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Trogonometrical identies question! Watch

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


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    I see you've managed to figure it out
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    (Original post by Libby18)
    I see you've managed to figure it out
    Yupp


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    (Original post by davros)
    that's what I got
    Did you then sub back and do:
    This is for 1/2

    7-6cos^2x = 2.5
    -6cos^2x = -4.5
    -Cos^2x = -0.75
    Cosx = sqroot 3/2
    Cos(inverse) = 30


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    (Original post by Libby18)
    I see you've managed to figure it out
    For the value of 1/2 do you sub back and get:

    7-6cos^2x = 2.5
    -6cos^2x = -4.5
    -Cos^2x = -0.75
    Cosx = sqroot 3/2
    Cos(inverse) = 30




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    sin is now 1/3 and 1/2 so you use those values and perhaps use your CAST diagram to help you out http://www.google.co.uk/imgres?imgur...9QEwAQ&dur=363

    If I've done it right, one of my values is 30 and I've got four values.

    is there a range given i.e between 0 and 180
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    (Original post by Annabel_lear)
    For the value of 1/2 do you sub back and get:

    7-6cos^2x = 2.5
    -6cos^2x = -4.5
    -Cos^2x = -0.75
    Cosx = sqroot 3/2
    Cos(inverse) = 30



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    doesn't work. I think you're over- complicating it.
    Also, I think you should go over the examples in the textbook (as I think there is alittle lack of understanding and more trial and error being used)
    Its the same as doing a quadratic
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    (Original post by Annabel_lear)
    For the value of 1/2 do you sub back and get:

    7-6cos^2x = 2.5
    -6cos^2x = -4.5
    -Cos^2x = -0.75
    Cosx = sqroot 3/2
    Cos(inverse) = 30




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    As Libby says you're overcomplicating it

    The object is to find the values of x. You have found that sin x = 1/2 or sin x = 1/3 so use your calculator to find the main (principal) angles where sin x = 1/2 or sin x = 1/3 and then look at the graph of sin x to read off any other values of x with the same value of sin within the range of angles you are given
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    (Original post by Libby18)
    doesn't work. I think you're over- complicating it.
    Also, I think you should go over the examples in the textbook (as I think there is alittle lack of understanding and more trial and error being used)
    Its the same as doing a quadratic
    @ davros & Libby
    Ok! I put the values in for 1/2 and got 30 and (180-30=) 150
    Then for 1/3 got
    19.5 and (180-19.5=) 160.5


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    (Original post by Annabel_lear)
    @ davros & Libby
    Ok! I put the values in for 1/2 and got 30 and (180-30=) 150
    Then for 1/3 got
    19.5 and (180-19.5=) 160.5


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    Yep I think thats what I got when I did it earlier.
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    (Original post by Libby18)
    Yep I think thats what I got when I did it earlier.
    Woooo thank you!
    And thanks to everyone else who helped


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    (Original post by Annabel_lear)
    @ davros & Libby
    Ok! I put the values in for 1/2 and got 30 and (180-30=) 150
    Then for 1/3 got
    19.5 and (180-19.5=) 160.5


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    That looks good. Remember that sin and cos repeat every 360 degrees, so there'll be even more solutions if there are no restrictions on the angle. looking back at your 1st post, you didn't say whether the question put any limits on the the range of \theta
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    (Original post by davros)
    That looks good. Remember that sin and cos repeat every 360 degrees, so there'll be even more solutions if there are no restrictions on the angle. looking back at your 1st post, you didn't say whether the question put any limits on the the range of \theta
    It was 0 < x < 360
    The method I used originally (subbing 1/2 and 1/3 back in) also got me the same answers, so I could use that method as well right?


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    (Original post by Annabel_lear)
    It was 0 < x < 360
    The method I used originally (subbing 1/2 and 1/3 back in) also got me the same answers, so I could use that method as well right?


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    Well you could but it's not very efficient because once you've found sin \theta you just need to invert the sine and find the angle(s). With your method you're having to work out what cos is as well, and then invert the cosine.
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    (Original post by davros)
    Well you could but it's not very efficient because once you've found sin \theta you just need to invert the sine and find the angle(s). With your method you're having to work out what cos is as well, and then invert the cosine.
    Ok thank you!


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