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Simultaneous equations? with trig? not sure watch

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    Eliminate theta from the equations: x=cos2theta , y=sectheta

    I have no clue what to do here. Is it meant to be like simultaneous equations because i tired and failed
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    Do you know the formulas relating:
    • cos(theta) and sec(theta)
    • cos(2theta) and cos(theta)
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    (Original post by rayquaza17)
    Do you know the formulas relating:
    • cos(theta) and sec(theta)
    • cos(2theta) and cos(theta)
    Well, i know the with cos2theta will be (most likely in this case) 2cos^2theta-1

    I am not sure by what you mean by: cos(theta) and sec(theta). I know they sec is 1/cos
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    (Original post by Nirm)
    Well, i know the with cos2theta will be (most likely in this case) 2cos^2theta-1

    I am not sure by what you mean by: cos(theta) and sec(theta). I know they sec is 1/cos
    Yep.

    Sec(theta)=1/cos(theta).
    So 1/y=cos(theta)
    Now if you know x=2cos^2(theta)-1, can you sub the above equation in?
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    (Original post by Nirm)
    Well, i know the with cos2theta will be (most likely in this case) 2cos^2theta-1

    I am not sure by what you mean by: cos(theta) and sec(theta). I know they sec is 1/cos
    That's good. So you know that

    y=\dfrac{1}{\cos \theta}

    and

    x=2 \cos^2( \theta)-1.

    Can you make any progress from there?
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    (Original post by rayquaza17)
    Yep.

    Sec(theta)=1/cos(theta).
    So 1/y=cos(theta)
    Now if you know x=2cos^2(theta)-1, can you sub the above equation in?
    (Original post by BuryMathsTutor)
    That's good. So you know that

    y=\dfrac{1}{\cos \theta}

    and

    x=2 \cos^2( \theta)-1.

    Can you make any progress from there?
    Yeah. So i ended up with (2/y^2)-1. Is this it? I thought there would be more to it haha ( if this is the answer then thank you for your help )
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    (Original post by Nirm)
    Yeah. So i ended up with (2/y^2)-1. Is this it? I thought there would be more to it haha
    Yes, x=\dfrac{2}{y^2}-1 but what values can x take?
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    (Original post by BuryMathsTutor)
    Yes, x=\dfrac{2}{y^2}-1 but what values can x take?
    Do i have to make y=0 then see what i get for x?If that is the case then i get x=-1
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    (Original post by Nirm)
    Do i have to make y=0 then see what i get for x?If that is the case then i get x=-1
    No, just note that -1\le x \le 1 and y \le -1 or y \ge 1.
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    (Original post by BuryMathsTutor)
    No, just note that -1\le x \le 1 and y \le -1 or y \ge 1.
    Ahh okay, thank you very much!
 
 
 
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