0:19
$begingroup$
Would you accept a solution where given known $,A,D,$ then $b$ is the solution of a quadratic equation in $sin(theta),,$ and $,c,$ the solution of a linear equation in $,b,sin(theta),$ and $,e,$ is linear in $,b,c,sin(theta).,$?
$endgroup$
– Somos
Feb 2 at 0:19
|
show 1 more comment
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2
$begingroup$
You only have 2 equations... The last equation is always satisfied and $theta$ can be eliminated. In the end you just have $frac ce = frac{D-e}{b-c}$ and $frac{c^2}{e^2} + frac{A^2}{(b+c)^2}=1$.
$endgroup$
– PierreCarre
Feb 1 at 23:11
$begingroup$
Thanks @JohnOmielan, i'll correct the comment.
$endgroup$
– PierreCarre
Feb 1 at 23:16
$begingroup$
@PierreCarre So the last equation is not independent of the others, is that right? I had a suspicion this was the case, but how does one determine this is the case?
$endgroup$
– Jens
Feb 1 at 23:26
$begingroup$
The last equation is an universal condition. You may as well consider $0=0$ as your last equation.
$endgroup$
– PierreCarre
Feb 1 at 23:27
$begingroup$
Would you accept a solution where given known $,A,D,$ then $b$ is the solution of a quadratic equation in $sin(theta),,$ and $,c,$ the solution of a linear equation in $,b,sin(theta),$ and $,e,$ is linear in $,b,c,sin(theta).,$?
$endgroup$
– Somos
Feb 2 at 0:19