Trigonometry problem on product of trig functions. Questions: 


*

*$\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=?$

*$\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}=?$

*$\tan\dfrac{\pi}{7}\tan\dfrac{2\pi}{7}\tan\dfrac{3\pi}{7}=?$


Approach: 
If $x=\dfrac{\pi}{7}\implies 7x=\pi\implies 4x=\pi-3x\implies \sin 4x = \sin 3x$.Upon expansion, $$8cos^3x-4cos²x-4cosx+1=0$$

"$\cos \dfrac{\pi}{7}, \cos \dfrac{3\pi}{7}, \cos \dfrac{5\pi}{7}$will satisfy this cubic."

Note: I'm not able to understand the above line. Why do these values satisfy this cubic and how did we get these values? 
These are therefore the roots of the cubic, and their product is: 
$$\cos\dfrac{\pi}{7}\cos\dfrac{3\pi}{7}\cos\dfrac{5\pi}{7}=\dfrac{-1}{8}$$
Finally,
$\cos\dfrac{5\pi}{7} = \cos\left(\pi-\dfrac{2\pi}{7}\right)=-\cos \dfrac{2\pi}{7}$. Replacing this in (1) gives $$\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=\dfrac{1}{8}$$
How can we apply this concept to find the values of 


*$\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}\sin\dfrac{3\pi}{7}?$

*$\tan\dfrac{\pi}{7}\tan\dfrac{2\pi}{7}\tan\dfrac{3\pi}{7}?$
I'm not able to form the equations for these two problems. Please Help. .
 A: Let $a=\pi/7$. 
We know $x=\cos a$ is a root of $8x^3-4x^2-4x+1=0$. This is because $$3a+4a=\pi\Rightarrow \sin(3a)=\sin(4a)$$ as you wrote.
Now, note that $3\cdot 3a+4\cdot 3a=3\pi\Rightarrow \sin(3\cdot 3a)=\sin(4\cdot 3a)$. This can be done for $5a$ as well. This is why $\cos (3a),\cos (5a)$ satisfy the cubic.
For 3, since $3a+4a=\pi\Rightarrow \tan(3a)+\tan(4a)=0$, we have
$$\frac{\tan a+\tan(2a)}{1-\tan a\tan(2a)}+\frac{2\tan(2a)}{1-\tan^2(2a)}=0,$$
i.e.
$$\tan a+3\tan(2a)-3\tan a\tan^2(2a)-\tan^3(2a)=0.$$
Setting $x=\tan a$ gives us
$$x+\frac{6x}{1-x^2}-\frac{12x^3}{(1-x^2)^2}-\frac{8x^3}{(1-x^2)^3}=0,$$
i.e.
$$x^6-21x^4+35x^2-7=0.$$
Here, note that $x=\tan a,\tan(2a),\tan(3a)$ are roots of this equation by the similar argument above. Hence, $\tan^2(ka)\ (k=1,2,3)$ are roots of
$$x^3-21x^2+35x-7=0.$$
So, we have
$$\tan^2a\tan^2(2a)\tan^2(3a)=7\Rightarrow \tan a\tan(2a)\tan(3a)=\sqrt 7.$$
Finally, we have
$$\sin a\sin(2a)\sin(3a)=\cos a\cos(2a)\cos (3a)\times \tan a\tan(2a)\tan(3a)=\frac{\sqrt 7}{8}.$$
A: As $\sin\left[7\left(\pm\dfrac{u\pi}7\right)\right]=0$ for any integer $u$
If $\sin7x=0,7x=n\pi$ where $n$ is any integer
$\implies x=\dfrac{n\pi}7$ where $n=0,\pm1,\pm2,\pm3\pmod7$
For $f(c)=8c^3-4c^2-4c+1=0$
$f'(c)=24c^2-8c-4=4(6c^2-2c-1)$
As $f'(c)=0,f(c)=0$ don't have any root in common, $f(c)=0$ does not have Repeated Root.
As $\cos(-A)=\cos A,\cos\left(\pm\dfrac{r\pi}7\right),r=1,2,3$ will be the roots of $f(c)=0$

For the Sine product, I would recommend this
and use $\sin(\pi-B)=\sin B$

For tangent Product, I would recommend method applied in
Sum of tangent functions where arguments are in specific arithmetic series and Prove that $\cot^2{(\pi/7)} + \cot^2{(2\pi/7)} + \cot^2{(3\pi/7)} = 5$
to find  the roots of $t^6-21t^4+35t^2-7=0--->(2)$ are $\tan\frac{r\pi}7$ where $r=1,2,3,4,5,6$  
and use $\tan(\pi-C)=-\tan C$
