Question

If A is not an integral multiple of $\pi$, prove that $\cos A \cdot \cos 2A \cdot \cos 4A \cdot \cos 8A = \frac{\sin 16A}{16\sin A}$ and hence deduce that $\cos \frac{2\pi}{15} \cdot \cos \frac{4\pi}{15} \cdot \cos \frac{8\pi}{15} \cdot \cos \frac{16\pi}{15} = \frac{1}{16}$

Answer 1

The identity $\cos A \cdot \cos 2A \cdot \cos 4A \cdot \cos 8A = \frac{\sin 16A}{16\sin A}$ is proven. Using this identity, it is deduced that $\cos \frac{2\pi}{15} \cdot \cos \frac{4\pi}{15} \cdot \cos \frac{8\pi}{15} \cdot \cos \frac{16\pi}{15} = \frac{1}{16}$.

Answer 2

The proof of the identity $\cos A \cdot \cos 2A \cdot \cos 4A \cdot \cos 8A = \frac{\sin 16A}{16\sin A}$ is achieved by multiplying the LHS by $\frac{16\sin A}{16\sin A}$ and repeatedly applying the double angle formula for sine ($2\sin x \cos x = \sin 2x$) to simplify the numerator. The condition that A is not an integral multiple of $\pi$ ensures $\sin A \neq 0$.

For the deduction, we identify $A = \frac{2\pi}{15}$ from the product $\cos \frac{2\pi}{15} \cdot \cos \frac{4\pi}{15} \cdot \cos \frac{8\pi}{15} \cdot \cos \frac{16\pi}{15}$, as the angles are in the ratio $1:2:4:8$. Substituting this $A$ into the proven identity and simplifying $\sin(\frac{32\pi}{15})$ using periodicity ($\sin(2\pi+x)=\sin x$) leads to the result $\frac{1}{16}$.