Our teacher assigned extra credit for which we are allowed to use any source (including this website) to find the answer. This material has not been covered in class or in the book we are using. We learned Taylor series using series of real numbers but we have not covered anything remotely similar to the problem below so I have no idea how to even begin this problem.
Let i =$\sqrt{−1}$
a) Use the Taylor series centered at x=0 to prove that, for all $\theta$,
$e^{i\theta} = cos(\theta) + isin(\theta)$
b) Use above answer to prove that $[cos(\theta)+ isin(\theta)]^n = cos(n\theta)+ isin(n\theta)]$
c) We may write $[cos(\theta)+ isin(\theta)]^n = cos^n(x)(1+itan(\theta))^n$
Use the 5th order Taylor polynomial for $(1+x)^\frac{1}{3}$ with $x = itan(\theta)$ to find an estimate for $sin\left(\frac{\theta}{3}\right)$.
Use this approximation to estimate the value of $sin\left(\frac{\pi}{18}\right)$.
Hint: Two complex numbers $a+bi$ and $c+di$ are equal if $a=c$ and $b=d$.
Best Answer
Take the series expansion of $e^z$, which is defined also for complex $z$ (and actually it is used to define the principal value of $e^z$, which is the usual acception for writing $e^z$). Split it into the even and odd components. $$ e^{\,z} = \sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,k} }} {{k!}}} = \sum\limits_{0\, \leqslant \,k} {\left( {\frac{{z^{\,\left( {2k} \right)} }} {{\left( {2k} \right)!}} + \frac{{z^{\,\left( {2k + 1} \right)} }} {{\left( {2k + 1} \right)!}}} \right)} = \sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,\left( {2k} \right)} }} {{\left( {2k} \right)!}}} + \sum\limits_{0\, \leqslant \,k} {\frac{{z^{\,\left( {2k + 1} \right)} }} {{\left( {2k + 1} \right)!}}} $$ Then putting $iy$ in place of $z$, you will notice that $$ \begin{gathered} e^{\,i\,y} = \sum\limits_{0\, \leqslant \,k} {\frac{{i^{\,\left( {2k} \right)} y^{\,\left( {2k} \right)} }} {{\left( {2k} \right)!}}} + \sum\limits_{0\, \leqslant \,k} {\frac{{i^{\,\left( {2k + 1} \right)} y^{\,\left( {2k + 1} \right)} }} {{\left( {2k + 1} \right)!}}} = \hfill \\ = \sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - 1} \right)^{\,k} y^{\,\left( {2k} \right)} }} {{\left( {2k} \right)!}}} + \sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - 1} \right)^{\,k} i\,y^{\,\left( {2k + 1} \right)} }} {{\left( {2k + 1} \right)!}}} = \hfill \\ = \sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - 1} \right)^{\,k} y^{\,\left( {2k} \right)} }} {{\left( {2k} \right)!}}} + i\sum\limits_{0\, \leqslant \,k} {\frac{{\left( { - 1} \right)^{\,k} \,y^{\,\left( {2k + 1} \right)} }} {{\left( {2k + 1} \right)!}}} = \hfill \\ = \cos y + i\sin y \hfill \\ \end{gathered} $$