Show that $\displaystyle \sin\left[\frac{\pi}{14}\right]\cdot\sin\left[\frac{3\pi}{14}\right]\sin\left[\frac{5\pi}{14}\right]=\frac 18$.
Would this happen to be in the "complex numbers" section of the textbook/exercise? If so, did it have a first part? Also, is that the question? Or does it say "hence show that..."
Because I have seen these questions in our 4U textbooks (highest level maths in high school), and recognise them, I might be able to give you a hand if there is a first part!
$\displaystyle 8\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\ pi}{14}=4\left(\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}\right)\sin\frac{5\pi}{14}=$
$\displaystyle 2\left(2\sin\frac{5\pi}{14}\cos\frac{\pi}{7}-2\sin\frac{5\pi}{14}\cos\frac{2\pi}{7}\right)=2\le ft(\sin\frac{\pi}{2}+\sin\frac{3\pi}{14}-\sin\frac{9\pi}{14}-\sin\frac{\pi}{14}\right)=$
$\displaystyle =2\left(1+\sin\frac{3\pi}{14}-\sin\left(\pi-\frac{9\pi}{14}\right)-\sin\frac{\pi}{14}\right)=2\left(1+\sin\frac{3\pi} {14}-\sin\frac{5\pi}{14}-\sin\frac{\pi}{14}\right)$
Now I will show that $\displaystyle \sin\frac{3\pi}{14}-\sin\frac{5\pi}{14}-\sin\frac{\pi}{14}=-\frac{1}{2}$ (*)
This is equivalent to $\displaystyle \sin\frac{\pi}{14}-\sin\frac{3\pi}{14}+\sin\frac{5\pi}{14}=\frac{1}{2 }$
Multiply both members by $\displaystyle 2\sin\frac{\pi}{7}$ then we transform products to sums.
$\displaystyle \cos\frac{\pi}{14}-\cos\frac{3\pi}{14}-\cos\frac{\pi}{14}+\cos\frac{5\pi}{14}+\cos\frac{3 \pi}{14}-\cos\frac{\pi}{2}=\sin\frac{\pi}{7}$
$\displaystyle \cos\frac{5\pi}{14}=\sin\frac{\pi}{7}$
$\displaystyle \cos\frac{5\pi}{14}=\cos\left(\frac{\pi}{2}-\frac{\pi}{7}\right)$
$\displaystyle \cos\frac{5\pi}{14}=\cos\frac{5\pi}{14}$
So, the identity (*) is true.