Polynomials with Trigonometric roots and a request (1 Viewer)

[Intilegience]

Hey, I've been stuck on this question for a while so was hoping if someone could help me with parts c and d. (parts a and b given for context)

a) obtain cos6θ as a polynomial in cosθ.
b) Hence show that x=cos ((2k+1) π)/12 where k=0,1,2,3,4,5 are the roots of the equation 32x^6-48x^4+18x^2-1=0
c)deduce that cos π/12 * cos5π/12=1/4
d) deduce that cos^2 5π/12 +cos^2 π/12=1

I also would like some advice on how to tackle these types of problems better, right now my approach is to just haphazardly use sum and product of roots to try to get a relationship, but If someone could maybe give some tips on how to approach these types of questions better, I would really appreciate it.

 

[cossine]

Hey, I've been stuck on this question for a while so was hoping if someone could help me with parts c and d. (parts a and b given for context)

a) obtain cos6θ as a polynomial in cosθ.
b) Hence show that x=cos ((2k+1) π)/12
c)deduce that cos π/12 * cos5π/12=1/4
d) deduce that cos^2 5π/12 +cos^2 π/12=1

I also would like some advice on how to tackle these types of problems better, right now my approach is to just haphazardly use sum and product of roots to try to get a relationship, but If someone could maybe give some tips on how to approach these types of questions better, I would really appreciate it.

I think I have seen similar questions in Terry Lee. Can you show your attempt for part c/d even if you have failed to do. I assume x=cos ((2k+1) π)/12 is the solution to polynomial equation

 

[ProGT408]

I think I have seen similar questions in Terry Lee. Can you show your attempt for part c/d even if you have failed to do. I assume x=cos ((2k+1) π)/12 is the solution to polynomial equation

I was trying to do this question too, all I was able to do for c was find the sums of couples of roots, (ab + cd + ae) and make them equal -3/2, then one of those terms is -cos π /12 * cos 5π/12 and using my calculator to compute the other terms i got that value as 1/4 but i feel there should be a more proper way since that is tedious and a bit sus

 

[ProGT408]

@Luukas.2 do you have any insights?

 

[liamkk112]

Hey, I've been stuck on this question for a while so was hoping if someone could help me with parts c and d. (parts a and b given for context)

a) obtain cos6θ as a polynomial in cosθ.
b) Hence show that x=cos ((2k+1) π)/12 where k=0,1,2,3,4,5 are the roots of the equation 32x^6-48x^4+18x^2-1=0
c)deduce that cos π/12 * cos5π/12=1/4
d) deduce that cos^2 5π/12 +cos^2 π/12=1

I also would like some advice on how to tackle these types of problems better, right now my approach is to just haphazardly use sum and product of roots to try to get a relationship, but If someone could maybe give some tips on how to approach these types of questions better, I would really appreciate it.

product of roots: cospi/12 cospi/4 cos5pi/12 cos7pi/12 cos3pi/4 cos11pi/12 = -1/32

then cospi/4 = 1/sqrt2, cos3pi/4 = -1/sqrt2:

so -0.5 cospi/12 cos5pi/12 cos7pi/12 cos11pi/12 = -1/32

-> cospi/12 cos5pi/12 cos7pi/12 cos11pi/12 = 1/16

from here, we notice that cos7pi/12 = cos(pi-5pi/12) = -cos5pi/12
similarly cos11pi/12 = -cospi/12
hence we have that (cos5pi/12 cospi/12)^2 = 1/16
then cos5pi/12 cospi/12 = 1/4 as required, positive sqrt only as cos5pi/12 >0, cospi/12 >0 due to quadrant of angle

 

[liamkk112]

product of roots: cospi/12 cospi/4 cos5pi/12 cos7pi/2 cos3pi/4 cos11pi/12 = -1/32

then cospi/4 = 1/sqrt2, cos3pi/4 = -1/sqrt2:

so -0.5 cospi/12 cos5pi/12 cos7pi/12 cos11pi/12 = -1/32

-> cospi/12 cos5pi/12 cos7pi/12 cos11pi/12 = 1/16

from here, we notice that cos7pi/12 = cos(pi-5pi/12) = -cos5pi/12
similarly cos11pi/12 = -cospi/12
hence we have that (cos5pi/12 cospi/12)^2 = 1/16
then cos5pi/12 cospi/12 = 1/4 as required, positive sqrt only as cos5pi/12 >0, cospi/12 >0 due to quadrant of angle

that's c. for d, you should be able to use product of roots two at a time or maybe five at a time, and use similar arguments as above, though as the polynomial is degree six there are a lot of terms lol. there might be a faster way to do this but thats all i can think of

