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Simultaneous equations? with trig? not sure

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
Do you know the formulas relating:

•

cos(theta) and sec(theta)

•

cos(2theta) and cos(theta)

Reply 2
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Nirm
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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 :s-smilie:
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 :s-smilie:


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 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 :s-smilie:


That's good. So you know that

y=1cos⁡θy=\dfrac{1}{\cos \theta}

and

x=2cos⁡2(θ)−1x=2 \cos^2( \theta)-1.

Can you make any progress from there?
Reply 5
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Nirm
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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=1cos⁡θy=\dfrac{1}{\cos \theta}

and

x=2cos⁡2(θ)−1x=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 :smile:)
(edited 9 years ago)
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=2y2−1x=\dfrac{2}{y^2}-1 but what values can x take?
Reply 7
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Nirm
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Original post by BuryMathsTutor
Yes, x=2y2−1x=\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
(edited 9 years ago)
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≤x≤1-1\le x \le 1 and y≤−1y \le -1 or y≥1y \ge 1.
Reply 9
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Nirm
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Original post by BuryMathsTutor
No, just note that −1≤x≤1-1\le x \le 1 and y≤−1y \le -1 or y≥1y \ge 1.

Ahh okay, thank you very much!

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