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 2017-11-09, 18:43 #1 bhelmes     Mar 2016 41910 Posts angle bisection A peaceful day for all, a, b, c element N are known, where a, b, c build a pyth. trippel with a^2+b^2=c^2 tan (alpha)=a/b is there a way to calculate tan (alpha/2) without using the angle ? if not, how accurate i have to choose alpha to get a good result / the rational number of tan (alpha/2) ? Thanks in advance, if you give me a hint. Greetings from the primes Bernhard
 2017-11-09, 18:53 #2 R. Gerbicz     "Robert Gerbicz" Oct 2005 Hungary 32·179 Posts
 2017-11-09, 20:05 #3 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 91F16 Posts To state the obvious the half tenant of alpha goes from 1/2 of the tan (alpha) for alpha=0° To 1 for alpha=90° So to get a good result for alpha=1°, just divide the tangent by 2 And for alpha= 89° Calculate the arctan With very little accuracy, Divide by 2 and calculate the tangent which should be very close to 1. Last fiddled with by a1call on 2017-11-09 at 20:07
 2017-11-09, 20:56 #4 WMHalsdorf     Feb 2005 Bristol, CT 33×19 Posts Where a, b, c build a pyth. trippel with a^2+b^2=c^2 tan (alpha)=a/b Code: tan(alpha/2)=(c-b)/a
 2017-11-09, 22:16 #5 Dr Sardonicus     Feb 2017 Nowhere 5×11×113 Posts $tan(\frac{\alpha}{2})=\frac{sin(\frac{\alpha}{2})}{cos(\frac{\alpha}{2})}= \frac{2*sin(\frac{\alpha}{2})*cos(\frac{\alpha}{2})}{2*cos^{2}(\frac{\alpha}{2})}= \frac{sin(\alpha)}{(1+cos(\alpha))}=\frac{sin(\alpha)-sin(\alpha)*cos(\alpha)}{sin^{2}(\alpha)} =cosec(\alpha)-cot(\alpha)$
 2017-11-12, 20:10 #6 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 5·467 Posts I have been thinking about this topic and find it quite interesting. For a tangent in the range of 45° to 90° (1 to ∞), the half tangent ranges from (0.41 to 1). You could think of this in many different ways. For example as an infinite increase in precision. Or, since bisecting an angle is s constructible process (it can be done using a compass and a straight edge), it is conceivable to have mechanical machine which with a theoretical 0 thickness parts can infinity magnify a force. Of course in practice the magnification will be limited by the >0 thickness of the parts. I haven't been able to think of an electronic equivalent but it could exist and would be quite interesting, at least in a theoretical sense. Last fiddled with by a1call on 2017-11-12 at 20:12
 2017-11-13, 12:15 #7 a1call     "Rashid Naimi" Oct 2015 Remote to Here/There 91F16 Posts I did some more thinking. A similar concept would be a bow with infinity non stretchable string. Such a (theoretical) string would exert a force to the bow that is infinite times of any pulling force applied to it at t0. However the gain would very quickly drop to a very low finite value exponentially, past t0.
2017-11-13, 12:33   #8
alpertron

Aug 2002
Buenos Aires, Argentina

2×32×83 Posts

Quote:
 Originally Posted by Dr Sardonicus $tan(\frac{\alpha}{2})=\frac{sin(\frac{\alpha}{2})}{cos(\frac{\alpha}{2})}= \frac{2*sin(\frac{\alpha}{2})*cos(\frac{\alpha}{2})}{2*cos^{2}(\frac{\alpha}{2})}= \frac{sin(\alpha)}{(1+cos(\alpha))}=\frac{sin(\alpha)-sin(\alpha)*cos(\alpha)}{sin^{2}(\alpha)} =cosec(\alpha)-cot(\alpha)$
It appears that you typed extra end of line characters inside the identities. After deleting these, we can see the formulas:

$tan(\frac{\alpha}{2})=\frac{sin(\frac{\alpha}{2})}{cos(\frac{\alpha}{2})}=\frac{2*sin(\frac{\alpha}{2})*cos(\frac{\alpha}{2})}{2*cos^{2}(\frac{\alpha}{2})}=\frac{sin(\alpha)}{(1+cos(\alpha))}=\frac{sin(\alpha)-sin(\alpha)*cos(\alpha)}{sin^{2}(\alpha)}=cosec(\alpha)-cot(\alpha)$

2017-11-13, 16:24   #9
Dr Sardonicus

Feb 2017
Nowhere

5·11·113 Posts

Quote:
 Originally Posted by alpertron It appears that you typed extra end of line characters inside the identities. After deleting these, we can see the formulas
Hmm, must be a difference in display settings, the two look exactly the same to me!

In any event, a theoretically more satisfactory approach, applicable to dividing an angle by any positive integer, is to use the multiple-angle formulas to get a polynomial equation for the required trig function. In the case of the tangent, let t = tan(\theta). Then, for k a positive integer, $tan(k\theta)$ may be expressed as Im(1+i*t)^k/Re(1 + i*t)^k. Setting this rational expression equal to a given value for $\tan(k\theta)$ gives a polynomial for t = $\tan(\theta)$. The zeroes of the polynomial are $\tan(\theta + j\pi/k)$, j = 0 to k-1.

In the case k = 2, the rational expression is 2t/(1 - t^2).

 2017-11-16, 19:54 #10 bhelmes     Mar 2016 1A316 Posts Thanks a lot for these informations. There is a question which follows immediately: If a, b, c element N and a² + b² = c² you can calculate tan (alpha) and tan (alpha/2) Is it possible to transform the tan (alpha/2) into a pyth. trippel with a*, b*, c* I thought that a*, b*, c* should be element N again, Example a=7, b=24, c=25 tan (alpha/2)=(c-b)/a=1/7 (very elegant solution! ) But 1²+7²=50 so that c=sqrt (50) and c is not element N On the other hand i thought that the pyth trippel form a group. Somewhere there is a logical mistake. Would be nice, if someone could give a hint. Greetings from the pyth. trippel Bernhard
 2017-11-16, 20:27 #11 Nick     Dec 2012 The Netherlands 34128 Posts We gave a complete parametrization of all primitive Pythagorean triples in the Number Theory discussion group here: http://www.mersenneforum.org/showthread.php?t=21850

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