Thread: angle bisection View Single Post
2017-11-13, 16:24   #9
Dr Sardonicus

Feb 2017
Nowhere

598610 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).