Area between a function and a rotation of its reflection.

It might be obvious or might be not, but the curve is a reflection of $f(x)$ in the line $y=\tan(\pi/8)x$ . To derive at this mathematically, I will assume you know about the rotation matrix $M_{\mathrm{rot}}$ of a rotation around $O$ by $\theta$ and the reflection matrix $M_{\mathrm{ref}}$ of a reflection around the line $y=\tan(\theta)x$ : $M_{\mathrm{rot}} = \begin{pmatrix} \cos{\theta} & -\sin{\theta} \\ \sin{\theta} & \cos{\theta} \end{pmatrix} \hskip 0.1em \mathrm{ and } \hskip 0.2em M_{\mathrm{ref}} = \begin{pmatrix} \cos{2\theta} & \sin{2\theta} \\ \sin{2\theta} & -\cos{2\theta} \end{pmatrix}.$ Consider a point $P(x_p,y_p)\in\mathbb{R}^2$ which corresponds to a vector $\mathbf{p}\in\mathbb{R}^2$ . The transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ given by a reflection in the line $y=0$ after which it is rotated around $O$ by 45 degrees will have the property that $\mathbf{p} \mapsto \begin{pmatrix} \cos(\pi/4) & -\sin(\pi/4) \\ \sin(\pi/4) & \cos(\pi/4) \end{pmatrix} \begin{pmatrix} \cos(0) & \sin(0) \\ \sin(0) & -\cos(0) \end{pmatrix} \mathbf{p} = \begin{pmatrix} \cos(\pi/4) & \sin(\pi/4) \\ \sin(\pi/4) & -\cos(\pi/4) \end{pmatrix} \mathbf{p}.$ Note that the last matrix is exactly the matrix corresponding to a reflection in the line $y=\tan(\pi/8)x$ . So indeed, the curve is a reflection of $f(x)$ in the line $y=\tan(\pi/8)x$ . Note that in general $\mathrm{Rot}(\theta) \circ \mathrm{Ref}(0) = \mathrm{Ref}(\theta/2),$ see Wikipedia .

Notice that the new information about the curve makes the problem much easier, because while we don't know what the formula is for the curve, we make the observation that $\mathcal{A} = 4\cdot\int_0^{\xi}(\sin(x)-\tan(\pi/8)x)dx,$ but before calculating this integral, note (thanks to Jon Haussman) that $\tan(t/2) = \frac{1-\cos(t)}{\sin(t)}$ , and so we see that $\tan(\pi/8) = \sqrt{2}-1.$ The method described by Anirudh Sreekumar in the comments will also work. Now back to integrating. We see that $\mathcal{A} = 4\cdot\int_0^{\xi}\left(\sin(x)+(1-\sqrt{2})x\right)dx = -4\cos{\xi}+2(1-\sqrt{2})\xi^2+4 = 4(1-\cos{\xi})+2(1-\sqrt{2})\xi^2.$ We can conclude that $a\cdot b \cdot c \cdot d \cdot e = 4\cdot 1\cdot 2\cdot 1\cdot 2 = \boxed{16}.$

For those who are curious, $\mathcal{A} \approx 2.36607545275$ .

Let $g$ be the new, reflected/rotated function. As stated by @Joël Ganesh , $g$ is the reflection of $f$ across the line $y=\left(\tan \frac{\pi}{8}\right)x$ . To understand why this is so, think about what it means to reflect and rotate $f$ : everything $f$ does relative to the $x$ -axis, $g$ does relative to the $45^\circ$ line, but in the opposite direction. Therefore, the two functions will literally meet halfway , on the line with half the angle from the $x$ -axis compared to the $45^\circ$ ( $\frac{\pi}{4}$ -radian) line--namely, the $22.5^\circ$ ( $\frac{\pi}{8}$ -radian) line, whose slope is $\tan \frac{\pi}{8}$ .

Using this information and the knowledge of the characteristics of sine curves, we can divide the area into four congruent regions as shown below:

All four regions have the same area, which you can calculate by integrating the difference between $f(x)$ and the line of reflection from $0$ to $\xi$ :

$\mathcal{A}=4\int_0^\xi \sin x - \left(\tan \frac{\pi}{8}\right)x dx=4(1-\cos \xi)-\frac{\tan \frac{\pi}{8}}{2}\xi^2$

Using the half-angle identity for the tangent function,

$\tan \frac{\pi}{8}=\sqrt{\frac{1-\cos \frac{\pi}{4}}{1+\cos \frac{\pi}{4}}}=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{1+\frac{\sqrt{2}}{2}}}=\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}\cdot \frac{2-\sqrt{2}}{2-\sqrt{2}}}=\sqrt{\frac{(2-\sqrt{2})^2}{4-2}}=\frac{2-\sqrt{2}}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{2(\sqrt{2}-1)}{2}=\bf \sqrt{2}-1$ .

Therefore,

$\mathcal{A}=\boxed{\bf 4(1-\cos \xi)+2(1-\sqrt{2})\xi^2}$ ,

with $a=4$ , $b=1$ , $c=2$ , $d=1$ , and $e=2$ .