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    ok so here is the question:


    4sin^2ϴ-1=0

    here is what i try to do:
    4(1-cos^2ϴ)-1=0
    =4-4cos^2ϴ-1=0
    = 3=4cos2ϴ
    3/4=cos2ϴ

    but not sure where to go from here?
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    (Original post by T_H_E)
    ok so here is the question:


    4sin^2ϴ-1=0

    here is what i try to do:
    4(1-cos^2ϴ)-1=0
    =4-4cos^2ϴ-1=0
    = 3=4cos2ϴ
    3/4=cos2ϴ

    but not sure where to go from here?
    What do you think? Is there an idea you've got but aren't sure whether to do it? Or are you completely stumped?
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    (Original post by Andy98)
    What do you think? Is there an idea you've got but aren't sure whether to do it? Or are you completely stumped?
    completely stuck..lol.
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    (Original post by T_H_E)
    completely stuck..lol.
    Is it the \cos^2(\theta) bit that's confusing you? If so, what would you do normally to get rid of a square?
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    (Original post by T_H_E)
    ok so here is the question:


    4sin^2ϴ-1=0

    here is what i try to do:
    4(1-cos^2ϴ)-1=0
    =4-4cos^2ϴ-1=0
    = 3=4cos2ϴ
    3/4=cos2ϴ

    but not sure where to go from here?
    I'd root it to get cosϴ = sqrt(3)/2 and then solve with whichever method you were taught (CAST diagram, drawing y = cos(x)...). p.s. examsolutions.net has a great explanation of both methods if you're not sure how to use them
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    (Original post by T_H_E)
    completely stuck..lol.
    Ok, first off, it's worrying that you mechanically used the sin^2 and cos^2 identity straight away without thinking about what you're trying to achieve with that.
    Why are you converting the sin^2 to a cos^2? How does that make it easier?

    You can simply rearrange the equation for sin^2, then obviously you need to take square roots of both sides and you should certainly be able to solve it from there.
    It's funny because this is no harder that solving something like 2x^2+1=9 or sin(theta) = 2/3. It's just both together. Too many students forget everything they know when they see a problem that's slightly different to what they've done before. Maths is not being taught well in schools...
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    (Original post by Andy98)
    Is it the \cos^2(\theta) bit that's confusing you? If so, what would you do normally to get rid of a square?
    umm well cos2=1-sin2ϴ
    but that then goes like this..

    3/4 = 1-sin2ϴ
    = 3/4 + sin2ϴ=1
    sin2ϴ=4/3
    and u r still left with a square lol

    edit: or u could sqrt lol
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    (Original post by vectorpi)
    I'd root it to get cosϴ = sqrt(3)/2 and then solve with whichever method you were taught (CAST diagram, drawing y = cos(x)...). p.s. examsolutions.net has a great explanation of both methods if you're not sure how to use them
    Well you're forgetting that you need to consider the negative square root too.
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    (Original post by PrimeLime)
    Ok, first off, it's worrying that you mechanically used the sin^2 and cos^2 identity straight away without thinking about what you're trying to achieve with that.
    Why are you converting the sin^2 to a cos^2? How does that make it easier?

    You can simply rearrange the equation for sin^2, then obviously you need to take square roots of both sides and you should certainly be able to solve it from there.
    It's funny because this is no harder that solving something like 2x^2+1=9 or sin(theta) = 2/3. It's just both together. Too many students forget everything they know when they see a problem that's slightly different to what they've done before. Maths is not being taught well in schools...
    lol or maybe i am just dumb:/
    edit: btw if u just rearrange
    4 sin2ϴ-1=0
    u get
    sin2ϴ=1/4, then u have a same problem, or can u square root both side before doin inverse sine? is it as easy as that?
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    (Original post by T_H_E)
    umm well cos2=1-sin2ϴ
    but that then goes like this..

