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 January 28th, 2014, 12:04 AM #1 Senior Member   Joined: Apr 2012 Posts: 799 Thanks: 1 irrational numbers prove: $sin 10^o \,\, and \,\, cos 10^o$ are all irrational numbers
January 28th, 2014, 03:07 AM   #2
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Joined: May 2013
From: EspaÃ±a

Posts: 151
Thanks: 4

Re: irrational numbers

Quote:
 Originally Posted by Albert.Teng prove: $sin 10^o \,\, and \,\, cos 10^o$ are all irrational numbers
Hello.

I will realize the cosine:

$\cos (30)=\dfrac{1}{2}$

$\dfrac{1}{2}=\cos (20) \cos (10)-\sin (20) \sin (10)$

$\dfrac{1}{2}=[\cos^2 (10)-\sin^2 (10)] \cos (10)-[2 \sin (10) \cos (10)] \sin (10)$

$\dfrac{1}{2}=\cos^3 (10)-\sin^2 (10) \cos (10)-2 \sin^2 (10) \cos (10)$

$\dfrac{1}{2}=\cos^3 (10)-3 \sin^2 (10) \cos (10)=\cos^3 (10)-3 [1-\cos^2 (10)] \cos (10)$

$\dfrac{1}{2}=\cos^3 (10)-3 \cos (10)+3 \cos^3 (10)$

$\dfrac{1}{2}=4 \cos^3 (10)-3 \cos (10)$

$8 \cos^3 (10)-6 \cos (10)-1=0$

$Let \ p, \ q \in{N} \ / \ p \ and \ q \ coprime \ / \cos(10)=\dfrac{p}{q} :$(*)

$8 \dfrac{p^3}{q^3}-6 \dfrac{p}{q}-1=0$

$8p^3-6pq^2-q^3=0 \rightarrow{}q=even \rightarrow{}p=odd$

1º) If "q" It is divisible only once by "2", then:

$8|8p^3 \ result \ odd$

$8|6pq^2 \ result \ odd$

$8|q^3 \ result \ odd$

Absurdity:

$odd \ - \ odd - \ odd \cancel{=}0$

2º)

$If \ 2^k|q \ / \ k>1$

$8|8p^3 \ result \ odd$

$8|9pq^2 \ result \ even$

$8|q^3 \ result \ even$

$odd \ - \ even - \ even \cancel{=}0$

The rest of the show, leave it to someone else.

Regards.

 January 28th, 2014, 05:50 AM #3 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: irrational numbers In general, it is provably true that $^{\sin(k\pi)}$ and $^k$ are both rational only for finitely many values of $^{\sin(k\pi)}$, i.e, one of 0, 1/2, or 1 (+ or -).
February 12th, 2014, 04:26 PM   #4
Senior Member

Joined: May 2013
From: EspaÃ±a

Posts: 151
Thanks: 4

Re: irrational numbers

[quote=mente oscura]
Quote:
 Originally Posted by "Albert.Teng":337yooi9 prove: $sin 10^o \,\, and \,\, cos 10^o$ are all irrational numbers
Hello.

I will realize the cosine:

$\cos (30)=\dfrac{1}{2}$

[/quote:337yooi9]

Hello.

I'm sorry. I had a mistake at the beginning.

$\cos (30)=\dfrac{\sqrt{3}}{2}$

$\dfrac{\sqrt{3}}{2}=\cos (20) \cos (10)-\sin (20) \sin (10)$

$\dfrac{\sqrt{3}}{2}=[\cos^2 (10)-\sin^2 (10)] \cos (10)-[2 \sin (10) \cos (10)] \sin (10)$

$\dfrac{\sqrt{3}}{2}=\cos^3 (10)-\sin^2 (10) \cos (10)-2 \sin^2 (10) \cos (10)$

$\dfrac{\sqrt{3}}{2}=\cos^3 (10)-3 \sin^2 (10) \cos (10)=\cos^3 (10)-3 [1-\cos^2 (10)] \cos (10)$

$\dfrac{\sqrt{3}}{2}=\cos^3 (10)-3 \cos (10)+3 \cos^3 (10)$

$\dfrac{\sqrt{3}}{2}=4 \cos^3 (10)-3 \cos (10)$

$8 \cos^3 (10)-6 \cos (10)=\sqrt{3}$

Suppose, $cos(10) \in{\mathbb{Q}}$:

$2 \cos(10) [4 \cos^2 (10)-3]=\sqrt{3}$

$[4 \cos^2 (10)-3]=k \ \cancel{\in{\mathbb{Q}}$

$\cos(10)=\sqrt{\dfrac{k+3}{4}}=\dfrac{\sqrt{k+3}}{ 2} \rightarrow{}$

$\rightarrow{} \sqrt{k+3} \in{\mathbb{Q}$,

Absurdity:

$\mathbb{Q}+\cancel{\mathbb{Q}} \ \cancel{=} \ \mathbb{Q}$

If we had started by: $\sin (10)$

$\sin(30)=\dfrac{1}{2}$

$\dfrac{1}{2}=\sin (20) \cos (10)+\cos (20) \sin (10)$

$\dfrac{1}{2}=[2 \sin (10) \cos (10)] \ \cos (10)+[\cos^2 (10)- \sin^2 (10)] \sin (10)$

$\dfrac{1}{2}=2 \sin (10) \cos^2 (10)+[1- \sin^2 (10)- \sin^2 (10)] \sin (10)$

$\dfrac{1}{2}=2 \sin (10) [1-\sin^2 (10)]+\sin(10)- 2 \ \sin^3 (10)$

$\dfrac{1}{2}=2 \sin (10)-2 \sin^3 (10)+\sin(10)- 2 \ \sin^3 (10)=3 \sin(10)-4 \sin^3(10)$

$4 \sin^3 (10)-3 \sin (10)=-\dfrac{1}{2}$

$8 \sin^3 (10)-6 \sin (10) + 1= 0$

$Let \ p, \ q \in{N} \ / \ p \ and \ q \ coprime \ / \sin(10)=\dfrac{p}{q} :$(*)

$8 \dfrac{p^3}{q^3}-6 \dfrac{p}{q}+1=0$

$8p^3-6pq^2+q^3=0 \rightarrow{}q=even \rightarrow{}p=odd$

1º) If "q" It is divisible only once by "2", then:

$8|8p^3 \ result \ odd$

$8|6pq^2 \ result \ odd$

$8|q^3 \ result \ odd$

Absurdity:

$odd \ - \ odd + \ odd \cancel{=}0$

2º)

$If \ 2^k|q \ / \ k>1$

$8|8p^3 \ result \ odd$

$8|6pq^2 \ result \ even$

$8|q^3 \ result \ even$

$odd \ - \ even + \ even \cancel{=}0$

Regards.

 February 12th, 2014, 05:55 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,659 Thanks: 964 Math Focus: Elementary mathematics and beyond Re: irrational numbers \begin{align*}\frac12\,=\,\sin(20\,+\,10)\,&=\,sin 20\cos10\,+\,sin10\cos20 \\ &=\,2\cos^210\sin10\,+\,\sin10\,-\,2\sin^310 \\ &=\,2\sin10\,-\,2\sin^310\,+\,\sin10\,-\,2\sin^310 \\ &=\,-4\sin^310\,+\,3\sin10\end{align*} $\Rightarrow\,8u^3\,-\,3u\,+\,1\,=\,0\text{ (with }u\text{ = \sin10)}$ The equation immediately above, by the rational root theorem, has no rational roots, hence sin(10) is irrational. A similar proof can be constructed for cos(10).

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