S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Tue, 22 Jul 2014 22:23:49 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 6 posts ] 
Author Message
 Post subject: trig product
PostPosted: Thu, 17 May 2012 04:34:29 UTC 
Offline
Member

Joined: Sun, 8 Apr 2012 09:36:59 UTC
Posts: 29
How can I calculate product of

\cos(1^0).\cos(3^0).\cos(5^0)..........\cos(89^0) =

where all angle are in Degree


Top
 Profile  
 
 Post subject: Re: trig product
PostPosted: Thu, 17 May 2012 07:53:46 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6775
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
jacks wrote:
How can I calculate product of

\cos(1^0).\cos(3^0).\cos(5^0)..........\cos(89^0) =

where all angle are in Degree


What have you tried? Where is your work?

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
 Post subject: Re: trig product
PostPosted: Fri, 18 May 2012 16:07:18 UTC 
Offline
Member

Joined: Sun, 8 Apr 2012 09:36:59 UTC
Posts: 29
Thanks outermeasure got it

Let A=\cos(1^0).\cos(2^0).\cos(3^0).........\cos(89^0)

Now using \cos(90-\theta^0)=\sin(\theta^0)

So A=\sin(1^0).\sin(2^0).\sin(3^0).........\sin(89^0)

and Let B=\sin(2^0).\sin(4^0).\sin(6^0).........\sin(88^0)

Now AB=(\sin(1^0).\sin(89^0)).(\sin(2^0).\sin(88^0)).........(\sin(44^0).\sin(46^0)).\sin(45^0)

So AB=(\sin(1^0).\cos(1^0)).(\sin(2^0).\cos(2^0)).........(\sin(44^0).\cos(44^0)).\sin(45^0)

So \displaystyle 2^{44}AB=\sin (2^0).\sin(4^0)........\sin(88^0).\frac{1}{\sqrt{2}}

So \displaystyle 2^{44}AB = \frac{1}{\sqrt{2}}B

So \displaystyle A=\frac{1}{\sqrt{2}.2^{44}}=\frac{\sqrt{2}}{2^{45}}

bcz B\neq 0


Top
 Profile  
 
 Post subject: Re: trig product
PostPosted: Fri, 18 May 2012 16:27:06 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6775
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
jacks wrote:
Thanks outermeasure got it

Let A=\cos(1^0).\cos(2^0).\cos(3^0).........\cos(89^0)

Now using \cos(90-\theta^0)=\sin(\theta^0)

So A=\sin(1^0).\sin(2^0).\sin(3^0).........\sin(89^0)

and Let B=\sin(2^0).\sin(4^0).\sin(6^0).........\sin(88^0)

Now AB=(\sin(1^0).\sin(89^0)).(\sin(2^0).\sin(88^0)).........(\sin(44^0).\sin(46^0)).\sin(45^0)

So AB=(\sin(1^0).\cos(1^0)).(\sin(2^0).\cos(2^0)).........(\sin(44^0).\cos(44^0)).\sin(45^0)

So \displaystyle 2^{44}AB=\sin (2^0).\sin(4^0)........\sin(88^0).\frac{1}{\sqrt{2}}

So \displaystyle 2^{44}AB = \frac{1}{\sqrt{2}}B

So \displaystyle A=\frac{1}{\sqrt{2}.2^{44}}=\frac{\sqrt{2}}{2^{45}}

bcz B\neq 0


I think you mean A:=\cos(1^\circ)\cos(3^\circ)\cdots\cos(89^\circ) instead.

Another way is to use symmetry of roots on a Chebyshev polynomial.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
 Post subject: Re: trig product
PostPosted: Fri, 18 May 2012 16:34:22 UTC 
Offline
Member

Joined: Sun, 8 Apr 2012 09:36:59 UTC
Posts: 29
Yes outermeasure

could you like to explain it to me using the Method of symmetry of roots on a Chebyshev polynomial.

Thanks


Top
 Profile  
 
 Post subject: Re: trig product
PostPosted: Fri, 18 May 2012 17:10:25 UTC 
Offline
Moderator
User avatar

Joined: Mon, 29 Dec 2008 17:49:32 UTC
Posts: 6775
Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
jacks wrote:
Yes outermeasure

could you like to explain it to me using the Method of symmetry of roots on a Chebyshev polynomial.

Thanks


Note that \cos(1^\circ),\cos(3^\circ),\dots,\cos(89^\circ),\cos(91^\circ)=-\cos(89^\circ),\dots,\cos(179^\circ)=-\cos(1^\circ) are the roots of the Chebyshev polynomial T_{90}(x)=\cos(90\arccos(x)). But \displaystyle T_{90}(x)=\frac{90}{2}\sum_{k=0}^{45}(-1)^k\dfrac{(89-k)!}{k!(90-2k)!}(2x)^{90-2k}=2^{89}x^{90}+\cdots+(-1), so the product of all roots is -[\cos(1^\circ)\cos(3^\circ)\cdots\cos(89^\circ)]^2=\dfrac{-1}{2^{89}}. Since \cos(1^\circ),\dots,\cos(89^\circ)>0, we have \cos(1^\circ)\cos(3^\circ)\cdots\cos(89^\circ)=\dfrac{1}{2^{89/2}}.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 6 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA