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AQA FP2 De Moivre's questions

Final question. from the June 13 paper


Part a is sin4x = 4(cosx)^3sinx - 4cosxsin^3x

I can do the first part of part b but don't know how to work out the other roots

And part c is just lost on me

Thanks
Original post by TheKian
Final question. from the June 13 paper


Part a is sin4x = 4(cosx)^3sinx - 4cosxsin^3x

I can do the first part of part b but don't know how to work out the other roots

And part c is just lost on me

Thanks

For the first part of (b), you noticed that t=tanπ/16t=\tan \pi /16 is a root of the quartic by recognising that it gives tan(4π/16)=1\tan (4 \pi/ 16) = 1.

But the same can be said for any t=tanθt=\tan \theta, where θ\theta is such that tan4θ=1\tan 4\theta = 1. Can you write down all such θ\theta (between 0 and π\pi - why is this enough?), and consequently all the remaining roots of the quartic?

For part (c), write down a quartic with roots t2t^2 by making a substitution for tt into the quartic, rearranging and squaring suitably. Then recall that you can determine the sum of the roots of a polynomial from it's coefficients.
Reply 2
Original post by Farhan.Hanif93
For the first part of (b), you noticed that t=tanπ/16t=\tan \pi /16 is a root of the quartic by recognising that it gives tan(4π/16)=1\tan (4 \pi/ 16) = 1.

But the same can be said for any t=tanθt=\tan \theta, where θ\theta is such that tan4θ=1\tan 4\theta = 1. Can you write down all such θ\theta (between 0 and π\pi - why is this enough?), and consequently all the remaining roots of the quartic?

For part (c), write down a quartic with roots t2t^2 by making a substitution for tt into the quartic, rearranging and squaring suitably. Then recall that you can determine the sum of the roots of a polynomial from it's coefficients.


I'm still stuck on part c. When you say write a quartic with roots t2t^2 by maing a substitution for tt into the quartic. Do you mean t24t64t1t2t^2 -4t -6 -4t^{-1} -t^{-2} ? if so, where do I go from there.
(edited 7 years ago)
Original post by TheKian
I'm still stuck on part c. When you say write a quartic with roots t2t^2 by maing a substitution for tt into the quartic. Do you mean t24t64t1t2t^2 -4t -6 -4t^{-1} -t^{-2} ? if so, where do I go from there.

Not quite. Let t = √u. Then, after some rearrangement followed by careful squaring (i.e. to get rid of the square roots), you will have another quartic in terms of u with roots u = t^2 for each t that is a root of the old quartic.
Reply 5
Original post by TheKian
Final question. from the June 13 paper


Part a is sin4x = 4(cosx)^3sinx - 4cosxsin^3x

I can do the first part of part b but don't know how to work out the other roots

And part c is just lost on me

Thanks


For part C it is just a disguised FP1 question where you find the relationship between the roots of the polynomial and the coefficients. If you remember that Σα2 = (Σαβ)2 -2Σαβ. Apply the same concept here and, and you will have to spot a symmetry between the roots you will get to the result.
(edited 7 years ago)
Original post by Farhan.Hanif93
Not quite. Let t = √u. Then, after some rearrangement followed by careful squaring (i.e. to get rid of the square roots), you will have another quartic in terms of u with roots u = t^2 for each t that is a root of the old quartic.


That sounds like hard work. I think it is easier to let the roots be t1,t2,t3,t4t_1, t_2, t_3, t_4 then form t12+t22+t32+t42t_1^2 + t_2^2 + t_3^2+t_4^2 by considering the square of the sum of the roots - you can then use the various relationships between the coefficients and sum, product, etc of the roots.

[Aside: latex still broken - FFS - someone please fix this]
Original post by atsruser
That sounds like hard work. I think it is easier to let the roots be t1,t2,t3,t4t_1, t_2, t_3, t_4 then form t12+t22+t32+t42t_1^2 + t_2^2 + t_3^2+t_4^2 by considering the square of the sum of the roots - you can then use the various relationships between the coefficients and sum, product, etc of the roots.

[Aside: latex still broken - FFS - someone please fix this]

It largely boils down to whether one is required to take the extra step in proving the identity relating the sum of the squares of the 4 roots to the sum and products of the roots at A-Level - I don't recall it as quotable when I did FP2 around 5 years ago but I may have misremembered.

If that is the case, I don't see how this is any easier than squaring both sides of u^2 - 6u + 1 = 4(1-u)√u and rearranging for a quartic that you may simply read the result off from; as opposed to reading two results from the old quartics, proving a relation (which requires squaring, too) and then computing the sum of the squares by substitution into said relation. [Although, I concede that this is probably the less obvious/unexpected approach.]
Reply 8
Original post by Farhan.Hanif93
It largely boils down to whether one is required to take the extra step in proving the identity relating the sum of the squares of the 4 roots to the sum and products of the roots at A-Level - I don't recall it as quotable when I did FP2 around 5 years ago but I may have misremembered.

If that is the case, I don't see how this is any easier than squaring both sides of u^2 - 6u + 1 = 4(1-u)√u and rearranging for a quartic that you may simply read the result off from; as opposed to reading two results from the old quartics, proving a relation (which requires squaring, too) and then computing the sum of the squares by substitution into said relation. [Although, I concede that this is probably the less obvious/unexpected approach.]


The Vieta's formulas can be assumed in FP1 as well as FP2.

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