3三重大学 2009年 数学 第3問解答・解説

$$\displaystyle \begin{aligned} &\text{(1) }\operatorname{slope}(QP)=\frac{\sin\theta}{1+\cos\theta}=\tan\frac{\theta}{2}.\text{ Since }QO\text{ is horizontal, }\angle RQO=\frac{\theta}{2},\quad t=\tan\frac{\theta}{2}.\\ &\text{(2) }QR:y=tx+t.\text{ Substituting in }x^2+y^2=1\text{ gives }(x+1)((1+t^2)x+t^2-1)=0.\text{ Hence }Q=(-1,0),\ P=\left(\frac{1-t^2}{1+t^2},\frac{2t}{1+t^2}\right).\text{ Therefore, }\quad \cos\theta=\frac{1-t^2}{1+t^2},\quad \sin\theta=\frac{2t}{1+t^2}.\\ &\text{(3) }t=\tan(\theta/2)\text{ and }\frac{dt}{d\theta}=\frac12\sec^2(\theta/2)=\frac{1+t^2}{2},\text{ so }\frac{d\theta}{dt}=\frac{2}{1+t^2}.\\ &\text{(4) Put }t=\tan(\theta/2).\text{ Then }1+\sin\theta-\cos\theta=\frac{2t(t+1)}{1+t^2},\quad d\theta=\frac{2\,dt}{1+t^2}.\text{ The limits become }\frac1{\sqrt3}\text{ and }1.\\ &\int_{\pi/3}^{\pi/2}\frac{d\theta}{1+\sin\theta-\cos\theta}=\int_{1/\sqrt3}^{1}\frac{dt}{t(t+1)}=\left[\ln\frac{t}{t+1}\right]_{1/\sqrt3}^{1}=\ln\frac{\sqrt3+1}{2}. \end{aligned}$$