$$96 \cos\frac{\pi}{33} \cos\frac{2\pi}{33} \cos\frac{4\pi}{33} \cos\frac{8\pi}{33} \cos\frac{16\pi}{33}$$ is equal to

We need to evaluate $$96 \cos\frac{\pi}{33} \cos\frac{2\pi}{33} \cos\frac{4\pi}{33} \cos\frac{8\pi}{33} \cos\frac{16\pi}{33}$$.

A well-known product formula for cosines with angles in geometric progression states that

$$\prod_{k=0}^{n-1}\cos(2^k\theta)=\frac{\sin(2^n\theta)}{2^n\sin\theta}$$ which follows by repeatedly applying the double-angle formula $$\sin(2\alpha)=2\sin\alpha\cos\alpha$$.

By setting $$\theta=\pi/33$$ and $$n=5$$ in this identity, it follows that

$$\cos\frac{\pi}{33}\cos\frac{2\pi}{33}\cos\frac{4\pi}{33}\cos\frac{8\pi}{33}\cos\frac{16\pi}{33} =\frac{\sin(2^5\cdot\pi/33)}{2^5\sin(\pi/33)} =\frac{\sin(32\pi/33)}{32\sin(\pi/33)}$$

Noting that $$\sin\frac{32\pi}{33}=\sin\bigl(\pi-\frac{\pi}{33}\bigr)=\sin\frac{\pi}{33}$$ by the identity $$\sin(\pi-x)=\sin x$$, the product simplifies to

$$\frac{\sin(\pi/33)}{32\sin(\pi/33)}=\frac{1}{32}$$

Finally, multiplying by 96 gives

$$96\times\frac{1}{32}=3$$ so the value of the given expression is 3.

Hence, the correct answer is Option 1: 3.