The simultaneous equations in x,y,
(cosθ)x  (sinθ)y = 2
(sinθ)x + (cosθ)y = 1
are solvable for: all values of θ between 0 and 2 pi; or all except 1, 2, or 3 in the same range.
What I've tried doing is redefining cosθ in terms of sinθ, and then working from there, but it doesn't seem to be taking me anywhere.

UCASLord
 Follow
 0 followers
 11 badges
 Send a private message to UCASLord
 Thread Starter
Offline11ReputationRep: Follow
 1
 10102017 14:16

 Follow
 2
 10102017 14:22
(Original post by UCASLord)
The simultaneous equations in x,y,
(cosθ)x  (sinθ)y = 2
(sinθ)x + (cosθ)y = 1
are solvable for: all values of θ between 0 and 2 pi; or all except 1, 2, or 3 in the same range.
What I've tried doing is redefining cosθ in terms of sinθ, and then working from there, but it doesn't seem to be taking me anywhere. 
DFranklin
 Follow
 56 followers
 17 badges
 Send a private message to DFranklin
Online17ReputationRep: Follow
 3
 10102017 14:26
(Original post by UCASLord)
The simultaneous equations in x,y,
(cosθ)x  (sinθ)y = 2
(sinθ)x + (cosθ)y = 1
are solvable for: all values of θ between 0 and 2 pi; or all except 1, 2, or 3 in the same range.
What I've tried doing is redefining cosθ in terms of sinθ, and then working from there, but it doesn't seem to be taking me anywhere.
Otherwise, just solve for x and y as you would any other simultaneous equations (effectively treating sin θ and cos θ as constants). 
artful_lounger
 Follow
 19 followers
 20 badges
 Send a private message to artful_lounger
 Community Assistant
Offline20ReputationRep: Follow
 4
 10102017 14:29
If you've done FM, it's a lot quicker to solve with matrices, and also "nicer" in the solution, I believe.
Otherwise it seems pretty ugly.Last edited by artful_lounger; 10102017 at 14:33. 
_gcx
 Follow
 110 followers
 18 badges
 Send a private message to _gcx
 Visit _gcx's homepage!
Online18ReputationRep: Follow
 5
 10102017 14:47
(Original post by UCASLord)
The simultaneous equations in x,y,
(cosθ)x  (sinθ)y = 2
(sinθ)x + (cosθ)y = 1
are solvable for: all values of θ between 0 and 2 pi; or all except 1, 2, or 3 in the same range.
What I've tried doing is redefining cosθ in terms of sinθ, and then working from there, but it doesn't seem to be taking me anywhere.Last edited by _gcx; 10102017 at 14:50. 
DFranklin
 Follow
 56 followers
 17 badges
 Send a private message to DFranklin
Online17ReputationRep: Follow
 6
 10102017 14:54
(Original post by artful_lounger)
If you've done FM, it's a lot quicker to solve with matrices, and also "nicer" in the solution, I believe. 
UCASLord
 Follow
 0 followers
 11 badges
 Send a private message to UCASLord
 Thread Starter
Offline11ReputationRep: Follow
 7
 10102017 15:31
(Original post by DFranklin)
If you know about determinants (of 2x2 matrices), this is the quickest approach here.
Otherwise, just solve for x and y as you would any other simultaneous equations (effectively treating sin θ and cos θ as constants).(Original post by artful_lounger)
If you've done FM, it's a lot quicker to solve with matrices, and also "nicer" in the solution, I believe.
Otherwise it seems pretty ugly.(Original post by _gcx)
To clarify what the others mean by "use matrices", write the system as . As DFranklin said, you can then use the determinant to check where the system is consistent.
In regards to this question, I tried to solve it by defining x, and then substituting the definition of x for x, and then trying to find y. I can post my full working if needed, but what I pretty much get to is this
[Realized my previous reply had incorrect working] When I multiply this out, it becomes Then So then the equations are solvable for any values where s =/=0?
In regards to my FP1, C3, C4: is it worth learning these units before doing the MAT? I've noticed quite a few of them offer faster solutions for a number of questions.Last edited by UCASLord; 10102017 at 15:40. 
DFranklin
 Follow
 56 followers
 17 badges
 Send a private message to DFranklin
Online17ReputationRep: Follow
 8
 10102017 15:39
(Original post by UCASLord)
Thanks for all your replies guys. I have a couple of questions here, one about this specific question, and one about the FP1/C3/C4 units.
