SOLUTION: Use this identify to solve the question: tan4Q = {{{(4tanQ-4(tanQ)^3)/(1-6(tanQ)^2+(tanQ)^4)}}} Find the polynomial of least degree that has zeroes {{{(cot(pi/24))^2}}}, {{{(

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: Use this identify to solve the question: tan4Q = {{{(4tanQ-4(tanQ)^3)/(1-6(tanQ)^2+(tanQ)^4)}}} Find the polynomial of least degree that has zeroes {{{(cot(pi/24))^2}}}, {{{(      Log On


   

Question 1165987: Use this identify to solve the question:
tan4Q = %284tanQ-4%28tanQ%29%5E3%29%2F%281-6%28tanQ%29%5E2%2B%28tanQ%29%5E4%29
Find the polynomial of least degree that has zeroes %28cot%28pi%2F24%29%29%5E2, %28cot%287pi%2F24%29%29%5E2, %28cot%2813pi%2F24%29%29%5E2, %28cot%2819pi%2F24%29%29%5E2%29
Thank you in advance!
Answer by ikleyn(52905) About Me  (Show Source):
You can put this solution on YOUR website!
.
Use this identify to solve the question:

tan4Q = %284tanQ-4%28tanQ%29%5E3%29%2F%281-6%28tanQ%29%5E2%2B%28tanQ%29%5E4%29

Find the polynomial of least degree that has zeroes %28cot%28pi%2F24%29%29%5E2, %28cot%287pi%2F24%29%29%5E2, %28cot%2813pi%2F24%29%29%5E2, %28cot%2819pi%2F24%29%29%5E2%29
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So,  according to the problem formulation,  I should accept the given identity  "as is",  and based on it,
to construct the polynomial.

Since this problem is designed/intendent for advanced students,  I will only show the major idea and the major steps
without going in details.


Let me start noticing that of four numbers  %28cot%28pi%2F24%29%29%5E2,  %28cot%287pi%2F24%29%29%5E2,  %28cot%2813pi%2F24%29%29%5E2,  %28cot%2819pi%2F24%29%29%5E2%29,
the first is equal to the third,  while the second is equal to the fourth.

So,  I will construct the polynomial which has two zeroes as the first and the second numbers of the four numbers listed:
then the third and the fourth numbers will be the roots of this polynomial automatically . . . 


Take Q = pi%2F24.  Then  4Q = pi%2F6  and  tan(4Q) = tan%28pi%2F6%29 = sqrt%283%29%2F3.


According to the given identity, I have then


    sqrt%283%29%2F3 =   or

    sqrt%283%29%2A%281-6%28tan%28Q%29%29%5E2%2B%28tan%28Q%29%29%5E4%29 = 3%2A%284tan%28Q%29-4%28tan%28Q%29%29%5E3%29.    (2)


Next, in (2), I will replace (tan(Q))^2  by x everywhere (at each appearance).

I will get then


    sqrt%283%29%2A%281+-+6x+%2B+x%5E2%29 = 12%2Atan%28Q%29%2A%281-x%29.    (3)


Now, for tan%28pi%2F24%29  there is the formula (the known expression)  

         tan%28pi%2F24%29 = -2+-+sqrt%283%29+%2B+2%2Asqrt%282+%2B+sqrt%283%29%29  (see the link  https://brainly.in/question/9263846 )

and/or   

         tan%28pi%2F24%29 = -2+%2B+sqrt%282%29+-+sqrt%283%29+%2B+sqrt%286%29  (see the link  https://mathworld.wolfram.com/TrigonometryAnglesPi24.html)


Replacing  tan%28pi%2F24%29  in  formula  (3)  by any of these constant expressions (they are equal (!)),

I get the polynomial of the degree 2


    sqrt%283%29%2A%281+-+6x+%2B+x%5E2%29 = 12c%281-x%29,           (4)


whose two roots are  %28tan%28pi%2F24%29%29%5E2  and  %28tan%287pi%2F24%29%29%5E2.


If you want to have the polynomial with the roots  %28cot%28pi%2F24%29%29%5E2  and  %28cot%287pi%2F24%29%29%5E2,

simply replace x in the polynomial (4) by 1%2Fx  and transform the obtained rational function to the form

of the ratio  P%28x%29%2FQ%28x%29  with polynomials  P(x) and Q(x).


Then the polynomial  P(x)  will be the answer to the problem's question.


Probably, it is more correctly in English to say "a polynomial P(x)", because every product of such a polynomial by a constant
satisfies the problem's conditions, too.

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Solved and explained.

In such problems,  knowing the way is more important than gettting the exact answer.