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C2 Trig question!

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    I have two questions:

    Question 1 (picture below)



    Question 2:

    Once you find out the two first values of say cos/sine/tan, for the remaining ones would you keep adding and minusing 360?

    i.e. say if you had 40, 120 as first two values, the third would be 40+360 and fourth would be 120+360 for all sine/tan/cos?
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    question 1: plot it in the A quadrant. 3x=120, then x=40. unless you want to plot 3x?

    and um CAST being positive or negative has nothing to do with the sign right in front of the degree. it has to do with the sign of the value after you put that degree in a trig function. just like in your question, 120 is positive, but if you find the value of cos 120, its -0.5.

    and mind you, 240 is also a solution here 3x=240, x=80.
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    question 2: yes. just check the range in which they ask you to find the values
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    If you need to leave your answer in general form (i.e. include every possible answer), it would be 60 +- 20 + 360n, where n is an integer. But i don't think you would have to do that in a question where you have to plot it.
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    For Q1, when you say Cos is positive (because it is 120) this demonstrates a complete lack of understanding

    Cos is negative as it = -0.5
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    You wouldn't plot it as 3x, you say Let 3x=X and then work out the solutions and then once you have the solutions say, oh but since 3x=X times the solutions by three.
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    (Original post by kitriviolet)
    question 1: plot it in the A quadrant. 3x=120, then x=40. unless you want to plot 3x?

    and um CAST being positive or negative has nothing to do with the sign right in front of the degree. it has to do with the sign of the value after you put that degree in a trig function. just like in your question, 120 is positive, but if you find the value of cos 120, its -0.5.

    and mind you, 240 is also a solution here 3x=240, x=80.
    Okay let's say if you were to plot x=40. This would go into the 'C' quadrant but as TenOfThem said it is negative so you can't plot x=40 in the 'C' quadrant as cos can only be positive there. What now?
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    (Original post by Kravez)
    Okay let's say if you were to plot x=40. This would go into the 'C' quadrant, then you would draw a line connecting it into the 'A' quadrant. What now?
    What on earth does this mean


    Which quadrant is 120 in

    Where else is Cos negative
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    (Original post by TenOfThem)
    What on earth does this mean


    Which quadrant is 120 in

    Where else is Cos negative


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    So you have the 2 positions for 3x

    and you can add 360 as many times as you need to for each of these


    Then, divide by 3 to get x
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    (Original post by Kravez)
    Have a look at this example:

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    (Original post by TenOfThem)
    So you have the 2 positions for 3x

    and you can add 360 as many times as you need to for each of these


    Then, divide by 3 to get x
    Yes, but you didn't read my question. What would you do first to find the second value after you have gotten the P.V which is 120. (Refer to picture in my previous post)
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    (Original post by Kravez)
    Yes, but you didn't read my question. What would you do first to find the second value after you have gotten the P.V which is 120.
    We got one value as 120deg.

    To find the other values, do 360-120 and 360+120, this gives 240 and 480. I don't know your interval, but 480 might be outside the interval, hence our solutions will be, 120 and 240.

    NB: All vaues are in degrees.
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    (Original post by Kravez)
    Yes, but you didn't read my question. What would you do first to find the second value after you have gotten the P.V which is 120. (Refer to picture in my previous post)
    If you do not understand CAST, and you clearly do not, I suggest that you use a different method

    You have a red line that represents 120 and you are saying that it is 180-120 ... why
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    (Original post by TenOfThem)
    If you do not understand CAST, and you clearly do not, I suggest that you use a different method

    You have a red line that represents 120 and you are saying that it is 180-120 ... why


    Now read below:

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    (Original post by Kravez)
    ...
    Angles are measured from the anticlockwise direction.
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    How can you go from writing x=-48.59 to x=48.59

    Surely you can see that your 2 red lines (which are correctly positioned) are -48.59 and -131.51

    Or that (if you want to go positive) the lines are 228.59 and 311.41
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    (Original post by TenOfThem)
    How can you go from writing x=-48.59 to x=48.59

    Surely you can see that your 2 red lines (which are correctly positioned) are -48.59 and -131.51

    Or that (if you want to go positive) the lines are 228.59 and 311.41
    My mistake for writing x = 48.59 (in a rush). So basically it doesn't matter which quadrant the line is in the (360 - theta), (180 - theta), and (180 + theta) is irrelevant?
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    (Original post by Kravez)
    My mistake for writing x = 48.59 (in a rush)
    Watch the videos on this page, http://examsolutions.co.uk/maths-rev.../example-1.php
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    you seem to think that 120 and -48.59 are theta

    they aren't

    In order to use the rules that you want to use you have to use the corresponding value in the first quadrant


    BUT

    I think you need to consider a different approach to these questions

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