Edexcel Further Maths FP2 June 2014

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A.M.C.
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Hi,
Can someone please help me explain the solution to part (b):
Its question 4 on the exam paper.

https://qualifications.pearson.com/c...-June-2014.pdf

Many thanks.

4. (a) Use de Moivre's theorem to show that

cos6theta = 32cos6theta - 48cos4theta +18cos2theta - 1

(b) Hence solve for 0 ≤ theta ≤ pi/2

64cos6theta - 96cos4theta+36cos2theta - 3 = 0

Giving your answer as exact multiples of pi.
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ghostwalker
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(Original post by A.M.C.)
Hi,
Can someone please help me explain the solution to part (b):
Its question 4 on the exam paper.

https://qualifications.pearson.com/c...-June-2014.pdf

Many thanks.

4. (a) Use de Moivre's theorem to show that

cos6theta = 32cos6theta - 48cos4theta +18cos2theta - 1

(b) Hence solve for 0 ≤ theta ≤ pi/2

64cos6theta - 96cos4theta+36cos2theta - 3 = 0

Giving your answer as exact multiples of pi.
Look at the coefficients of the powers of cos theta. How do they compare between what you've shown in (a), and what you want to solve in (b)? How might you substitute one into the other?
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A.M.C.
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(Original post by ghostwalker)
Look at the coefficients of the powers of cos theta. How do they compare between what you've shown in (a), and what you want to solve in (b)? How might you substitute one into the other?
Hi,
So far, this is what I have got for part b:
I know that 64cos6theta-96cos4theta+36cos2theta-3 = 0 is equal to 2cos6theta-1 comparing it to my answer from part a.
With 2cos6theta-1 = 0;
6theta = cos-1(1/2)
6theta = pi/3

I'm not sure how to elaborate from here.
Could you please provide some steer?
Many Thanks.
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old_engineer
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(Original post by A.M.C.)
Hi,
So far, this is what I have got for part b:
I know that 64cos6theta-96cos4theta+36cos2theta-3 = 0 is equal to 2cos6theta-1 comparing it to my answer from part a.
With 2cos6theta-1 = 0;
6theta = cos-1(1/2)
6theta = pi/3

I'm not sure how to elaborate from here.
Could you please provide some steer?
Many Thanks.
You solution 6theta = (pi)/3 should really be 6theta = (pi)/3 + 2k(pi) where k is any integer, to account for the repeating nature of the cosine function. You can solve that to find those values of theta that are within the required interval. But there is a second solution to cos(6theta) = 1/2, which you have ignored so far. You can find the second solution from a CAST diagram or from knowledge of the shape of the cosine curve. Once you have found the second solution, you need to handle it in the same way as the first solution above.
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A.M.C.
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(Original post by old_engineer)
You solution 6theta = (pi)/3 should really be 6theta = (pi)/3 + 2k(pi) where k is any integer, to account for the repeating nature of the cosine function. You can solve that to find those values of theta that are within the required interval. But there is a second solution to cos(6theta) = 1/2, which you have ignored so far. You can find the second solution from a CAST diagram or from knowledge of the shape of the cosine curve. Once you have found the second solution, you need to handle it in the same way as the first solution above.
Hi,
Could you please explain how you got 6theta = pi/3+2k(pi)?
Also, I am not quite sure how you've approached towards the second solution, could you please elaborate upon that too please?

After having a look on the mark-scheme, as well as pi/3, pi/5 and pi/7 are also solutions. However, there is no explanation towards how those answers were received.
Could you please explain?

Many Thanks.
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old_engineer
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(Original post by A.M.C.)
Hi,
Could you please explain how you got 6theta = pi/3+2k(pi)?
Also, I am not quite sure how you've approached towards the second solution, could you please elaborate upon that too please?

After having a look on the mark-scheme, as well as pi/3, pi/5 and pi/7 are also solutions. However, there is no explanation towards how those answers were received.
Could you please explain?

Many Thanks.
The cosine function is cyclic, repeating every 2(pi) radians. Thus cos(pi/3) = 1/2 but also cos(7pi/3) = 1/2, cos(13pi/3) = 1/2 etc etc. So if cos(x) = 1/2, then x = (pi)/3 + 2k(pi) are all solutions. But, separately, x = -(pi)/3 is also a solution, as are all the possible x = -(pi)/3 + 2k(pi).
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A.M.C.
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(Original post by old_engineer)
The cosine function is cyclic, repeating every 2(pi) radians. Thus cos(pi/3) = 1/2 but also cos(7pi/3) = 1/2, cos(13pi/3) = 1/2 etc etc. So if cos(x) = 1/2, then x = (pi)/3 + 2k(pi) are all solutions. But, separately, x = -(pi)/3 is also a solution, as are all the possible x = -(pi)/3 + 2k(pi).
That's wonderful thank you so much.
Do you think you can provide me with some reading material about cosine/sin graphs, so I can grasp an better understanding of how to tackle such questions?
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old_engineer
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(Original post by A.M.C.)
That's wonderful thank you so much.
Do you think you can provide me with some reading material about cosine/sin graphs, so I can grasp an better understanding of how to tackle such questions?
This is an AS level topic. It is covered, for example, by Chapter 10 of the Pearson Year 1/AS textbook.
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A.M.C.
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(Original post by old_engineer)
This is an AS level topic. It is covered, for example, by Chapter 10 of the Pearson Year 1/AS textbook.
Thank you once again.
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