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Forget about the $\sin$ and $\cos$ functions, are there possibly some brilliant way to show that $$\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^2+ \left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)^2=1$$

?

I've thought for a long time, without making much progress. Can someone help me? Thanks.

JSCB
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    The terms in the parenthesis are equal to $\sin(x)$ and $\cos(x)$ respectively. – Daniel Montealegre Aug 27 '13 at 14:05
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    Can you see "Forget about the sin and cos functions" on the first line? :) – JSCB Aug 27 '13 at 14:06
  • I think there are formulas for the coefficients. Assuming there are, this would be one way to brute force calculate the limit. Show the coefficients on the left equal the coefficients on the right, so when you subtract, you only have $1$ left over. Certainly note that this is not a brilliant answer :) – Clayton Aug 27 '13 at 14:06
  • @ᴊᴀsᴏɴ sorry i tend to read only the titles for a lot of posts :P – Daniel Montealegre Aug 27 '13 at 14:07

3 Answers3

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If we denote $$f(x)=\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots\right)^2+ \left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots\right)^2$$

then by differentiation term by term we see that $f'(x)=0$ so $f(x)=f(0)=1$.

2

Both power series are even.

The coefficient of $x^{2n}$ in the first power series is $$ \sum_{k=0}^{n}\frac{1}{(2k-1)!(2n-2k+1)!} $$ The coefficient of $x^{2n}$ in the second power series is $$ \sum_{k=0}^n \frac{1}{(2k)!(2n-2k)!} $$

So the difference between the two coefficients is

$$ \sum_{k=0}^{2n}(-1)^k \frac{1}{k!(2n-k)!}=\frac{1}{(2n)!}(1-1)^{2n} \, . $$

Micah
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When I ask Mathematica for the series expansion for Sin[x]^2, I get:

x^2 - x^4/3 + (2 x^6)/45 - x^8/315 + (2 x^10)/14175-...

And Cos[x]^2 is:

1 - x^2 + x^4/3 - (2 x^6)/45 + x^8/315 - (2 x^10)/14175+...

So add those two expressions together, and you're left with 1.

I'm too lazy to do it, but clearly, if you manually expanded your squared terms, you ought to be left with closed form expressions for the expansions I just listed, with the immediate consequence that added together they equal 1.

Nathan McKenzie
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