Find the value of $k$ in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$

Question

Find the value of 'k' in the equation $\sin {1^\circ}\sin {3^\circ}.......\sin {179^\circ} = \frac{1}{{{2^k}}}$

My approach is as follow
$\sin {1^\circ} = \sin {179^\circ}$

$T = {\sin ^2}{1^\circ}{\sin ^2}{3^\circ}..{\sin ^2}{89^\circ}$

$\left( {\frac{{1 - \cos {2^\circ}}}{2}} \right) = {\sin ^2}{1^\circ}$

$T = \left( {\frac{{1 - \cos {2^\circ}}}{2}} \right)\left( {\frac{{1 - \cos {6^\circ}}}{2}} \right)\left( {\frac{{1 - \cos {{10}^\circ}}}{2}} \right)....\left( {\frac{{1 - \cos {{178^\circ}}}}{2}} \right)$

$2 + \left( {n - 1} \right)4 = 178 \Rightarrow n = 45$

$T = \frac{1}{{{2^{45}}}}\left( {1 - \cos {2^\circ}} \right)\left( {1 - \cos {6^\circ}} \right)\left( {1 - \cos {{10}^\circ}} \right)....\left( {1 - \cos {{178^\circ}}} \right)$

$\left( {1 - \cos {{178}^\circ}} \right) = \left( {1 + \cos {2^\circ}} \right)$

$T = \frac{1}{{{2^{45}}}}\left( {1 - {{\cos }^2}{2^\circ}} \right)\left( {1 - {{\cos }^2}{6^\circ}} \right)\left( {1 - {{\cos }^2}{10^\circ}} \right)....\left( {1 - {{\cos }^2}{{86}^\circ}} \right)\left( {1 - \cos {{90}^\circ}} \right)$

$T = \frac{{{{\sin }^2}{2^\circ}.{{\sin }^2}{6^\circ}.{{\sin }^2}{{10}^\circ}......{{\sin }^2}{{86}^\circ}}}{{{2^{45}}}}$

Not able to proceed from here

Answer 1

In your approach, there's a shorter way to arrive at T, but anyway taking off from there :

$$ T = \frac{{{{\sin }^2}{2^\circ}.{{\sin }^2}{6^\circ}.{{\sin }^2}{{10}^\circ}......{{\sin }^2}{{86}^\circ}}}{{{2^{45}}}} \\ = \frac{({\sin }{2^\circ}.{\sin }{6^\circ}.{\sin }{{10}^\circ}......{\sin }{{86}^\circ})^2}{{{2^{45}}}} \\ = \frac{({\cos }{4^\circ}.{\cos }{8^\circ}....{\cos }{{84}^\circ}.{\cos }{{88}^\circ})^2}{{{2^{45}}}} $$ Now let $p= {\cos }{4^\circ}.{\cos }{8^\circ}....{\cos }{{84}^\circ}.{\cos }{{88}^\circ} \\ $

And let $q = {\sin }{4^\circ}.{\sin }{8^\circ}....{\sin }{{84}^\circ}.{\sin }{{88}^\circ}$

So $pq = {(1/2^{22})}.{{\sin }{8^\circ}.{\sin }{16^\circ}...{\sin }{{176}^\circ}}\\ = {(1/2^{22})}.{\sin }{4^\circ}.{\sin }{8^\circ}....{\sin }{{84}^\circ}.{\sin }{{88}^\circ} {(\because \sin(180-x)=\sin x)}\\ = {(1/2^{22})}q $

So $p={(1/2^{22})}$

Therefore $T={(1/2^{89})}$

So $k=89$

Answer 2

As $\cos(90^\circ-A)=\sin A,$

$$\prod_{k=0}^{44}\sin(2k+1)^\circ=\prod_{k=0}^{44}\cos(2k+1)^\circ$$

As $\cos180(2k+1)^\circ=-1$

as $\cos180x=2^{180-1}c^{180}x+\cdots+(-1)^{90}$

and for $\cos180x=\cos180^\circ,180x=360^\circ n\pm180^\circ$

$x=(2 n+1)^\circ$ where $0\le n\le179$

If $\cos180x=-1,$ the roots of $$2^{180-1}c^{180}x+\cdots+(-1)^{90}=-1$$ are

$\cos(2 n+1)^\circ$ where $0\le n\le179$

$$\implies\prod_{n=0}^{179}\cos(2 n+1)^\circ=\dfrac2{2^{179}}$$

Now as $\cos(360^\circ-y)=+\cos y$

$$\implies\prod_{n=0}^{179}\cos(2 n+1)^\circ=\prod_{n=0}^{89}\cos^2(2 n+1)^\circ$$

Now $90<2n+1<180\iff 45\le n\le89\implies$ there are $45$(odd) number of multiplicands $<0$

$$\prod_{n=0}^{89}\cos(2 n+1)^\circ<0$$

$$\implies\prod_{n=0}^{89}\cos(2 n+1)^\circ=-\sqrt{\dfrac2{2^{179}}}$$

Finally as $\cos(180^\circ-z)=-\cos z,$

$$\implies\prod_{n=0}^{89}\cos(2 n+1)^\circ=-\prod_{n=0}^{44}\cos^2(2 n+1)^\circ$$

Finally use $$\prod_{n=0}^{44}\cos(2 n+1)^\circ>0$$

Answer 3

The connection of these kind of products with Chybshev polynomials in my opinion is the most important. Because, these polynomials show up everywhere like a unification of many things. Some nice properties of them are listed here. I know so few of them!: https://en.wikipedia.org/wiki/Chebyshev_polynomials

$T_n(x)=2^{n-1}\prod_{k=0}^{n-1}(x-\cos(\frac{2k+1}{2n}\pi))$ and $T_n(0)=(-1)^{\frac{n}{2}}$ when $n$ is even. So we get $2^{n-1}\prod_{k=0}^{n-1}\cos(\frac{2k+1}{2n}\pi)=(-1)^{\frac{n}{2}}$ when $n$ is even. Now let $n=90$. Then $2^{89}\prod_{k=0}^{89}\cos(\frac{2k+1}{180}\pi)=-1$ and thus $2^{89}\prod_{k=0}^{44}\cos^2(\frac{2k+1}{180}\pi)=2^{89}\prod_{k=0}^{44}\sin^2(\frac{2k+1}{180}\pi)=2^{89}\prod_{k=0}^{89}\sin(\frac{2k+1}{180}\pi)=1$.