 

[liamkk112]

Hey, I've been stuck on this question for a while so was hoping if someone could help me with parts c and d. (parts a and b given for context)

a) obtain cos6θ as a polynomial in cosθ.
b) Hence show that x=cos ((2k+1) π)/12 where k=0,1,2,3,4,5 are the roots of the equation 32x^6-48x^4+18x^2-1=0
c)deduce that cos π/12 * cos5π/12=1/4
d) deduce that cos^2 5π/12 +cos^2 π/12=1

I also would like some advice on how to tackle these types of problems better, right now my approach is to just haphazardly use sum and product of roots to try to get a relationship, but If someone could maybe give some tips on how to approach these types of questions better, I would really appreciate it.

in terms of how to approach this question, you can guess pretty well about which form of vieta's formulas (the sum of roots thingies) would apply. eg for c - there is one term, that suggests product of roots. all these questions pretty much use the same tricks that cos(pi-x) = -cosx or something of that form, and de moivres' theorem for obtaining cos or sin in terms of powers.

 

[liamkk112]

that's c. for d, you should be able to use product of roots two at a time or maybe five at a time, and use similar arguments as above, though as the polynomial is degree six there are a lot of terms lol. there might be a faster way to do this but thats all i can think of

thought of a better way: cos(x-pi/2) = sinx
now we have that LHS = cos^2(5pi/12) + cos^2(pi/12) = cos^2(11pi/12 - pi/2) +cos^2(pi/12)
= sin^2(11pi/12) + cos^2(pi/12)

now sin 11pi/12 = sinpi/12:
hence LHS = sin^2(pi/12) + cos^2(pi/12) = 1 = RHS bc sin^2x + cos^2x = 1
much easier than root stuff

 

[Intilegience]

thought of a better way: cos(x-pi/2) = sinx
now we have that LHS = cos^2(5pi/12) + cos^2(pi/12) = cos^2(11pi/12 - pi/2) +cos^2(pi/12)
= sin^2(11pi/12) + cos^2(pi/12)

now sin 11pi/12 = sinpi/12:
hence LHS = sin^2(pi/12) + cos^2(pi/12) = 1 = RHS bc sin^2x + cos^2x = 1
much easier than root stuff

thanks for that, really appreciate it.

 

[tywebb]

You can do c too like how liamkk112 used complementary angles in d.



For a and b you know it is the Tchebyshev polynomial of the first kind defined by



with (whose roots are the Tchebyshev nodes of the first kind)

and can much more quickly be found using the hypergeometric

So with n=6 in wolframalpha you can put

Hypergeometric2F1(-6,6,1/2,(1-x)/2)

and the answer comes out straight away like this:


or you can use the formula



with n=6 so



which is not as fast as the hypergeometric method but still much faster than using de Moivre's theorem.

 

[Luukas.2]

@Luukas.2 do you have any insights?

Noting that the polynomial is even, the roots being

​

are more conveniently taken as

​

and hence are

​

From this, it becomes much clearer which formulae for roots to use. So, my first piece of advice with this type of question is to simplify the roots before seeking relationships between them.

Recognising that the root is

​

is not difficult, but needing to carry out such conversions can be avoided by selecting the roots efficiently in the first place.

From these, the result in (c) can be deduced from the product of the roots:

​

Some other approaches available here include:

• the roots at arise from being a factor of the degree 6 polynomial, and so a degree 4 polynomial can be formed with only the other four roots.
• this polynomial could be transformed into a quadratic using as a substitution, which would have only two roots, , allowing the desired results to be found easily.
• Another approach would be to recognise that (c) allows a sums-to-products approach for part (d), available if the question in (d) said "hence, deduce that ...":

​