My Fifteenth Problem

$\frac{\pi}{2} \leq \theta \leq \pi$ and $\frac{\pi}{2} \leq \phi \leq \pi$ .

$a=1+sin(\theta -\frac{\pi}{2})=1+sin(\theta)*0-cos(\theta)*1=1-cos(\theta)$ . $b=1+cos(\theta -\frac{\pi}{2})=1+cos(\theta)*0+sin(\theta)*1=1+sin(\theta)$ .

Hence $f^{-1}(\frac{a}{b})=\theta$ and $f^{-1}(\frac{b}{a})=\phi$ . Also the interior angles of a pentagon sum to $3\pi$ . So $\theta + \phi + \frac{3\pi}{2} = 3\pi$ . $\rightarrow \theta + \phi = \frac{3\pi}{2}$ .

$\int_{-1/2}^{1/2} f^{-1} \circ g(t) \, dt = \frac{1}{2} \int_{-1/2}^{1/2} f^{-1} \circ g(t) + f^{-1} \circ g(-t) \, dt + \frac{1}{2} \int_{-1/2}^{1/2} f^{-1} \circ g(t) - f^{-1} \circ g(-t) \, dt$ $= \frac{1}{2} \int_{-1/2}^{1/2} f^{-1} \circ g(t) + f^{-1} \circ g(-t) \, dt + 0$ $= \frac{1}{2} \int_{-1/2}^{1/2} f^{-1} ( g(t) ) + f^{-1} ( \frac{1}{g(t)} ) \, dt$

$= \frac{1}{2} \int_{-1/2}^{1/2} \frac{3\pi}{2} \, dt = \frac{3\pi}{4} \approx 2.356$