C3 Trig

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    How do I solve sin(3X-53.1) = -1 without using cast? This was my solution but the answer is 107.7.

    sin(3X-53.1)=-1
    inverse sin both sides: 3x-53.1 = -90, 3x = -36.9, x=-12.3
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    (Original post by doglover123)
    How do I solve sin(3X-53.1) = -1 without using cast? This was my solution but the answer is 107.7.

    sin(3X-53.1)=-1
    inverse sin both sides: 3x-53.1 = -90, 3x = -36.9, x=-12.3
    Use general solutions.

    \sin(x)=\alpha \Rightarrow x=360n+\arcsin(\alpha), x=360n+180-\arcsin(\alpha), \forall n \in \mathbb{Z}
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    (Original post by RDKGames)
    Use general solutions.

    \sin(x)=\alpha \Rightarrow x=360n+\arcsin(\alpha), x=360n+180-\arcsin(\alpha), \forall n \in \mathbb{Z}

    I don't quite understand what you did. Why did you include the 360n and 180?
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    (Original post by doglover123)
    How do I solve sin(3X-53.1) = -1 without using cast? This was my solution but the answer is 107.7.

    sin(3X-53.1)=-1
    inverse sin both sides: 3x-53.1 = -90, 3x = -36.9, x=-12.3
    sin 270 also equals -1 so if you use that, using the same method, you get 107.7
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    (Original post by miless090)
    sin 270 also equals -1 so if you use that, using the same method, you get 107.7
    Ah I see - is that something i'd just have to know?
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    (Original post by doglover123)
    Ah I see - is that something i'd just have to know?
    There is usually a range given in the question e.g. 0 < x < 360

    but because you are solving 3x - 53 you need to adjust the range.

    0 < 3x < 1080 multiplying inequality by 3,

    -53 < x < 1027 so use that for CAST.
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    (Original post by doglover123)
    I don't quite understand what you did. Why did you include the 360n and 180?
    They're to do with symmetry of sine.

    If you think about your problem, you essentially have a horizontal line intersecting some curve of sine. If you imagine you have \frac{1}{2}=\sin(x) then if you were to sketch the line and the curve between 0 and 2pi then you would see that the line intersects the sine curve at x=\arcsin(1/2) as well as some point later on. Due to symmetry of sine around x=\frac{\pi}{2}, you get that other point of intersection by going \pi radians along the x-axis from 0, then turning around and going -\arcsin(1/2) radians. So your two solutions here would be x=\arcsin(1/2) and x=\pi-\arcsin(1/2).

    Since sine repeats the same pattern every 2\pi, we can generalise by saying intersections happen every 2\pi which turns our equations into x=2\pi n + \arcsin(1/2) and x=2\pi n + \pi - \arcsin(1/2) where of course n can be any integer as it determines which period of sine you are referring to. You should choose it accordingly to your range.

    Anyway, that's just a stupid long explanation for something very simple, and avoids the nonsense of CAST. Applies in the same way for degrees using conversion.
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    (Original post by Muttley79)
    There is usually a range given in the question e.g. 0 < x < 360

    but because you are solving 3x - 53 you need to adjust the range.

    0 < 3x < 1080 multiplying inequality by 3,

    -53 < x < 1027 so use that for CAST.
    I completely forgot to use that method! I think i should brush up on C2 trig! Thank you!

    (Original post by RDKGames)
    They're to do with symmetry of sine.

    If you think about your problem, you essentially have a horizontal line intersecting some curve of sine. If you imagine you have \frac{1}{2}=\sin(x) then if you were to sketch the line and the curve between 0 and 2pi then you would see that the line intersects the sine curve at x=\arcsin(1/2) as well as some point later on. Due to symmetry of sine around x=\frac{\pi}{2}, you get that other point of intersection by going \pi radians along the x-axis from 0, then turning around and going -\arcsin(1/2) radians. So your two solutions here would be x=\arcsin(1/2) and x=\pi-\arcsin(1/2).

    Since sine repeats the same pattern every 2\pi, we can generalise by saying intersections happen every 2\pi which turns our equations into x=2\pi n + \arcsin(1/2) and x=2\pi n + \pi - \arcsin(1/2) where of course n can be any integer as it determines which period of sine you are referring to. You should choose it accordingly to your range.

    Anyway, that's just a stupid long explanation for something very simple, and avoids the nonsense of CAST. Applies in the same way for degrees using conversion.
    I had to read that 3 times to understand it but thank you, I see where your coming from now!
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    (Original post by RDKGames)
    They're to do with symmetry of sine.

    If you think about your problem, you essentially have a horizontal line intersecting some curve of sine. If you imagine you have \frac{1}{2}=\sin(x) then if you were to sketch the line and the curve between 0 and 2pi then you would see that the line intersects the sine curve at x=\arcsin(1/2) as well as some point later on. Due to symmetry of sine around x=\frac{\pi}{2}, you get that other point of intersection by going \pi radians along the x-axis from 0, then turning around and going -\arcsin(1/2) radians. So your two solutions here would be x=\arcsin(1/2) and x=\pi-\arcsin(1/2).

    Since sine repeats the same pattern every 2\pi, we can generalise by saying intersections happen every 2\pi which turns our equations into x=2\pi n + \arcsin(1/2) and x=2\pi n + \pi - \arcsin(1/2) where of course n can be any integer as it determines which period of sine you are referring to. You should choose it accordingly to your range.

    Anyway, that's just a stupid long explanation for something very simple, and avoids the nonsense of CAST. Applies in the same way for degrees using conversion.
    Your method is overly complicated for A level Maths where general solutions are NOT required - there is nothing wrong with CAST ...
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    (Original post by Muttley79)
    Your method is overly complicated for A level Maths where general solutions are NOT required - there is nothing wrong with CAST ...
    Well he DID ask for a method without CAST so I provided one for him, it isn't too hard to understand. I find CAST as too much of a fuss as opposed to general solutions.
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    (Original post by RDKGames)
    Well he DID ask for a method without CAST so I provided one for him, it isn't too hard to understand. I find CAST as too much of a fuss as opposed to general solutions.
    Why offer the general solution method then? It's not appropriate ...
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    (Original post by Muttley79)
    Why offer the general solution method then? It's not appropriate ...
    It is a perfectly acceptable method within the A-Level exams as it is covered in FP1, I used it for C2, C3 and C4, so I'd say it is an appropriate substitute for CAST if OP understands it and prefers it.
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    (Original post by RDKGames)
    It is a perfectly acceptable method within the A-Level exams as it is covered in FP1, I used it for C2, C3 and C4, so I'd say it is an appropriate substitute for CAST if OP understands it and prefers it.
    It is not advised for A level Maths - it's not in the spec.

    Just because you prefer it does not make it appropriate for everyone.
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    (Original post by Muttley79)
    It is not advised for A level Maths - it's not in the spec.

    Just because you prefer it does not make it appropriate for everyone.
    Just because it is not on the spec it does not make it inappropriate. I used a method in Mechanics 3 for a question in Further Pure 4, so obviously it was not on the mark scheme and I wasn't sure whether I would get the marks, and when my teacher reached out to AQA they said it is fine as long as it is clearly displayed and the correct solution is obtained. Same idea applies here.

    Yes I prefer it, and I put it forward for someone who may prefer it as well, I'm not forcing them to use it.
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    (Original post by RDKGames)
    Just because it is not on the spec it does not make it inappropriate. I used a method in Mechanics 3 for a question in Further Pure 4, so obviously it was not on the mark scheme and I wasn't sure whether I would get the marks, and when my teacher reached out to AQA they said it is fine as long as it is clearly displayed and the correct solution is obtained. Same idea applies here.

    Yes I prefer it, and I put it forward for someone who may prefer it as well, I'm not forcing them to use it.
    It requires memorising the formulas for general solution - not a great idea.

    Please don't over-complicate methods for people ... you could have suggested they talk to their teacher who will know better methods than your approach.
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    (Original post by Muttley79)
    It requires memorising the formulas for general solution - not a great idea.

    Please don't over-complicate methods for people ... you could have suggested they talk to their teacher who will know better methods than your approach.
    Memorising formulae is not difficult, they are not difficult to derive either.

    This is not an over-complicated method, I simply explained everything that is going on with the method, OP can feel free to ignore it, and as I said, he/she is not forced to use it. It's completely up to OP whether they choose to ignore it, or look more into it, I merely provide it with my own explanation.
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    (Original post by RDKGames)
    Memorising formulae is not difficult, they are not difficult to derive either.

    This is not an over-complicated method, I simply explained everything that is going on with the method, OP can feel free to ignore it, and as I said, he/she is not forced to use it. It's completely up to OP whether they choose to ignore it, or look more into it, I merely provide it with my own explanation.
    You are assuming a massive amount - most people find memorising formulas hard.

    If you are going to offer help then please think hard about whether the approach you are suggesting is appropriate for that poster.
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    (Original post by Muttley79)
    You are assuming a massive amount - most people find memorising formulas hard.

    If you are going to offer help then please think hard about whether the approach you are suggesting is appropriate for that poster.
    General solutions are fine, you have to learn so many formulae anyway... and its another tool you can use so I dont see why this is a problem
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    (Original post by Muttley79)
    You are assuming a massive amount - most people find memorising formulas hard.

    If you are going to offer help then please think hard about whether the approach you are suggesting is appropriate for that poster.
    Memorising formulae may be hard for some, yes, but anyone can derive them if he/she investigates where they come from. Nothing more than a simple trig sketch is needed. This shouldn't be too hard for somebody doing C3 which is why I believe it is appropriate, and OP understood it so I do not see anything wrong here. Some students will understand, others wont, but mentioning this method can be worth it in the long-term.
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    (Original post by 123Master321)
    General solutions are fine, you have to learn so many formulae anyway... and its another tool you can use so I dont see why this is a problem
    There are no formulae that you need to learn in C1 to C4 - why learn this complicated one?
 
 
 
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