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Trogonometrical identies question! Watch

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    Help pleaase

     7 - 6cos^2 \theta = 5sin\theta

    Thank you!


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    You know an identity involving \cos^2 \theta.
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    use cos^2+ sin^2 =1 to help
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    (Original post by Libby18)
    use cos^2+ sin^2 =1 to help
    I understand that you have to use that but I don't know how, I rearranged it so it was equal to 7


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    (Original post by Annabel_lear)
    I understand that you have to use that but I don't know how, I rearranged it so it was equal to 7


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    erm you rearrange it to get what cos^2 is equal to, then substitute that into the equation above.



    If you type your working, I can help you. I'm not allowed to give answers (we're suppose to help OP work it out).
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    (Original post by Libby18)
    erm you rearrange it to get what cos^2 is equal to, then substitute that into the equation above.



    If you type your working, I can help you. I'm not allowed to give answers (we're suppose to help OP work it out).
    When you make 6cos^2theta the subject you can't jut divide by 6 and theta right?


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    Has anyone besides the OP actually tried solving it? I'm reaching a quadratic with no real solutions ...
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    (Original post by Annabel_lear)
    When you make 6cos^2theta the subject you can't jut divide by 6 and theta right?


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    nope

    right first step rearrange cos^2 +sin^2 =1 ?

    tell me what you get and we'll take it from that.

    step at a time
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    (Original post by Annabel_lear)
    When you make 6cos^2theta the subject you can't jut divide by 6 and theta right?


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    How could you divide by \theta nothing is multiplied by \theta
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    (Original post by Big-Daddy)
    Has anyone besides the OP actually tried solving it? I'm reaching a quadratic with no real solutions ...
    Nope I haven't but following the basic rules, I would get a quadratic and then factorise that (if its possible) if not, maybe the quadratic formula? if not, then I dont know :P
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    (Original post by Big-Daddy)
    Has anyone besides the OP actually tried solving it? I'm reaching a quadratic with no real solutions ...
    No, you've made a mistake. The question is fine.
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    (Original post by Libby18)
    nope

    right first step rearrange cos^2 +sin^2 =1 ?

    tell me what you get and we'll take it from that.

    step at a time
    It's not just 1 - sin^2 theta Is it?


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    (Original post by Mr M)
    No, you've made a mistake. The question is fine.
    Yes you're right it's fine, I misread 5sin(x) as sin(x).

    To the OP: yes it is just cos2(x)=1-sin2(x), substitute in and solve
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    (Original post by Big-Daddy)
    Yes you're right it's fine, I misread 5sin(x) as sin(x).

    To the OP: yes it is just cos2(x)=1-sin2(x), substitute in and solve
    I subbed it back, got:
    6sin^2 theta - 5sin theta - 1 = 0
    Turned it into:
    6x^2 -5x -1
    Factorised:
    (X-1)(6x+1)
    Does that make sinx = -1/6 , 1
    And then do I put that back into the
    7 - 6cos^2 theta = 5sin theta ?



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    (Original post by Annabel_lear)
    It's not just 1 - sin^2 theta Is it?


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    Exactly that
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    (Original post by Annabel_lear)
    I subbed it back, got:
    6sin^2 theta - 5sin theta - 1 = 0
    Turned it into:
    6x^2 -5x -1
    Factorised:
    (X-1)(6x+1)
    Does that make sinx = -1/6 , 1
    And then do I put that back into the
    7 - 6cos^2 theta = 5sin theta ?



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    You've made a mistake in substituting, it shouldn't be negative 1
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    (Original post by Annabel_lear)
    I subbed it back, got:
    6sin^2 theta - 5sin theta - 1 = 0
    Turned it into:
    6x^2 -5x -1
    Factorised:
    (X-1)(6x+1)
    Does that make sinx = -1/6 , 1
    And then do I put that back into the
    7 - 6cos^2 theta = 5sin theta ?



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    The bit in bold is where I stopped reading. Check arithmetic!
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    (Original post by Annabel_lear)
    I subbed it back, got:
    6sin^2 theta - 5sin theta - 1 = 0

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    check your algebra - I got
    6sin^2 \theta - 5sin \theta + 1 = 0
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    (Original post by davros)
    check your algebra - I got
    6sin^2 \theta - 5sin \theta + 1 = 0
    Yupp!
    Does that make it:
    (2x-1)(3x-1)

    So sinx = 1/2 , 1/3



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    (Original post by Annabel_lear)
    Yupp!
    Does that make it:
    (2x-1)(3x-1)

    So sinx = 1/2 , 1/3



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    that's what I got
 
 
 
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