If $f(x)=16\cos^2(3x)\cos^2(6x)\cos^2(12x)\cos^2(24x)$, what is the
value of $f(\frac{\pi}{36})$ ?$1)\frac{6-3\sqrt3}{16}\qquad\qquad2)\frac{6-\sqrt3}{16}\qquad\qquad3)\frac{6+\sqrt3}{16}\qquad\qquad4)\frac{6+3\sqrt3}{16}$
I'm looking for alternative approaches to solve this problem.
Here is my approach:
$$f(\frac{\pi}{36})=16\cos^2(\frac{\pi}{12})\cos^2(\frac{\pi}6)\cos^2(\frac{\pi}{3})\cos^2(\frac{2\pi}{3})=\frac34\cos^2(\frac{\pi}{12})=\frac38(1+\cos(\frac{\pi}6))=\frac{6+3\sqrt3}{16}$$
I'm looking for other ideas to solve the problem, so can you please solve it differently?
Best Answer
We can simplify $f(x)$ by using the identity $\sin 2x=2\sin x\cos x$ multiple times.
$f(x)=\dfrac{16\sin^2(3x)\cos^2(3x)\cos^2(6x)\cos^2(12x)\cos^2(24x)}{\sin^2(3x)}=\dfrac{\sin^2 48x}{16\sin^2 3x}=\dfrac{\sin^2 48x}{8(1-\cos 6x)}$
$f(\frac{\pi}{36})=\dfrac{2\sin^2\frac{4\pi}{3}}{1-\cos \frac{\pi}{6}}=\dfrac{ \frac{3}{4}}{8(1-\frac{\sqrt 3}{2})}=\dfrac{6+3\sqrt3}{16}$