Find the value of $\sin10^0 \sin 20^0 \cdots \sin 80^0 .$

Solution:

We have proved the identity,

$\sin \frac {\pi}{n} \sin\frac {2\pi}{n} \cdots \sin \frac{(n-1)\pi} {n}=\frac{n}{2^{n-1}}$

$n=18,$

$\sin \frac{\pi}{18} \sin \frac{2\pi}{18} \cdots \sin \frac{17\pi}{18} =\frac{18}{2^{17}}$

Observation:

(1) In the expression, $9^{th}$ term=$1$ ,$\quad$ put k=9, $\sin \frac {k\pi}{18} = \sin \frac {\pi}{2} =1$

(2) $\sin \frac {(18-k)\pi}{18}\ = \sin \frac {k\pi}{18}$

$[\sin (n - k) \pi = \sin\frac{k \pi}{n} ;$ $k = 1, \cdots ,n]$

$\therefore \sin \frac{\pi}{18} \sin \frac{2\pi}{18}\ \cdots \sin \frac{17\pi}{18}\ =\frac{18}{2^{17}}$

$\sin ^2\frac{\pi}{18}\ \sin ^2\frac{2\pi}{18}\ \cdots \sin ^2\frac{8\pi}{18}\ =\frac{9}{2^{16}}$

$\sin 10^0\ \sin 20^0 \cdots \sin 80^0\ =\frac{3}{2^{8}}$

Show that

(1) For $n=2m, \cos \frac {\pi}{n}\ \cos \frac{2\pi}{n}…\cos \frac {(m-1)\pi}{n} =\frac {\sqrt m}{2^{m-1}}$

(2) For $n=2m+1, \cos \frac {\pi}{n}\cos\frac{2\pi}{n}…\cos\frac {m\pi}{n} =\frac {1}{2^{m-1}}$

(3) For $n=2m, \tan\frac {\pi}{n}\tan\frac{2\pi}{n}…\tan\frac {(m-1)\pi}{n} =1$

(4) For $n=2m+1, \tan\frac {\pi}{n} \tan\frac{2\pi}{n}…\tan\frac {m\pi}{n} =\sqrt {2m+1}$

W.L.G. $P_k =$ cis$(\frac{2k\pi}{n})=\alpha^k,$
$k=0,\cdots ,n-1$

$z^{n-1} +z^{n-2}+ \cdots +z+1=\Pi_{k=1}^{n-1}(z-\alpha^k)$

$z = 1,$ $n=\Pi_{k=1}^{n-1}(1-\alpha^k)$

$z = -1,$

$|1-\alpha^k|^2 =4\biggl(\sin\frac{k\pi}{n}\biggl)^2, \quad k=0, \cdots n-1$

$\Rightarrow|1-\alpha^k| =2\biggl|\sin\frac{k\pi}{n}\biggl|, \quad 0\le\frac{\pi}{n}\le \frac{k\pi}{n}\le\frac{(n-1)\pi}{n}\le\pi$

$P_0P_k\ = 2 \sin\frac{k\pi}{n}$

$\Rightarrow 2 \sin\frac{\pi}{n}\times2\ \sin\frac{2\pi}{n} \cdots 2\ \sin\frac{(n-1)\pi}{n}=n$

$\ \sin\frac{\pi}{n}\times\ \sin\frac{2\pi}{n} \cdots \sin\frac{(n-1)\pi}{n}=\frac {n}{2^{n-1}}$

Using this hint, we can show the result.

$\ n=3, \quad z^3=1$

$\rightarrow\ \biggl\{ 1,\ \text{cis}\frac {2\pi}{3},\quad \text{cis}\frac {4\pi}{3}\biggl\}$

$\rightarrow\quad \omega=\ \text{cis} \frac{2\pi}{3} =\frac {-1}{2}+i\frac{\sqrt3}{2}$

$\rightarrow\quad \omega^2=\ \text{cis} \frac{4\pi}{3} =\biggl(\text{cis}\frac{2\pi}{3}\biggl)^2=\omega^2=\bar{\omega}$

$1, \omega,\omega^2$ are cube roots of unity.

$(1)\ 1+\omega+\omega^2 =0$

$(2)\ \omega^3 =1=\biggl(\text{cis}\frac{2\pi}{3}\biggl)^3=\text{cis}2\pi =1$

$\ \omega^{3n} =(\omega^3)^n =1 \quad n\in \Z$

Find the value of

$4+5\biggl(\frac {-1}{2}+i\ \frac {{\sqrt3}}{2}\biggl)^{334} + 3\biggl(\frac {-1}{2}+i\ \frac{{\sqrt3}}{2}\biggl)^{365}$

$4+5\ \omega^{334}\ +3\ \omega^{365}$

$=4+5\cdot \ \omega^{333}\ \cdot \omega+3\ \cdot \omega^{363} \cdot \omega^2$

$=4+5 \omega+3\omega^2 \quad \quad (\therefore \omega^2=-\omega-1)$

$=4+5 \omega-3\omega-3$

$=1+2 \biggl(\frac {-1}{2}+i\ \frac {{\sqrt3}}{2}\biggl)$

$=i{\sqrt3}$

Prove that $\biggl(\frac {\sqrt3+i}{{\sqrt3-i}}\biggl)^{3(2n+1)} =-1,\quad$ for all $n \in \Z.$

