Thread: angle bisection
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Old 2017-11-13, 16:24   #9
Dr Sardonicus
 
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Feb 2017
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Quote:
Originally Posted by alpertron View Post
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).
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