Krdytkyu

Evaluation of $$sin frac{pi}{7}cdot sin frac{2pi}{7}cdot sin frac{3pi}{7} = $$

$bf{My; Try::}$ I have solved Using Direct formula::

$$sin frac{pi}{n}cdot sin frac{2pi}{n}cdot......sin frac{(n-1)pi}{n} = frac{n}{2^{n-1}}$$

Now Put $n=7;,$ We get
$$sin frac{pi}{7}cdot sin frac{2pi}{7}cdot sin frac{3pi}{7}cdot sin frac{4pi}{7}cdot sin frac{5pi}{7}cdot sin frac{6pi}{7}=frac{7}{2^{7-1}}$$

So $$sin frac{pi}{7}cdot sin frac{2pi}{7}cdot sin frac{3pi}{7} =frac{sqrt{7}}{8}$$

Now my question is how can we solve it without using Direct Formula, Help me

Thanks

$$
begin{align}
prod_{k=1}^3sinleft(frac{kpi}7right)^2
&=prod_{k=1}^6sinleft(frac{kpi}7right)tag{1}\
&=-frac1{64}prod_{k=1}^6left(e^{ikpi/7}-e^{-ikpi/7}right)tag{2}\
&=frac1{64}prod_{k=1}^6left(1-e^{-i2kpi/7}right)tag{3}\
&=frac1{64}lim_{zto1}prod_{k=1}^6left(z-e^{-i2kpi/7}right)tag{4}\
&=frac1{64}lim_{zto1}frac{z^7-1}{z-1}tag{5}\[6pt]
&=frac7{64}tag{6}
end{align}
$$
Explanation:

$(1)$: $sin(x)=sin(pi-x)$

$(2)$: $sin(x)=frac{e^{ix}-e^{-ix}}{2i}$

$(3)$: pull $prodlimits_{k=1}^6e^{ikpi/7}=-1$ out of the product

$(4)$: $1=limlimits_{zto1}z$

$(5)$: $prodlimits_{k=1}^6left(z-e^{-i2kpi/7}right)=frac{z^7-1}{z-1}$

$(6)$: evaluate limit

Therefore, taking the square root of $(6)$, we get
$$
prod_{k=1}^3sinleft(frac{kpi}7right)=frac{sqrt7}8tag{7}
$$

Using $2sin asin b=cos(a-b)-cos(a+b)$ and $2sin acos b=sin(a+b)+sin(a-b)$, write

$$sin frac{pi}7cdotsin frac{2pi}7cdot sin frac{3pi}7 = frac12left(cosfrac{pi}7-cosfrac{3pi}7right)sinfrac{3pi}7=frac14left(sinfrac{4pi}7+sinfrac{2pi}7-sinfrac{pi}7right)\=frac14left(sinfrac{2pi}7+sinfrac{4pi}7+sinfrac{8pi}7right)$$

Then have a look at this question: Trigo Problem : Find the value of $sinfrac{2pi}{7}+sinfrac{4pi}{7}+sinfrac{8pi}{7}$

This is a Gauss sum in disguise, but it can be computed through elementary geometry, without resorting to algebraic number theory. Let us consider a regular heptagon $P_1 P_2 P_3 P_4 P_5 P_6 P_7$ with center $O$ and take $A=P_1, B=P_5, C=P_6$. The angles of $ABC$ are $frac{pi}{7},frac{2pi}{7},frac{4pi}{7}$. By Euler's formula and the sine theorem

$$ [ABC] = frac{abc}{4R} = 2R^2 sin(A)sin(B)sin(C) $$
but we also have
$$[ABC] = [AOB]+[BOC]-[AOC] = frac{R^2}{2}left(sinfrac{4pi}{7}+sinfrac{2pi}{7}-sinfrac{6pi}{7}right) $$
hence the claim is equivalent to:
$$ sinfrac{2pi}{7}+sinfrac{4pi}{7}+sinfrac{8pi}{7} = frac{1}{2}sqrt{7}. $$
and since the values of the sine function at $frac{pi}{7},frac{2pi}{7},frac{4pi}{7}$ are positive, the claim is also equivalent to:
$$ left(1-cosfrac{2pi}{7}right)left(1-cosfrac{4pi}{7}right)left(1-cosfrac{8pi}{7}right)=frac{7}{8}. $$
Chebyshev polynomials ensure that $cosfrac{2pi}{7},cosfrac{4pi}{7},cosfrac{8pi}{7}$ are algebraic conjugates over $mathbb{Q}$, and they are the roots of $p(x)=8x^3+4x^2-4x-1$. By factoring such polynomial over $mathbb{R}$ it follows that
$$ p(x) = 8x^3+4x^2-4x-1 = 8left(x-cosfrac{2pi}{7}right)left(x-cosfrac{4pi}{7}right)left(x-cosfrac{8pi}{7}right) $$
and by evaluating at $x=1$ we get
$$ 7 = p(1) = 8left(1-cosfrac{2pi}{7}right)left(1-cosfrac{4pi}{7}right)left(1-cosfrac{8pi}{7}right) $$
QED.

Hint;

Assume
$$ theta = (npi)/7$$

$$implies 4theta = npi -3theta$$

Take the sine on both the sides
An apply the required manipulations

You should get a cubic in $sin^2 theta$
Since we have defined out $theta$, we are quite aware what the roots are going to be

So the numerical part in the end divided by the coefficient of $x^3$ gives you

$$ {{[sin frac{pi}{7}cdot sin frac{2pi}{7}cdot sin frac{3pi}{7}}]}^2 = 7/64$$

That should be it