polynomial question (1 Viewer)

[Jaydels]

Blah! This isnt an EXT2 question!

it was in my EXT2 text book

 

[Trev]

Another Poly. Q.

heres a q i dont kno what to do with.....
(exercise 4.4 q. 9) in arnold arnold)
show that cos4x = 8cos^4(x) - 8cos^2(x) + 1. (i can do this part)
hence..
a) solve the equation 8x^4 - 8x^2 + 1 = 0 and deduce the exact values of cos(pi/8) and cos(5.pi/8)
any1 bothered to help me?

 

[Templar]

a) solve the equation 8x^4 - 8x^2 + 1 = 0 and deduce the exact values of cos(pi/8) and cos(5.pi/8)

Substitute cos x for x on the LHS. Then you know cos 4x = 0. The two expressions are the solutions to the equation, so go from there.

 

[Trev]

i did that... but i dont kno how to get the exact values...help again plz? :|

 

[TimeAndTide]

it was in my EXT2 text book

Craphouse...should've done ext.2

 

[Templar]

You know the solutions are cos pi/8, cos 3pi/8, cos 5pi/8 and cos 7pi/8, and that cos pi/8 = -cos 7pi/8 and cos 3pi/8 = -cos 5pi/8. Use product of roots etc to work it out.

Otherwise just solve the quantic.

 

[Trev]

lol omg im so stoopd i forgot u could workout wat that equals lol *runs and hides*
thanx!

 

[Captain pi]

heres a q i dont kno what to do with.....
(exercise 4.4 q. 9) in arnold arnold)
show that cos4x = 8cos^4(x) - 8cos^2(x) + 1. [1] (i can do this part)
hence..
a) solve the equation 8y^4 - 8y^2 + 1 = 0 [2] and deduce the exact values of cos(pi/8) and cos(5.pi/8)
any1 bothered to help me?

Note the numbering of the equations and the change in variable.

As Templar said, substituing cos(x) for y in [2] will give the RHS of [1]. Therefore, the sol'ns of cos(4x) = 0 will be the same as [2] (but for y).

Solving cos(4x) = 0 gives x = pi/8 and 5pi/8 (inter alia). Now, since y is equivalent to cos(x), cos(pi/8) and cos(5pi/8) are sol'n of y in [2].

Use the quadratic formula twice to give four solutions to the equation in [2]. Test each one with the calculator values of cos(pi/8) and cos(5pi/8). I think this is a valid method because it said "deduce" and not "derive" or "prove" or any other stronger verb.

I believe that solves your question.

 

[Trev]

ahh thanx, ps. i should hav posted the answers too (being: x= (+/-) root(2 +/- root{2})/2 if that makes any sense how i typed it?).
'Captain pi' - don't ruin maths by referring to the stoopd verbs we need to kno heh

 

[Trev]

mm another q. ppl (yes - i am stoopd)
show that cos5x = 16cos^5x - 20cos^3x + 5cosx. hence
solve the equation 16x^5 - 20x^3 + 5x - 1 = 0 and deduce the exact values of cos(2pi/5) and cos(4pi/5)
i did the first part and found the roots r cosx = 1, +/- 2pi/5, +/- 4pi/5.
but... i cant solve the equation, found since (x-1) is a factor, by division of the poly. it come out as; 16x^4 + 16x^3 - 4x^2 - 4x + 1.
but i can't solve that.. . any help plz ppl? [answers for x = 1, (+/-)1/4[+/-1 + root5]

 

[mojako]

16x^4 + 16x^3 - 4x^2 - 4x + 1
has roots a, b, c, d
a = cos 2pi/5
b = cos 4pi/5 = 2a^2 - 1 (double angle identity)
c = cos -2pi/5 = a
d = cos -4pi/5 = b

sum of roots:
a + b + c + d = -16/16 = -1
i.e. 2a + 4a^2 - 2 = -1
then you can find a, and b.

 

[Trev]

told u i was stoopd, i didn't even think of using sum of roots method

 

[Captain pi]

told u i was stoopd, i didn't even think of using sum of roots method

Don't be to hard on yourself, big Trev. You'll acquire these skills the more questions you do.

PS: When you (and others) are writing maths question or answers, can you write in full English? When I see things like "r" and "u" implicated in a maths question, I think that they are variables, not pronouns and substantive verbs. Thank you.

 

[Trev]

Don't be to hard on yourself, big Trev. You'll acquire these skills the more questions you do.

PS: When you (and others) are writing maths question or answers, can you write in full English? When I see things like "r" and "u" implicated in a maths question, I think that they are variables, not pronouns and substantive verbs. Thank you.

haha sure, it'll take more effort but i'm sure i can do it! have you read the dictionary, or parts of? i don't know what substantive means.... *whoosh*