1. ## Trigonometry

Show that $\sin\left[\frac{\pi}{14}\right]\cdot\sin\left[\frac{3\pi}{14}\right]\sin\left[\frac{5\pi}{14}\right]=\frac 18$.

2. 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!

3. I'd appreciate your help. There are no other parts. This is the only question but I don't know from where did they get this.. It's between a collection of exercises xD and no complex numbers nearby.

4. $8\sin\frac{\pi}{14}\sin\frac{3\pi}{14}\sin\frac{5\ pi}{14}=4\left(\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}\right)\sin\frac{5\pi}{14}=$

$2\left(2\sin\frac{5\pi}{14}\cos\frac{\pi}{7}-2\sin\frac{5\pi}{14}\cos\frac{2\pi}{7}\right)=2\le ft(\sin\frac{\pi}{2}+\sin\frac{3\pi}{14}-\sin\frac{9\pi}{14}-\sin\frac{\pi}{14}\right)=$

$=2\left(1+\sin\frac{3\pi}{14}-\sin\left(\pi-\frac{9\pi}{14}\right)-\sin\frac{\pi}{14}\right)=2\left(1+\sin\frac{3\pi} {14}-\sin\frac{5\pi}{14}-\sin\frac{\pi}{14}\right)$

Now I will show that $\sin\frac{3\pi}{14}-\sin\frac{5\pi}{14}-\sin\frac{\pi}{14}=-\frac{1}{2}$ (*)

This is equivalent to $\sin\frac{\pi}{14}-\sin\frac{3\pi}{14}+\sin\frac{5\pi}{14}=\frac{1}{2 }$

Multiply both members by $2\sin\frac{\pi}{7}$ then we transform products to sums.

$\cos\frac{\pi}{14}-\cos\frac{3\pi}{14}-\cos\frac{\pi}{14}+\cos\frac{5\pi}{14}+\cos\frac{3 \pi}{14}-\cos\frac{\pi}{2}=\sin\frac{\pi}{7}$

$\cos\frac{5\pi}{14}=\sin\frac{\pi}{7}$

$\cos\frac{5\pi}{14}=\cos\left(\frac{\pi}{2}-\frac{\pi}{7}\right)$

$\cos\frac{5\pi}{14}=\cos\frac{5\pi}{14}$

So, the identity (*) is true.