Comments on: Tan = sin/cos http://trevorpythag.co.uk/2008/mathematics/trigonometry/tan-sincos/ Maths help and revision for GCSE, A/AS Level and Further Maths Thu, 17 May 2012 08:44:32 +0000 hourly 1 http://wordpress.org/?v=3.3.2 By: Brandon http://trevorpythag.co.uk/2008/mathematics/trigonometry/tan-sincos/comment-page-1/#comment-98 Brandon Sat, 07 Mar 2009 21:32:42 +0000 http://trevorpythag.wordpress.com/?p=46#comment-98 I am fully agree with you antoni I am fully agree with you antoni

]]> By: antoni http://trevorpythag.co.uk/2008/mathematics/trigonometry/tan-sincos/comment-page-1/#comment-97 antoni Mon, 09 Feb 2009 01:30:18 +0000 http://trevorpythag.wordpress.com/?p=46#comment-97 I've just known about tan(t)=sin(2t)/[1+cos(2t)], and of course mathematician like masteranza can prove easily using trig relation. But, how to create it using SMT. Sorry until now, I can't not do it altough I've read the SMT's explanation from this address : http://eqworld.ipmnet.ru/forum/viewtopic.php?f=2&t=34.Thx. I’ve just known about tan(t)=sin(2t)/[1+cos(2t)], and of course mathematician like masteranza can prove easily using trig relation. But, how to create it using SMT. Sorry until now, I can’t not do it altough I’ve read the SMT’s explanation from this address : http://eqworld.ipmnet.ru/forum/viewtopic.php?f=2&t=34.Thx.

]]> By: Robert http://trevorpythag.co.uk/2008/mathematics/trigonometry/tan-sincos/comment-page-1/#comment-96 Robert Sat, 20 Dec 2008 22:22:23 +0000 http://trevorpythag.wordpress.com/?p=46#comment-96 Thx for an excelent discussion Thx for an excelent discussion

]]> By: Juliani http://trevorpythag.co.uk/2008/mathematics/trigonometry/tan-sincos/comment-page-1/#comment-95 Juliani Fri, 19 Dec 2008 19:26:30 +0000 http://trevorpythag.wordpress.com/?p=46#comment-95 Nadya, Denaya, and Rohedi. I believe the proof of tan=sin/cos and creating another expression of the tangent function tan(t)=sin(2t)/[1+cos(2t)] only can be performed if the arctangent differential equation is solved by a new technique. There is a new method so-called SMT (shortened from stable modulation technique). I know the smart method of solving first order ODE after visitng to this address : http://eqworld.ipmnet.ru/forum/viewtopic.php?f=2&t=34. I also saw at this address : http://masteranza.wordpress.com/2007/12/30/useful-tangent-triangle-identities/#comment-265 about the equality of 0/0=1/0 as the proof that mathematics ever inexactly but it be saved by L'Hospital theory. Nadya, Denaya, and Rohedi.
I believe the proof of tan=sin/cos and creating another expression of the tangent function tan(t)=sin(2t)/[1+cos(2t)] only can be performed if the arctangent differential equation is solved by a new technique. There is a new method so-called SMT (shortened from stable modulation technique). I know the smart method of solving first order ODE after visitng to this address : http://eqworld.ipmnet.ru/forum/viewtopic.php?f=2&t=34.
I also saw at this address :
http://masteranza.wordpress.com/2007/12/30/useful-tangent-triangle-identities/#comment-265
about the equality of 0/0=1/0 as the proof that mathematics ever inexactly but it be saved by L’Hospital theory.

]]> By: Denaya Lesa http://trevorpythag.co.uk/2008/mathematics/trigonometry/tan-sincos/comment-page-1/#comment-94 Denaya Lesa Fri, 19 Dec 2008 12:25:14 +0000 http://trevorpythag.wordpress.com/?p=46#comment-94 Apologize, I also met a similar question on this address from miss Nadya Fermega http://blueollie.wordpress.com/2008/12/13/how-to-succeed-in-calculus-class-and-why-it-is-worth-it/#comment-30790 The question is how to prove that tan(x)=sin(x)/cos(x) from solution of dy/dx=1+y^2, with x(0)=0, and y(0)=0, but they did not give a respon. I am so very interested to this topic, but like miss Nadya I have also no idea to prove it. Thx for your help. Apologize, I also met a similar question on this address from miss Nadya Fermega

http://blueollie.wordpress.com/2008/12/13/how-to-succeed-in-calculus-class-and-why-it-is-worth-it/#comment-30790

The question is how to prove that tan(x)=sin(x)/cos(x) from solution of dy/dx=1+y^2, with x(0)=0, and y(0)=0, but they did not give a respon. I am so very interested to this topic, but like miss Nadya I have also no idea to prove it. Thx for your help.

]]> By: rohedi http://trevorpythag.co.uk/2008/mathematics/trigonometry/tan-sincos/comment-page-1/#comment-93 rohedi Fri, 19 Dec 2008 07:38:27 +0000 http://trevorpythag.wordpress.com/?p=46#comment-93 Yes, I agree with your explanation about the usual definition of tangent function as tan(t) =sin(t)/cos(t) that obtained by using opp and adj of a triangle, and we find tan(pi/2)=1/0. But, another expression of the tangent function tan(t)=sin(2t)/[1+cos(2t)] can be created from arctangent differential equation dy/dt=1+y^2, with t(0)=0, and y(0)=0. Let substitute t=pi/2, we will find that tan(pi/2)=0/0. So, without L'Hospital I believe mathematics ever inexactly. How about you to the last statement. Yes, I agree with your explanation about the usual definition of tangent function as tan(t) =sin(t)/cos(t) that obtained by using opp and adj of a triangle, and we find tan(pi/2)=1/0. But, another expression of the tangent function tan(t)=sin(2t)/[1+cos(2t)] can be created from arctangent differential equation dy/dt=1+y^2, with t(0)=0, and y(0)=0. Let substitute t=pi/2, we will find that tan(pi/2)=0/0. So, without L’Hospital I believe mathematics ever inexactly. How about you to the last statement.

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