    3/4 = 1-sin2ϴ
    = 3/4 + sin2ϴ=1
    sin2ϴ=4/3
    and u r still left with a square lol

    edit: or u could sqrt lol
    See you don't even understand when to use that identity. It's just what you've been taught and you're using it without understanding what you're doing.
    When doing a maths question (I hesitate to call it a 'problem'), just take a second to think what you're trying to accomplish:
    "I want to be able to solve for theta"
    "I guess I'll need something like sin(theta)=k, which I know how to solve"
    "So I'll get rid of the square by square rooting"
    Please don't routinely applying formulae to a problem in hopes that it works. Understanding questions will allow you to answer a lot more and a lot more easily.
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    (Original post by T_H_E)
    umm well cos2=1-sin2ϴ
    but that then goes like this..

    3/4 = 1-sin2ϴ
    = 3/4 + sin2ϴ=1
    sin2ϴ=4/3
    and u r still left with a square lol

    edit: or u could sqrt lol
    Try the square root

    (Original post by T_H_E)
    lol or maybe i am just dumb:/
    He is right, secondary schools teach maths to a poor standard (generally speaking).
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    (Original post by T_H_E)
    lol or maybe i am just dumb:/
    edit: btw if u just rearrange
    4 sin2ϴ-1=0
    u get
    sin2ϴ=1/4, then u have a same problem?
    No, there's no problem there. The sin(theta) term is clearly putting you off for some reason, since you definitely (I hope, anyway) know how to solve x^2=1/4.
    Don't know how to work with sin^2? Square rooting is a good solution to that.
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    (Original post by PrimeLime)
    See you don't even understand when to use that identity. It's just what you've been taught and you're using it without understanding what you're doing.
    When doing a maths question (I hesitate to call it a 'problem', just take a second to think what you're trying to accomplish:
    "I want to be able to solve for theta"
    "I guess I'll need something like sin(theta)=k, which I know how to solve"
    "So I'll get rid of the square by square rooting"
    Please don't routinely applying formulae to a problem in hopes that it works. Understanding questions will allow you to answer a lot more and a lot more easily.
    You're right, but you could probably tone it down a little....
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    (Original post by Andy98)
    Try the square root



    He is right, secondary schools teach maths to a poor standard (generally speaking).
    Plus, the style of the A-Level exams completely allow for blind memorisation of formulae and methods with virtually no understanding required. It's really quite sad, because people get the complete wrong idea of what maths is.
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    for some reason i see new posts a bit late lol, but thanks will try that
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    (Original post by Andy98)
    You're right, but you could probably tone it down a little....
    Trying to be as honest as possible. I think people need to be aware of what exactly they're doing wrong, even if it is a little harsh. It's only help.
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    (Original post by PrimeLime)
    Plus, the style of the A-Level exams completely allow for blind memorisation of formulae and methods with virtually no understanding required. It's really quite sad, because people get the complete wrong idea of what maths is.
    Indeed, I'm currently on A-level FM - which is a little better, but not much
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    (Original post by PrimeLime)
    Trying to be as honest as possible. I think people need to be aware of what exactly they're doing wrong, even if it is a little harsh. It's only help.
    True, it's just usually I find people seem to learn better when you're patient with them
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    (Original post by T_H_E)
    ok so here is the question:


    4sin^2ϴ-1=0

    here is what i try to do:
    4(1-cos^2ϴ)-1=0
    =4-4cos^2ϴ-1=0
    = 3=4cos2ϴ
    3/4=cos2ϴ

    but not sure where to go from here?
    Just square root cos^2 so you get
    Cos x = +/- root3/2
    The use the usual cast diagram to solve for your angles!
    So if you had the interval 0<x<360
    the angles would be: 30,150,210,330
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    (Original post by Andy98)
    Indeed, I'm currently on A-level FM - which is a little better, but not much
    Yeah further maths is bit better in that regard. FP1 is certainly getting a lot better at testing understanding, but D1... XD that's LITERALLY just algorithms.
 
 
 
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