Question

...............................
.​

Answer 1

$(1+ix)^n = C_n^0+C_n^1(ix)+C_n^2(ix)^2+C_n^3(ix)^3+...\\ (1-ix)^n = C_n^0-C_n^1(ix)+C_n^2(ix)^2-C_n^3(ix)^3+...$

$(1+ix)^n - (1-ix)^n = 2C_n^1ix-2C_n^3ix^3+2C_n^5ix^5-2C_n^7ix^7+...$

$= 2ix\cdot (C_n^1-C_n^3x^2+C_n^5x^4-C_n^7x^6+...$

$\text{Notez: }x = \sqrt a$

$(1+i\sqrt{a})^n - (1-i\sqrt{a})^n = 2i\sqrt{a}\cdot (C_n^1-aC_n^3+a^2C_n^5-a^3C_n^7+...$

$C_n^1-aC_n^3+a^2C_n^5-a^3C_n^7+... = \dfrac{(1+i\sqrt{a})^n - (1-i\sqrt{a})^n}{2i\sqrt{a}} =$

$= \dfrac{(1+i\tan b)^n - (1-i\tan b)^n}{2i\tan b} = \dfrac{\left(1+i\dfrac{\sin b}{\cos b} \right)^n - \left(1-i\dfrac{\sin b}{\cos b}\right)^n}{2i\tan b} =$

$=\dfrac{\left(\dfrac{\cos b+i\sin b}{\cos b}\right)^n - \left(\dfrac{\cos b-i\sin b}{\cos b}\right)^n}{2i\tan b} =$

$=\dfrac{\dfrac{(\cos b+i\sin b)^n}{(\cos b)^n} - \dfrac{\left[\cos (-b)+i\sin (-b\right)]^n}{(\cos b)^n}}{2i\tan b} =$

$=\dfrac{(\cos b+i\sin b)^n - \left[\cos (-b)+i\sin (-b\right)]^n}{2i\tan b\cdot (\cos b)^n} =$

$=\dfrac{\cos (nb)+i\sin(nb) - \left[\cos (-nb)+i\sin (-nb)\right]}{2i\tan b \cos^nb}=$

$=\dfrac{\cos (nb)+i\sin(nb) - \cos (-nb)-i\sin (-nb)}{2i\tan b \cos^nb}=$

$=\dfrac{\cos (nb)+i\sin(nb) - \cos (nb)+i\sin (nb)}{2i\tan b \cos^n b}=$

$=\dfrac{2i\sin(nb)}{2i\tan b \cos^nb}= \dfrac{\sin(nb)}{\tan b \cos^nb}$

$\Rightarrow \boxed{S = \dfrac{\sin(nb)}{\tan b \cos^nb}}$