I have to prove the following identity $\sin^2 (x) + \cos^2(x)=1$. I can easily prove this, but this exercise is given in the section introducing the series expansions for $\sin(x)$ and $\cos(x)$ and I should use these in the proof but am not quite sure how to do this.

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    $\begingroup$ Prove that the statement is true for $x=0$, then prove that the derivative of the left hand side is zero - thus, the left hand side is constant. $\endgroup$ – Mark McClure Mar 9 '14 at 22:58
  • $\begingroup$ Yeah, but he needs to incorporate the expansions somehow... $\endgroup$ – recursive recursion Mar 9 '14 at 23:01
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    $\begingroup$ @recursiverecursion Right. Those steps are very easy using series. That's rather the point. $\endgroup$ – Mark McClure Mar 9 '14 at 23:03
  • $\begingroup$ To prove the result directly from the series expansions, we could use the Cauchy Product. $\endgroup$ – preferred_anon Mar 9 '14 at 23:07
  • $\begingroup$ @zyx What's the combinatorics tag for? $\endgroup$ – draks ... Mar 9 '14 at 23:07

There is no need for complicated Cauchy products. We will simply use $\Big(f^2(x)\Big)'=2f(x)f'(x)$.



Now, $(\sin^2x+\cos^2x)'=(\sin^2x)'+(\cos^2x)'=2\sin x\sin'x+2\cos x\cos'x=2\sin x\cos x$ $+2\cos x(-\sin x)=0\iff\sin^2x+\cos^2x=C$. Now, to determine the value of the constant of integration, let us compute $\sin^20+\cos^20$ using the above definitions. It follows immediately that $\sin0=0\iff\sin^20=0$ and $\cos0=1\iff\cos^20=1$. QED.


Series are absolutely convergent, so we can expand using Cauchy products:

$$\begin{align}\cos^2 x&=\left(\sum_{n=0}^\infty(-1)^n\frac{x^{2n}}{(2n)!}\right)^2=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{n-k}\frac{x^{2n-2k}}{(2n-2k)!}(-1)^k\frac{x^{2k}}{(2k)!}\\&=\sum_{n=0}^\infty x^{2n}\sum_{k=0}^n\frac{(-1)^n}{(2n-2k)!(2k)!}=1+\sum_{n=1}^\infty x^{2n}\sum_{k=0}^n\frac{(-1)^n}{(2n-2k)!(2k)!}\\\sin^2 x&=\left(\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{(2n+1)!}\right)^2=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^{n-k}\frac{x^{2n-2k+1}}{(2n-2k+1)!}(-1)^k\frac{x^{2k+1}}{(2k+1)!}\\&={\sum_{n=0}^\infty x^{2n+2}\sum_{k=0}^n\frac{(-1)^n}{(2n-2k+1)!(2k+1)!}}=\sum_{n=1}^\infty x^{2n}\sum_{k=0}^{n-1}\frac{(-1)^{n-1}}{(2n-2k-1)!(2k+1)!}\end{align}$$

It remains to show that for $n\geq1$: $$\begin{align}&\sum_{k=0}^n\frac1{(2n-2k)!(2k)!}-\sum_{k=0}^{n-1}\frac1{(2n-2k-1)!(2k+1)!}=0\Longleftrightarrow\\&\sum_{k=0}^{2n}(-1)^k\frac{1}{(2n-k)!k!}=0\Longleftrightarrow\sum_{k=0}^{2n}\binom{2n}{k}(-1)^k=0\end{align}$$

Which is just an expansion of $(1-1)^{2n}$, so we are done.


I realise this thread has been well-resolved for some time, but I arrived here from another thread where I realised that there's some middle-of-the-road-like approach.

It just requires $$ \exp(z)=\sum_{n=0}^{\infty} \frac{z^n}{n!} $$ for any $z\in\mathbb{C}$. Evaluating at $z=ix$ for $x\in\mathbb{R}$ and comparing with the series expansions of $\sin$ and $\cos$ clearly yields $\exp(ix)=\cos(x)+i\sin(x)$. Furthermore, by the Binomial Theorem, we see that $$ \exp(z+w)=\sum_{n=0}^{\infty} \frac{(z+w)^n}{n!}=\sum_{n=0}^{\infty}\sum_{j=0}^n {n \choose j}\frac{z^j w^{n-j}}{n!}=\sum_{n=0}^{\infty} \sum_{j=0}^n \frac{z^j}{j!}\frac{w^{n-j}}{(n-j)!}=\sum_{j=0}^{\infty}\frac{z^j}{j!}\sum_{k=0}^{\infty} \frac{w^k}{k!} $$ and the last expression is equal to $\exp(z)\exp(w)$.

This simplies things a lot as, clearly, $\exp(0)=1$ and so $\exp(ix)=(\exp(-ix))^{-1}$ and, by continuity of complex conjugation, $\exp(-ix)$ is the complex conjugate of $\exp(ix)$. However, this happens if and only if $|\exp(ix)|=1$. And now, $$ 1=|\exp(ix)|=(\Re\exp(ix))^2+(\Im\exp(ix))^2=\cos^2(x)+\sin^2(x) $$


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