Let $a$ be a rational number with $0\leq a\leq 1$. Suppose that \[cos 3\pi a+2 cos 2\pi a=0\].
(angles are in radians)Determine,with proof,the value of $a$.
Re: ISL 91
Posted: Sat Jun 02, 2012 1:48 am
by SANZEED
My steps:
Derive an equation. consider two cases. One will satisfy the claim.Prove the other to be contradictory.
Re: ISL 91
Posted: Sat Jun 02, 2012 2:30 pm
by sakibtanvir
Converting radian into degree,we get,\[cos(540a)+2cos(360a)=0\]
\[Or,cos(180a)+2cos0=0\]
\[Or,cos(180a)=-2\]
Contradiction.....
Re: ISL 91
Posted: Sat Jun 02, 2012 5:07 pm
by Phlembac Adib Hasan
sakibtanvir wrote:Converting radian into degree,we get,\[cos(540a)+2cos(360a)=0\]
\[Or,cos(180a)+2cos0=0\]
\[Or,cos(180a)=-2\]
Contradiction.....
Who said $a$ is an integer?It's rational only.
Re: ISL 91
Posted: Sat Jun 02, 2012 5:10 pm
by Phlembac Adib Hasan
SANZEED wrote:Let $a$ be a rational number with $0\leq a\geq 1$. Suppose that \[cos 3\pi a+2 cos 2\pi a=0\].
(angles are in radians)Determine,with proof,the value of $a$.
Edit this.I'm surprised that this problem was posted a long time ago.Why nobody noticed this?(I've visited this topic just now.)
Re: ISL 91
Posted: Sat Jun 02, 2012 7:26 pm
by *Mahi*
Phlembac Adib Hasan wrote:
SANZEED wrote:Let $a$ be a rational number with $0\leq a\geq 1$. Suppose that \[cos 3\pi a+2 cos 2\pi a=0\].
(angles are in radians)Determine,with proof,the value of $a$.
Edit this.I'm surprised that this problem was posted a long time ago.Why nobody noticed this?(I've visited this topic just now.)
Edited.
Re: ISL 91
Posted: Sat Jun 02, 2012 10:52 pm
by SANZEED
My solution(in brief):
Set $x=cos \pi a$. Then the equation is equivalent to $4x^{3}+4x^{2}-3x-2=0$. This turns to $(2x+1)(2x^{2}+x-2)=0$. For the case $2x+1=0\Rightarrow cos \pi a=-\frac{1}{2}\Rightarrow a=\frac{2}{3}$. Now it remains,from further calculation, to prove that for every integer $n\geq 0,cos (2^{n}\pi a)=\frac{a_{n}+b_{n}\sqrt 17}{4}$. This is easily proved by induction.
Re: ISL 91
Posted: Sun Jun 03, 2012 7:02 am
by Phlembac Adib Hasan
SANZEED wrote:My solution(in brief):
Set $x=cos \pi a$. Then the equation is equivalent to $4x^{3}+4x^{2}-3x-2=0$. This turns to $(2x+1)(2x^{2}+x-2)=0$. For the case $2x+1=0\Rightarrow cos \pi a=-\frac{1}{2}\Rightarrow a=\frac{2}{3}$. Now it remains,from further calculation, to prove that for every integer $n\geq 0,cos (2^{n}\pi a)=$$\frac{a_{n}+b_{n}\sqrt 17}{4}$. This is easily proved by induction.
What's $a_n$ and $b_n$?
Re: ISL 91
Posted: Sun Jun 03, 2012 10:53 am
by SANZEED
They are odd integers. Sorry for eliminating.
Re: ISL 91
Posted: Sun Jun 03, 2012 5:25 pm
by sakibtanvir
Phlembac Adib Hasan wrote:
sakibtanvir wrote:Converting radian into degree,we get,\[cos(540a)+2cos(360a)=0\]
\[Or,cos(180a)+2cos0=0\]
\[Or,cos(180a)=-2\]
Contradiction.....
Who said $a$ is an integer?It's rational only.
\[-1<cos\angle x<1\] for any value of $x$ .
So there is no real value of $a$.
Derive an equation.