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UCASLord
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 10102017 13:16
Last edited by UCASLord; 16122017 at 19:06. 
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 10102017 13:22
(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
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 10102017 13:26
(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
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 10102017 13:29
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 13:33. 
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 10102017 13:47
(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 13:50. 
DFranklin
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 10102017 13:54
(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
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 10102017 14:31
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Last edited by UCASLord; 16122017 at 19:07. 
DFranklin
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 10102017 14:39
(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
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 10102017 14:48
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 15:05. 
DFranklin
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 10102017 15:11
(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
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 10102017 15:55
(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
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 10102017 15:56
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Last edited by UCASLord; 16122017 at 19:07. 
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 10102017 16:00
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Last edited by UCASLord; 16122017 at 19:07. 
DFranklin
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 10102017 16:01
(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
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 10102017 16:11
(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. 
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 10102017 16:27
(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
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