Sine and cosine/Real/Properties/Fact/Proof

Proof

(1) and (2) follow directly from the definitions of the series.
(3). The ${\displaystyle {}2n}$-th summand (the term which refers to the power with exponent ${\displaystyle {}2n}$) in the cosine series (the coefficients referring to ${\displaystyle {}x^{i}}$, ${\displaystyle {}i}$ odd, are ${\displaystyle {}0}$) of ${\displaystyle {}x+y}$ is

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

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

{\displaystyle {}{\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 ${\displaystyle {}0}$. 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 ${\displaystyle {}x^{i}y^{j}}$ with ${\displaystyle {}i,j}$ 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 ${\displaystyle {}x^{i}y^{j}}$ with ${\displaystyle {}i+j}$ even occur. Therefore terms of the form ${\displaystyle {}x^{i}y^{j}}$ with ${\displaystyle {}i+j}$ odd occur neither on the left nor on the right-hand side. The addition theorem for sine is proved in a similar way.
(4). From the addition theorem for cosine, applied to ${\displaystyle {}y:=-x}$, and because of (2), we get

{\displaystyle {}{\begin{aligned}1&=\cos 0\\&=\cos {\left(x-x\right)}\\&=\cos x\cdot \cos {\left(-x\right)}-\sin x\cdot \sin {\left(-x\right)}\\&=\cos x\cdot \cos x+\sin x\cdot \sin x.\end{aligned}}}