Note that this function has symmetry about the \(y\)-axis since \(\cos \) is an even function.
When \(x = 0\), \(y = 1 + \sqrt {1} = 2\). When \(x = \pm \frac {\pi }{4}\), \(y = \frac {1}{\sqrt {2}}\).
We investigate the gradient: \begin {align*} \DiffFrac {y}{x} & = - \sin x - 2 \sin 2x \cdot \frac {1}{2} \cdot \frac {1}{\sqrt {\cos 2x}} \\ & = - \sin x - \frac {\sin 2x}{\sqrt {\cos 2x}}, \end {align*}
so \(\DiffFrac {y}{x}\) takes opposite sign as \(x\), which means that \(y\) is decreasing when \(x > 0\), and \(y\) is increasing when \(x < 0\), and \(x = 0\) gives a maximum.
Also, note that \[ \lim _{x \to \frac {\pi }{4}^{-}} \DiffFrac {y}{x} = -\infty , \lim _{x \to -\frac {\pi }{4}^{+}} \DiffFrac {y}{x} = \infty , \] which means the tangent to the graph at those points are vertical.
Hence, the graph looks as follows:
The graph looks as follows.
By solving the quadratic, we have \[ r = \frac {2\cos \theta \pm \sqrt {4 \cos ^2 \theta - 4 \sin ^2 \theta }}{2} = \cos \theta \pm \sqrt {\cos 2 \theta }. \]
Hence, at \(\theta = \pm \frac {1}{4}\pi , r = \frac {1}{\sqrt {2}}\).
When \(r\) is small, we must have that \(r = \cos \theta - \sqrt {\cos 2 \theta }\) and \(\theta \) is small, and \begin {align*} -2r \cos \theta + \sin ^2 \theta & \approx 0 \\ r & \approx \frac {\sin ^2 \theta }{2 \cos \theta } \\ r & \approx \frac {1}{2} \sin \theta \tan \theta \\ r & \approx \frac {1}{2} \theta ^2, \end {align*}
as desired.
The curve will look as follows. At \(\theta = \pm \frac {1}{4}\pi \), the curve is tangent to the lines. At \(r = 0\), the curves are tangent to the initial line.
The area between \(C_2\) and \(\theta = 0\) above the line is given by \begin {align*} A & = \frac {1}{2} \int _{0}^{\frac {\pi }{4}} \left [\left (\cos \theta + \sqrt {\cos 2\theta }\right )^2 - \left (\cos \theta - \sqrt {\cos 2\theta }\right )^2\right ]\Diff \theta \\ & = \frac {1}{2} \int _{0}^{\frac {\pi }{4}} 4 \cos \theta \sqrt {\cos 2\theta } \Diff \theta \\ & = 2 \int _{0}^{\frac {\pi }{4}} \cos \theta \sqrt {\cos 2\theta } \Diff \theta \\ & = 2 \int _{0}^{\frac {\pi }{4}} \cos \theta \sqrt {1 - 2 \sin ^2 \theta } \Diff \theta \\ & = 2 \int _{0}^{\frac {\pi }{4}} \sqrt {1 - 2 \sin ^2 \theta } \Diff \sin \theta \\ & = 2 \int _{0}^{\frac {1}{\sqrt {2}}} \sqrt {1 - 2x^2} \Diff x \\ & = \sqrt {2} \int _{0}^{1} \sqrt {1 - y^2} \Diff y \\ & = \sqrt {2} \cdot \frac {\pi }{4} \\ & = \frac {\pi }{2 \sqrt {2}}, \end {align*}
as desired, the final integral being because this is \(\frac {1}{4}\) of the area of the unit circle, which is \(\pi \).