Can we prove $\sin(x)^2 + \cos(x)^2 = 1$ using just their series?

Question

In particular I mean:

$$\sin(x)^2 + \cos(x)^2$$

$$=\left(\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}\right)^2 + \left(\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}\right)^2$$

However I am not sure how you're supposed to correctly expand and recombine terms when dealing with the sum of two squared series, especially when there are factorials involved.

Edit: To be clear, I am asking about manipulating the series I have just stated in order to show that they sum to $1$.

Answer 1

Absolutely. Power series actually multiply just like polynomials do: $$(a_0+a_1x+a_2x^2+a_3x^3+\ldots)(b_0+b_1x+b_2x^2+\ldots)=\sum_{n=0}^{\infty}\left(\sum_{c=0}^na_cb_{n-c}\right)x^n.$$

Let $$\alpha(x)=\left(\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}\right)^2$$ $$\beta(x)=\left(\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!}\right)^2.$$ First, it is clear that the constant term of $\alpha(x)+\beta(x)$ is indeed $1$, as we can check directly. Thus, we merely need to check that every other coefficient vanishes.

Note that, either from the formula for products or by noting that both are even functions, all the coefficients of odd powers of $x$ in both $\alpha$ and $\beta$ and thus $\alpha+\beta$ vanish. Now, consider the coefficient of $x^{2n}$ in either. In $\beta$, the formula gives the coefficient of $x^{2n}$ as, where we use the variable $k$ to count only the even (non-zero) coefficients of $\cos(x)$: $$\sum_{k=0}^n(-1)^c\cdot (-1)^{n-k}\cdot \frac{1}{(2n-2k)!}\cdot \frac{1}{(2k)!}=(-1)^n\cdot \sum_{k=0}^n\frac{1}{(2k)!(2n-2k)!}.$$ The same can be done to find the coefficient of $x^{2n}$ in $\alpha$, using $k$ to enumerate odd coefficients of $\sin(x)$: $$\sum_{k=0}^{n-1}(-1)^k\cdot (-1)^{n-k-1}\cdot \frac{1}{(2n-2k-1)!}\cdot \frac{1}{(2k+1)!}=(-1)^{n-1}\cdot \sum_{k=0}^{n-1}\frac{1}{(2k+1)!(2n-2k-1)!}.$$ We are trying to show that the coefficient of $x^n$ in $\alpha+\beta$ is zero for $n>0$. This amounts to showing the following equality for all $n>0$: $$\sum_{k=0}^{n-1}\frac{1}{(2k+1)!(2n-2k-1)!}=\sum_{k=0}^n\frac{1}{(2k)!(2n-2k)!}.$$ Multiplying through by $(2n)!$ on both sides reduces this to a combinatorial equality: $$\sum_{k=0}^{n-1}{2n\choose 2k+1} = \sum_{k=0}^n{2n\choose 2k}.$$ This just says that the number of subsets of $2n$ with an odd number of elements equals the number of subsets of $2n$ with an even number of elements - but this is easy to show: We can define a bijection $\pi$ which takes a set $S\subseteq \{1,\ldots,2n\}$ and takes it to $S\cup \{1\}$ if $1\not\in S$ and $S\setminus \{1\}$ if $1\in S$. This places sets of odd and even parity into bijection, showing the desired equality.

Answer 2

Note that just using their series

$$\sin^2 x + \cos^2 x=(\cos x+i\sin x)(\cos x-i\sin x)\stackrel{\text{by series}}=e^{ix}e^{-ix}=1$$

Answer 3

Trying it out on my own using some points made in Milo's post (not going to accept my own answer, this is just for my own benefit):

$$\sin(x)^2 + \cos(x)^2$$

$$=\left(\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{(2n+1)!}\right)^2 + \left(\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{(2n)!}\right)^2$$

$$=\left(\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{n-k}\frac{x^{2n-2k+1}}{(2n-2k+1)!}(-1)^k\frac{x^{2k+1}}{(2k+1)!}\right) + \left(\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{n-k}\frac{x^{2n-2k}}{(2n-2k)!}(-1)^k\frac{x^{2k}}{(2k)!}\right)$$

$$=\left(\sum_{n=1}^\infty x^{2n}\sum_{k=0}^{n-1}\frac{(-1)^{n-1}}{(2n-2k-1)!(2k+1)!}\right) + \left(1+\sum_{n=1}^\infty x^{2n}\sum_{k=0}^n\frac{(-1)^n}{(2n-2k)!(2k)!}\right)$$

$$=1 + \sum_{n=1}^\infty x^{2n}(-1)^{n-1}\left(\sum_{k=0}^{n-1}\frac{1}{(2n-2k-1)!(2k+1)!} - \sum_{k=0}^n\frac{1}{(2n-2k)!(2k)!}\right)$$

$$=1 + \sum_{n=1}^\infty x^{2n}(-1)^{n-1}\frac{1}{(2n)!}\left(\sum_{k=0}^{n-1}\binom{2n}{2k+1} - \sum_{k=0}^n\binom{2n}{2k}\right)$$

$$=1 + \sum_{n=1}^\infty x^{2n}(-1)^{n-1}\frac{1}{(2n)!}\left(2^{2n-1} - 2^{2n-1}\right)$$

$$=1$$

Answer 4

The essential result here is that $e^z e^{-z} = 1$ (cf. gimusi's answer).

In general, for $x$ with the radius of convergence of the series $\sum_k a_k z^k$ and $\sum_k b_k z^k$, a small amount of work shows that $(\sum_k a_k z^k) (\sum_k b_k z^k) = \sum_k c_k z^k$, where $c_k = \sum_l a_l b_{k-l}$ (the series convolution).

In the above case, take $a_k = {1 \over k!}, b_k = (-1)^k {1 \over k!}$ for $k \ge 0$, and zero otherwise. Then $c_k = 0$ for $k <0$, $c_0 = 1$ and for $k >0$, we have $c_k = \sum_{l=0}^k {1 \over k!} (-1)^{k-l} {1 \over (l-k)!} = \sum_{l=0}^k (-1)^{k-l} \binom{k}{l} {1 \over k!} = (1-1)^k {1 \over k!} = 0$.

Hence $e^z e^{-z} = 1$.

Then $\cos^2 x + \sin^2 x =(\cos x + i \sin x) (\cos x - i \sin x) = e^{ix} e^{-ix} = 1$.