In regards to this question, I tried to solve it by defining x, and then substituting the definition of x for x, and then trying to find y. I can post my full working if needed, but what I pretty much get to is this
When you wish to solve simultaneous equations, particularly ones with nonnumeric coefficients (i.e. things like x sin t + y cos t = 1, rather than 3 sin t + 4 cos 4 = 1), you should generally avoid dividing until you absolutely have to.Folliowing is a big hint about the correct approach, but not a full solutionMultiply the first equation by cos and the 2nd by sin and add the two resultant equations to find x. Do something similar to find y.In regards to my FP1, C3, C4: is it worth learning these units before doing the MAT? I've noticed quite a few of them offer faster solutions for a number of questions. 
OnceUponAThyme
 Follow
 0 followers
 5 badges
 Send a private message to OnceUponAThyme
Offline5ReputationRep: Follow
 9
 10102017 15:48
First of all, the question is not asking you to solve the equations. The question is asking you about whether or not it is solvable for a range of values of theta.
Think about it. You have two lines. If there is a solution, then the lines must intersect (as the point of intersection is the solution). Any pair of lines will intersect unless they are parallel. Therefore, you just need to work out the gradients of the two lines and equate them (i.e. make the lines parallel) and the values of theta that solve that equation (if there are any solutions) will be the values of theta for which the lines don't intersect and thus don't have a solution.Last edited by OnceUponAThyme; 10102017 at 16:05. 
DFranklin
 Follow
 56 followers
 17 badges
 Send a private message to DFranklin
Online17ReputationRep: Follow
 10
 10102017 16:11
(Original post by OnceUponAThyme)
First of all, the question is not asking you to solve the equations. The question is asking you about whether or not it is solvable for a range of values of theta.
Think about it. You have two lines. If there is a solution, then the lines must intersect (as the point of intersection is the solution). Any pair of lines will intersect unless they are parallel. Therefore, you just need to work out the gradients of the two lines and equate them (i.e. make the lines parallel) and the values of theta that solve that equation (if there are any solutions) will be the values of theta for which the lines don't intersect and thus don't have a solution.
The other point I'd make is that the OP should definitely make sure he does know how to solve equations of this form before taking the MAT. So at this point I recommend he press on with that approach. 
OnceUponAThyme
 Follow
 0 followers
 5 badges
 Send a private message to OnceUponAThyme
Offline5ReputationRep: Follow
 11
 10102017 16:55
(Original post by DFranklin)
The one issue with this is that although it doesn't happen here, for many problems of this type, you may need to distinguish between the case where the lines are parallel and there is no solution and the case where they are parallel and there are infinite solutions. At which point you're back to solving the equations.
The other point I'd make is that the OP should definitely make sure he does know how to solve equations of this form before taking the MAT. So at this point I recommend he press on with that approach.
And for the MAT, I think it is very useful to think about the problem from different angles and carefully think about what the problem is actually asking. Of course, both approaches are valid. 
UCASLord
 Follow
 0 followers
 11 badges
 Send a private message to UCASLord
 Thread Starter
Offline11ReputationRep: Follow
 12
 10102017 16:56
(Original post by OnceUponAThyme)
First of all, the question is not asking you to solve the equations. The question is asking you about whether or not it is solvable for a range of values of theta.
Think about it. You have two lines. If there is a solution, then the lines must intersect (as the point of intersection is the solution). Any pair of lines will intersect unless they are parallel. Therefore, you just need to work out the gradients of the two lines and equate them (i.e. make the lines parallel) and the values of theta that solve that equation (if there are any solutions) will be the values of theta for which the lines don't intersect and thus don't have a solution.
EDIT: to clarify, the gradient of the first line is cos(t)/sin(t) and the gradient of the second line is sin(t)/cost(t)
cos(t)/sin(t) = sin(t)/cos(t)
cos^2(t) = sin^2(t)
cos^2(t) + sin^2(t) = 0
Now can you spot the answer?
However, I do have a question in regards to how you arrived at the gradient of the first line being cos(theta)/sin(theta). From what I know, there's two main ways to find the gradient. One of them is m=difference in y/difference in x, and the other method is differentiation. I can't quite tell how you've found cos(t)/sin(t) being the gradient. Could you explain a bit more on this point? 
