\mathrm{tg}\,\phi=\frac{d}{r-d\,\mathrm{ctg}{\theta\over 2}}=\frac{2r(1-\cos\theta)}{3\theta}\cdot\frac{3\theta}{3\theta r-2r(1-\cos\theta)\mathrm{ctg}\,{\theta\over 2}}=\frac{2(1-\cos\theta)}{3\theta-2(1-\cos\theta)\mathrm{ctg}\,{\theta\over 2}}=\frac{1-\cos\theta}{{3\over 2}\theta-\sin\theta}