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Suppose that humanity haven't discovered Taylor Series Expansion of trigonometric functions or of any function that would help us on this. Which means we are not allowed to replace the given infinite series sums with the corresponding $\sin x$ and $\cos x$ functions.

Could we still show that the following identity is true for all real $x$?

$$ 1 - \left( x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \dots \right)^2 = \left( 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \dots \right)^2 $$

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Yes. Expand both sides and compare term by term. This would show that the expressions coincide formally. Standard results allow you to show that the formal expansion of the square is indeed the square, and that the series converge everywhere. –  Andres Caicedo Apr 27 '13 at 15:55
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If $S(x)$ is the series on the left and $C(x)$ is the series on the right, one can (after proving the relevant facts) show that the function $C^2+S^2$ has zero derivative everywhere. Thus $C^2+S^2$ is a constant, which can be seen to be $1$ by evaluating $ C^2(0)+S^2(0)$ –  David Mitra Apr 27 '13 at 15:59
    
In some treatments of the subject, the trigonometric functions are introduced through series, and verifying your identity is one of the first steps in deriving the basic properties of the trig functions from the series definition (periodicity, addition laws). –  André Nicolas Apr 27 '13 at 16:26
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3 Answers

up vote 13 down vote accepted

As Andres says in his comment, you can compare the coefficients of $x^{2k}$ in both quantities.

On the left-hand side, this coefficient is $(-1)^k \sum_{i+j=k-1} \frac 1 {(2i+1)!(2j+1)!}$, while on the right-hand side, it is $(-1)^k \sum_{i+j=k} \frac 1{(2i)!(2j)!}$.

Multiplying both coefficients by $(2k)!$, you are left to prove $\sum \binom {2k}{2i} = \sum \binom {2k}{2j+1}$, i.e. that the number of even subsets of a set of size $2k$ is the number of odd subsets.

This is in fact true for any nonempty set, not only those of even size. To show this combinatorially, pick a distinguished element $x$ of a set $S$ of size $n$. Then given a subset $X \subset S$, map it either to $X \setminus \{ x \}$ or to $X \cup \{x \}$ according to wether $x \in S$ or not. Then you can check that this is a bijection between odd-sized subsets of $S$ and even-sized subsets of $S$.

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This boils down to showing $$\tag1\exp(x+y)=\exp(x)\exp(y),$$ where $\exp(x)=\sum_{n=0}^\infty \frac{x^n}{n!}$, which is more or less straightforward: On the left hand side, the coefficient of $x^ny^m$ is ${n+m\choose n}\cdot\frac1{(n+m)!}$ and on the right it is $\frac1{n!}\cdot\frac1{m!}$, and these expressions are equal. Now we find that for real (or in fact arbitrary) $x$ we have $$\tag2\exp(ix)\exp(-ix)=\exp(0)=1.$$ If we denote the expression in parentheses on the left of your equation with $s(x)$ and on the right with $c(x)$, we see that $\exp(\pm ix)=c(x)\pm is(x)$, so that $(2)$ becomes $$c^2(x)+s^2(x)=1.$$

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you've forgot about the assumption about not knowing about taylor series, which is strongly related to knowledge, that there are exponential functions... the task was related to ugly polynomials which are assumed to not have any better expression... –  V-X Apr 27 '13 at 22:20
    
@V-X I did not use Taylor series anywhere, specifically I made no claim about the function I innocouosly named $\exp$ that relates to taking derivatives (specifically, I did not define it as solution of $y'=y$) and used the Cauchy product to show the homomorphism property instead. –  Hagen von Eitzen Apr 28 '13 at 10:36
    
and you used not relation to the question. thus your answer look to me like off-topic... –  V-X Apr 28 '13 at 20:18
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A note on mercio's solution:

The proof depends on showing that $\sum_{i=0}^k \binom {2k}{2i} = \sum_{j=0}^{k-1} \binom {2k}{2j+1}$.

This can be rewritten as $\sum_{i=0}^{2k} (-1)^i \binom {2k}{i} =0 $ and this is the expansion of $0 = (1-1)^{2k}$ by the binomial theorem.

As mercio says, this is true for any $k$, not just even $k$, since $0 = (1-1)^{k}$.

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(+1) This is how I would have shown the equality. However, it is not true for $k=0$ since $1\ne0$. –  robjohn Aug 27 '13 at 14:53
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