$\biggl(\frac {\sqrt3+i}{{\sqrt3-i}}\biggl)\cdot\biggl(\frac {\sqrt3+i}{{\sqrt3+i}}\biggl)$ $=\frac {(\sqrt3+i)^2}{4}$ $=i^2\ \biggl(\frac {1}{2}- \frac {{\sqrt3\ }i}{2}\biggl)^2=-\omega^2$

$[\because \omega=\frac{-1}{2}+i\ \frac{{\sqrt3}}{2}]$

$\biggl(\frac {{\sqrt3+i}}{{\sqrt3-i}}\biggl )^{3(2n+1)}=(-\omega^2)^{3(2n+1)}$ $=(-1)^{6n+3}(\omega^6)^{(2n+1)} =-1$

Find the least positive integer ’n’ such that

$(1+\omega^2)^n = (1+\omega^4)^n$

$1+\omega^2=-\omega$

$\omega^4=\omega^3\cdot \omega=\omega$

$(1+\omega^2)^n=(1+\omega^4)^n$

$\Leftrightarrow (-\omega)^n=(-\omega^2)^n$

$\Leftrightarrow \omega^n=\omega^{2n}$

$\Leftrightarrow \omega^n=1$

$n=3$

Solve for $z \in \mathbb{C}:$ $z^2 +z|z|+|z|^2=0$

$z^2 +z|z|+|z|^2=0 \rightarrow (1)$

$z=0 \quad (1)$ is satisfied

Let $z \ne\ 0$

$\frac {z^2}{|z|^2}+ \frac {z}{|z|}+1=0$

$\biggl (\frac {z}{|z|}\biggl)^2+ \biggl(\frac {z}{|z|}\biggl)+1=0$

$\omega=\frac {z}{|z|}, \quad \omega^2+\omega+1=0\quad$ It has two solution $\text{cis}\frac{2\pi}{3},\ \text{cis}\frac{4\pi}{3}$

$z=\lambda\omega, \quad \lambda \omega^2,\quad \lambda\ge 0$ are solution to (1)

$|1-\omega| =|\omega-\omega^2|$
$|1-\omega| =|\omega(1-\omega)|$
$|1-\omega| =|1-\omega|$
$|1-\omega| =|1-\omega^2|$

Equilateral Triangle

$a=b=c$

$\alpha=\beta=x$

Proposition

Let T be a triangle with 15 vertices a,b,c in $\mathbb{C}$ (in anticlockwise direction).

Then the following are equivalent:

(i) T is equilateral triangle

(ii) $a+\omega b+\omega^2c=0, \quad \omega=\text{cis}\frac {2\pi}{3}$

(iii) $a^2+b^2+c^2 = ab+bc+ca$

$(i)\Leftrightarrow(ii)$

$(i) \Rightarrow(ii) \quad a+\omega b+\omega^2c\ =0\rightarrow(1)$

suppose (1) is satisfied

$(a-z),(b-z),(c-z)$

$a-z+\omega(b-z)+\omega^2(c-z)$

$=a+\omega b+\omega^2c-z\ (1+\omega+\omega^2)$

$=0\rightarrow(2)$

$(1) \Leftrightarrow(2)$

T is equilateral $\Leftrightarrow c-b=\omega(b-a)$

$\Leftrightarrow c+a\omega-b(1+\omega)=0$

$\Leftrightarrow a\omega+b\omega^2+c=0$

$\Leftrightarrow a+b\omega+c\omega^2=0$

(ii) $\Leftrightarrow$ (iii) (or) (i)$\Leftrightarrow$ (iii)

$a^2+b^2+c^2= ab+bc+ca$

$\Leftrightarrow$ $(a-z)^2+(b-z)^2+(c-z)^2=(a-z)(b-z)+(b-z)(c-z)+(c-z)(a-z)$

$z=a \quad$ shift the triangle T such that one of vertex is ‘0’.

W.l.g. O,B,C are vertices of T.

$b^2+c^2=bc \Leftrightarrow$ $\omega b+\omega^2c=0$

$b^2+c^2=bc \Leftrightarrow$ $\omega=\frac {-b}{c}$

$\frac{b}{c}+\frac{c}{b}+1$ $\Rightarrow \frac{b}{c}=1-\frac{c}{b}$

claim $\quad (\frac{-b}{c})^3=1$

$\biggl(\frac{-b}{c}\biggl)^2=\frac {b}{c}\cdot\biggl(1-\frac {c}{b}\biggl) = \frac {b}{c}-1=-\frac {c}{b}$

$\biggl(\frac{-b}{c}\biggl)^3=\biggl(\frac {-b}{c}\biggl)^2\ \biggl(\frac {-b}{c}\biggl)= \frac {-c}{b}\ \cdot\ \frac {-b}{c}=1$

Thank you