Tan = sin/cos
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This is often useful when solving trig equations so i thought i’d include it
basically:
sin = opp/hyp
and
cos=adj/hyp
so
sin/cos = (opp/hyp)/(adj/hyp)
so if we cancel the hyp’s we get
sin/cos = opp/adj
and since tan = opp/adj
tan = sin/cos
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.
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.
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.
Thx for an excelent discussion
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 am fully agree with you antoni