Question

$(cos\theta +isin\theta {)}^n=cos\theta +isinn\theta$

Answer 1

Let P(n) be true for $n=m$
$\therefore (cos\theta +isin\theta {)}^{m+1}$
$=(cos\theta +isin\theta {)}^m(cos\theta +isin\theta )$
$=(cosm\theta -isinm\theta )(cos\theta +sin\theta )$
$=(cosm\theta cos\theta -sinm\theta sin\theta )+i\left(sinmcosm\theta sin\theta \right)$
$=cos(m+1)\theta +isin(m+1)\theta$
Above relation shows that P(n) is true for $n=m+1$
Again when $n=1$
$(cos\theta +isin\theta {)}^1=cos1.0+isin\theta$
when $n=2,(cos\theta +isin\theta {)}^2$
$co{s}^2\theta +i^{2}sin\theta +2isin\theta cos\theta$
$(co{s}^2\theta +si{n}^{\theta })+i\left(2sin\theta cos\theta \right)$
$=cos2\theta +isin2\theta$
Above relation shows that P(n) is true for $n=1,2$
Hence P(n) is universally true