Cosine/R/Addition theorem/Power series direct/Fact/Proof

Proof

The ${}2n$-th summand (the term which refers to the power with exponent ${}2n$) in the cosine series (the coefficients referring to ${}x^{i}$, ${}i$ odd, are ${}0$) of ${}x+y$ is

${}{\begin{aligned}{\frac {(-1)^{n}(x+y)^{2n}}{(2n)!}}&={\frac {(-1)^{n}}{(2n)!}}\sum _{i=0}^{2n}{\binom {2n}{i}}x^{i}y^{2n-i}\\&=(-1)^{n}\sum _{i=0}^{2n}{\frac {1}{i!(2n-i)!}}x^{i}y^{2n-i}\\&=(-1)^{n}\sum _{j=0}^{n}{\frac {x^{2j}y^{2n-2j}}{(2j)!(2n-2j)!}}+(-1)^{n}\sum _{j=0}^{n-1}{\frac {x^{2j+1}y^{2n-2j-1}}{(2j+1)!(2n-2j-1)!}},\end{aligned}}$

where in the last step we have split up the index set into even and odd numbers.

The ${}2n$-th summand in the Cauchy product of ${}\cos x$ and ${}\cos y$ is

${}{\begin{aligned}\sum _{j=0}^{n}{\frac {(-1)^{j}(-1)^{n-j}}{(2j)!(2(n-j))!}}x^{2j}y^{2(n-j)}&=(-1)^{n}\sum _{j=0}^{n}{\frac {x^{2j}y^{2(n-j)}}{(2j)!(2(n-j))!}}\\\end{aligned}}$

and the ${}2n$-th summand in the Cauchy product of ${}\sin x$ and ${}\sin y$ is

${}{\begin{aligned}\sum _{j=0}^{n-1}{\frac {(-1)^{j}(-1)^{n-1-j}}{(2j+1)!(2(n-1-j)+1)!}}x^{2j+1}y^{2(n-j)+1}&=(-1)^{n-1}\sum _{j=0}^{n-1}{\frac {x^{2j+1}y^{2(n-1-j)+1}}{(2j+1)!(2(n-1-j)+1)!}}\\\end{aligned}}$

Hence, both sides of the addition theorem coincide in the even case. For an odd index the left-hand side is ${}0$. Since in the cosine series only even exponents occur, it follows that in the Cauchy product of the two cosine series only exponents of the form ${}x^{i}y^{j}$ with ${}i,j$ even occur. Since in the sine series only odd exponents occur, it follows that in the Cauchy product of the two sine series only exponents of the form ${}x^{i}y^{j}$ with ${}i+j$ even occur. Therefore terms of the form ${}x^{i}y^{j}$ with ${}i+j$ odd occur neither on the left nor on the right-hand side.

To fact