Identify if some special name is given to polynomials Is there some special name given to the two polynomials below:
$${n \choose 1}x - {n \choose 3}x^3 + \ldots \label{} \tag{1}$$ and $$1 - {n \choose 2}x^2 + {n \choose 4}x^4 - \ldots \label{} \tag{2}$$.
 A: HInt:try $$(1+ix)^n$$ wher $i^2=-1$
$$(1+ix)^n=\\\left(\begin{array}{c}n\\ 0\end{array}\right)1^{n}(ix)^{0}+\left(\begin{array}{c}n\\ 1\end{array}\right)1^{n-1}(ix)^{1}+
\left(\begin{array}{c}n\\ 2\end{array}\right)1^{n-2}(ix)^{2}+
\left(\begin{array}{c}n\\ 3\end{array}\right)1^{n-3}(ix)^{3}+
\left(\begin{array}{c}n\\ 4\end{array}\right)1^{n-4}(ix)^{4}+...=\\\left(\begin{array}{c}n\\ 0\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)(ix)+\left(\begin{array}{c}n\\ 20\end{array}\right)(ix)^{2}+\left(\begin{array}{c}n\\ 3\end{array}\right)(ix)^{3}+\left(\begin{array}{c}n\\ 4\end{array}\right)(ix)^{4}+...=\\(
\left(\begin{array}{c}n\\ 0\end{array}\right)-\left(\begin{array}{c}n\\ 2\end{array}\right)x^2+\left(\begin{array}{c}n\\ 4\end{array}\right)x^4-...)+\\i(
\left(\begin{array}{c}n\\ 1\end{array}\right)x-\left(\begin{array}{c}n\\ 3\end{array}\right)x^3+...)$$
A: Define
$$P_n(x)=1 - {n \choose 2}x^2 + {n \choose 4}x^4 - \ldots$$
$$Q_n(x)={n \choose 1}x - {n \choose 3}x^3 + \ldots$$
I have never heard of these polynomials before, but the polynomials $P_n$ bare some vague resemblances to Chebyschev polynomials $T_n(x)$.
Since
$${n\choose 1}x - {n \choose 3}x^3 + \ldots=\Im(1+ix)^n,$$
$$1 - {n \choose 2}x^2 + {n \choose 4}x^4 - \ldots=\Re(1+ix)^n,$$
writing $(1+ix)$ in polar form yields
$$
\begin{align}
(1+ix)^n=
&
(\sqrt{1+x^2}e^{i\arctan(x)})^n
\\
=&(1+x^2)^{\frac n2}e^{in\arctan(x)}
\end{align}.
$$
This shows that 
$$P_n(x)=(1+x^2)^{\frac n2}\cos\left(n \arctan(x)\right)
\\
Q_n(x)=(1+x^2)^{\frac n2}\sin\left(n \arctan(x)\right).$$
Compare 
$$T_n(x)=\cos(n\arccos(x))$$
$$P_n(x)=(1+x^2)^\frac n2 \cos( n\arctan(x)).$$
Also, the family $P_n$ has the property that
$$P_n(\tan(\theta))=(1+\tan^2(\theta))^{\frac n2}\cos(n\theta)=\frac{\cos(n\theta)}{\cos^{n}(\theta)},$$
mirroring 
$$T_n(\cos(\theta))=\cos(n\theta).$$
Through this relation, we can also express our polynomials as a function of Chebyschev polynomials $T_n$:
$$P_n(\tan(\theta))=\frac{T_n(\cos(\theta))}{\cos^n(\theta)}.$$
