The value of $$\sin 4^o \sin 8^o \sin 16^o \sin 20^o \sin 28^o \sin 32^o\cdots \sin 76^o \sin 80^o \sin 88^o$$ equals $N$. Find the value of $[2^{15}N]$.
This seems like I can use complex numbers real parts but not sure how. Multiplying by $\cos 4^o$ only complicates stuff. Please help.
Hint. Consider the identity (see e.g. Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$ ) $$\prod_{k=1}^{n-1}\sin \left( \frac{k\pi}{n}\right) = \frac{n}{2^{n-1}}$$ and note that, since $\sin(x^{\circ})=\sin(180^{\circ} -x^{\circ})$, it follows that the square of your product is equal to $$N^2=\frac{\prod_{k=1}^{44}\sin \left( \frac{4\pi}{180}\right)}{\prod_{k=1}^{14}\sin \left( \frac{12\pi}{180}\right)}.$$ Can you take it from here? What is the value of $2^{15}N$?