Answer: Using \(P=\theta\) and \(Q=\theta-\frac{\pi}{4}\),

\(2\sin\theta\cos\left(\theta-\frac{\pi}{4}\right)=\sin\left(2\theta-\frac{\pi}{4}\right)+\sin\left(\frac{\pi}{4}\right).\)

Since \(\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\) and \(\sin x=\cos\left(\frac{\pi}{2}-x\right)\),

\(\sin\left(2\theta-\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{2}-2\theta+\frac{\pi}{4}\right)=\cos\left(2\theta-\frac{3\pi}{4}\right).\)

Hence

\(2\sin\theta\cos\left(\theta-\frac{\pi}{4}\right)\equiv \cos\left(2\theta-\frac{3\pi}{4}\right)+\frac{1}{\sqrt{2}}.\)

The curve is a closed loop through the origin, symmetric about the line \(\theta=\frac{3\pi}{8}\).

The enclosed area is \(\frac{3}{8}(\pi+1)\).

Apply the identity \(2\sin P\cos Q\equiv \sin(P+Q)+\sin(P-Q)\) with \(P=\theta\) and \(Q=\theta-\frac{\pi}{4}\):

\(2\sin\theta\cos\left(\theta-\frac{\pi}{4}\right)=\sin\left(\theta+\theta-\frac{\pi}{4}\right)+\sin\left(\theta-\theta+\frac{\pi}{4}\right).\)

So

\(2\sin\theta\cos\left(\theta-\frac{\pi}{4}\right)=\sin\left(2\theta-\frac{\pi}{4}\right)+\sin\left(\frac{\pi}{4}\right).\)

Now \(\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\), and

\(\sin\left(2\theta-\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{2}-\left(2\theta-\frac{\pi}{4}\right)\right)=\cos\left(\frac{3\pi}{4}-2\theta\right)=\cos\left(2\theta-\frac{3\pi}{4}\right).\)

Therefore

\(r=2\sin\theta\cos\left(\theta-\frac{\pi}{4}\right)=\cos\left(2\theta-\frac{3\pi}{4}\right)+\frac{1}{\sqrt{2}}.\)

This is of the form \(r=\cos(2\theta-\alpha)+\text{constant}\), so the curve is symmetric about the line where the cosine argument is zero:

\(2\theta-\frac{3\pi}{4}=0\quad\Rightarrow\quad \theta=\frac{3\pi}{8}.\)

Hence the line of symmetry is \(\theta=\frac{3\pi}{8}\). The sketch is a single closed loop through the pole, lying in the sector \(0\le \theta\le \frac{3\pi}{4}\), with symmetry about \(\theta=\frac{3\pi}{8}\).

For the area enclosed by a polar curve,

\(A=\frac12\int r^2\,d\theta.\)

So

\(A=\frac12\int_0^{3\pi/4}\left(\cos\left(2\theta-\frac{3\pi}{4}\right)+\frac{1}{\sqrt{2}}\right)^2 d\theta.\)

Expanding,

\(A=\frac12\int_0^{3\pi/4}\left\{\cos^2\left(2\theta-\frac{3\pi}{4}\right)+\sqrt{2}\cos\left(2\theta-\frac{3\pi}{4}\right)+\frac12\right\}d\theta.\)

Using \(\cos^2 x=\frac12(1+\cos 2x)\), this becomes

\(A=\frac12\int_0^{3\pi/4}\left\{\frac12\cos\left(4\theta-\frac{3\pi}{2}\right)+\sqrt{2}\cos\left(2\theta-\frac{3\pi}{4}\right)+1\right\}d\theta.\)

Integrating term by term,

\(A=\left[\frac{1}{16}\sin\left(4\theta-\frac{3\pi}{2}\right)+\frac{1}{2\sqrt{2}}\sin\left(2\theta-\frac{3\pi}{4}\right)+\frac{\theta}{2}\right]_0^{3\pi/4}.\)

At \(\theta=\frac{3\pi}{4}\), this gives

\(-\frac{1}{16}+\frac14+\frac{3\pi}{8}.\)

At \(\theta=0\), this gives

\(\frac{1}{16}-\frac14.\)

Therefore

\(A=\left(-\frac{1}{16}+\frac14+\frac{3\pi}{8}\right)-\left(\frac{1}{16}-\frac14\right)=\frac{3}{8}(\pi+1).\)