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# Exam question

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1. Exam question

If you substitute 4 in place of x in the equation x = 8cos t, you get t = (1/3)pi

however if you substitute 2root3 in place of the y in the equation y=4sin2t, you get:

2root3/4 = sin 2t
root3/2 = sin 2t

2t = pi/3
t =pi/6

Why are the 2 t values different?
2. Re: Exam question
Ignore, accidental double post.
Last edited by Xarren; 20-06-2012 at 14:05.
3. Re: Exam question
(Original post by sabre2th1)

If you substitute 4 in place of x in the equation x = 8cos t, you get t = (1/3)pi

however if you substitute 2root3 in place of the y in the equation y=4sin2t, you get:

2root3/4 = sin 2t
root3/2 = sin 2t

2t = pi/3
t =pi/6

Why are the 2 t values different?
cos-1(1/2) = pi/3.

4sin(2pi/3) = 2rt3.

sin-1(0.5rt3)=pi/3

Whats the problem?
4. Re: Exam question
(Original post by Xarren)
cos-1(1/2) = pi/3.

4sin(2pi/3) = 2rt3.

sin-1(0.5rt3)=pi/3

Whats the problem?
The part in bold finds 2t, so don't you divide by 2, to find t ?
5. Re: Exam question
(Original post by sabre2th1)
The part in bold finds 2t, so don't you divide by 2, to find t ?
The solutions to that are pi/3, 2pi/3, (2pi+pi/3, 2pi +2pi/3 etc. but out of range)

So you divide the second solution by two and get pi/3.

Edit: Always look at the range, that is usually important and stated for a reason.
Last edited by Xarren; 20-06-2012 at 14:13.
6. Re: Exam question
(Original post by Xarren)
The solutions to that are pi/3, 2pi/3, (2pi+pi/3, 2pi +2pi/3 etc. but out of range)

So you divide the second solution by two and get pi/3.

Edit: Always look at the range, that is usually important and stated for a reason.
I understand that there is a range and that there are more solutions. However, the range is between 0 and pi/2, and pi/6 would fall in between this range. So why isn't t = pi/6? (I am not doubting you, just trying to understand your justification) thanks
7. Re: Exam question
(Original post by sabre2th1)
I understand that there is a range and that there are more solutions. However, the range is between 0 and pi/2, and pi/6 would fall in between this range. So why isn't t = pi/6? (I am not doubting you, just trying to understand your justification) thanks
Because pi/6 doesn't work for the other co-ordinate. The range is 0 < 2t < pi. In which you get 2t = 2pi/3 which does work for the other co-ordinate. pi/6 gives you a solution which is on the same level as P, but on the other side of the curve, right where the letter C is. Both give valid points, one just gives a different point to the one you are looking for.
Last edited by Xarren; 20-06-2012 at 14:37.

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Last updated: June 20, 2012
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