Answer: \(\displaystyle \frac{9\pi^2}{128}\).

Answer: \(\displaystyle \frac{9\pi^2}{128}\).

At \(A(a,0)\), the curve satisfies

\(\sin(4a-\pi)=0.\)

Therefore

\(4a-\pi=n\pi,\)

where \(n\) is an integer. Hence

\(a=\frac{(n+1)\pi}{4}.\)

Since

\(\frac{\pi}{2}\lt a\lt \pi,\)

we get

\(a=\frac{3\pi}{4}.\)

Differentiate the curve:

\(\frac{dy}{dx}=4\cos(4x-\pi).\)

At \(x=\frac{3\pi}{4}\),

\(4x-\pi=2\pi,\)

so

\(\frac{dy}{dx}=4\cos2\pi=4.\)

The tangent gradient is \(4\), so the normal gradient is

\(-\frac14.\)

The normal passes through

\(\left(\frac{3\pi}{4},0\right),\)

so its equation is

\(y=-\frac14\left(x-\frac{3\pi}{4}\right).\)

At the \(y\)-axis, \(x=0\), giving

Thus

\(B=\left(0,\frac{3\pi}{16}\right).\)

The triangle \(OAB\) is right-angled with

\(OA=\frac{3\pi}{4},\qquad OB=\frac{3\pi}{16}.\)

Therefore its area is