UCASLord
 Follow
 0 followers
 11 badges
 Send a private message to UCASLord
 Thread Starter
Offline11ReputationRep: Follow
 13
 10102017 17:00
(Original post by DFranklin)
The solution you get should not be anything like this messy. I'm not saying what you've done is wrong, but I can't recognize the correct answer from it, which doesn't look good.
When you wish to solve simultaneous equations, particularly ones with nonnumeric coefficients (i.e. things like x sin t + y cos t = 1, rather than 3 sin t + 4 cos 4 = 1), you should generally avoid dividing until you absolutely have to.Folliowing is a big hint about the correct approach, but not a full solutionMultiply the first equation by cos and the 2nd by sin and add the two resultant equations to find x. Do something similar to find y.
I think it can't hurt, but only if you have the time. I think one of the reasons people struggle with the MAT is that although it's "only" C1, C2 material, you don't generally get a good understanding of the material until you've used it in the context of the later modules.
Edit: It seems so easy now that I know how to do it...Last edited by UCASLord; 10102017 at 17:09. 
DFranklin
 Follow
 56 followers
 17 badges
 Send a private message to DFranklin
Online17ReputationRep: Follow
 14
 10102017 17:01
(Original post by OnceUponAThyme)
Surely if the lines are parallel and there are infinite solutions then the lines are the same? In which case, you would just have to show if there's a common point or not (e.g. by letting x = 0 and finding y for both lines).And for the MAT, I think it is very useful to think about the problem from different angles and carefully think about what the problem is actually asking. Of course, both approaches are valid. 
DFranklin
 Follow
 56 followers
 17 badges
 Send a private message to DFranklin
Online17ReputationRep: Follow
 15
 10102017 17:11
(Original post by UCASLord)
Hey, only just saw your reply. Thanks for your advice. I've managed to confirm the solution to the equation using the "big hint." (plus it's easier when you already know what the answer should be :3)
If you have a pair of simultaneous equations to solve for X and Y:
aX + bY = P
cX + dY = Q
then rather than doing something like writing X = (PbY) / a (which involves dividing by a, which might be 0), the thing to do is multiply the 1st equation by d and the 2nd by b so the "Y" terms are the same:
ad X + bd Y = dP
bc X + bd Y = bQ
then subtracting the 2nd from the first gives
(adbc) X = dP  bQ
Now I've said" don't divide", but now you've eliminated Y, you need to "sit or get off the pot". More specifically, at this point, either (adbc) = 0 (and then if dP  BQ = 0, any value of X will work, otherwise no value of X will work), or (adbc) =/= 0 and you can divide by it to find X.
If you do similar calculations for Y, you'll find you have the same (adbc) factor.
adbc is the "determinant" we were talking about earlier, if it's nonzero, then the equations will always have a unique solution. Note in particular that it's possible for any one of a, b, c, d to be 0 and there to still be a unique solution, so dividing by any of them when you don't know if they might be 0 is not a good idea.
I doubt you'd get this in the MAT, but If you had 3 variables X, Y, Z, you would use a similar approach to first eliminate Z and then Y. 
OnceUponAThyme
 Follow
 0 followers
 5 badges
 Send a private message to OnceUponAThyme
Offline5ReputationRep: Follow
 16
 10102017 17:27
(Original post by UCASLord)
Made it all too easy there, but thanks in any case. :P By the end point, I can spot the answer: it's solvable for any value of theta.
However, I do have a question in regards to how you arrived at the gradient of the first line being cos(theta)/sin(theta). From what I know, there's two main ways to find the gradient. One of them is m=difference in y/difference in x, and the other method is differentiation. I can't quite tell how you've found cos(t)/sin(t) being the gradient. Could you explain a bit more on this point?
then m is the gradient
Reply
Submit reply
Related discussions:
TSR Support Team
We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.
This forum is supported by:
 SherlockHolmes
 Notnek
 charco
 Mr M
 TSR Moderator
 Nirgilis
 usycool1
 Changing Skies
 James A
 rayquaza17
 RDKGames
 randdom
 davros
 Gingerbread101
 Kvothe the Arcane
 The Financier
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 Reality Check
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 LeCroissant
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Leviathan1741
 Moltenmo
Updated: October 10, 2017
Share this discussion:
Tweet