Proving the sum of squares of sine and cosine using the Cauchy product formula

Just to see how ugly it is, you actually can do it 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^{2(n-k)}}{(2(n-k))!}(-1)^k\frac{x^{2k}}{(2k)!}=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^n\binom{2n}{2k}\frac{x^{2n}}{(2n)!}\\ \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^{2(n-k)+1}}{(2(n-k)+1)!}(-1)^k\frac{x^{2k+1}}{(2k+1)!}\\&=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^n\binom{2n+2}{2k+1}\frac{x^{2n+2}}{(2n+2)!}=\sum_{n=1}^\infty\sum_{k=0}^{n-1}(-1)^{n-1}\binom{2n}{2k+1}\frac{x^{2n}}{(2n)!}\end{align*} Now utilize the binomial theorem: \begin{align*}\cos^2 x+\sin^2 x&=1+\sum_{n=1}^\infty(-1)^n\left(\sum_{k=0}^n\binom{2n}{2k}-\sum_{k=0}^{n-1}\binom{2n}{2k+1}\right)\frac{x^{2n}}{(2n)!}\\&=1+\sum_{n=1}^\infty(-1)^n\sum_{k=0}^{2n}(-1)^k\binom{2n}{k}\frac{x^{2n}}{(2n)!}=1+\sum_{n=1}^\infty(-1)^n(1-1)^{2n}\frac{x^{2n}}{(2n)!}=1+\sum_{n=0}^\infty 0\\&=1\end{align*}

This doesn't use the Cauchy product formula, but you can resolve this identity using the power series themselves.

Using the power series we find that $$\frac{d}{dx}\sin(x) = \cos(x)$$ and $$\frac{d}{dx} \cos(x) = -\sin(x).$$

Note also that: $$2\sin(x)\cos(x)-2\cos(x)\sin(x)=0$$

Integrating both sides gives us $$(\sin(x))^2 + (\cos(x))^2 = C$$ for some constant $C$. Finally we can determine $C$ by plugging in $x=0$. This gives us $(0)^2 + 1^2 = C$. Thus we can conclude that $$(\sin(x))^2+(\cos(x))^2=1.$$

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As a side note you can also determine the power series for $f(x)=(\sin(x))^2$ as well as for $(\cos(x))^2$ without using the Cauchy formula.

Note that $f(0)=0$ and $$f'(x)=2\sin(x)\cos(x)=\sin(2x)=\sum_{n=0}^\infty \frac{(-1)^n2^{2n+1}x^{2n+1}}{(2n+1)!}$$ thus $$(\sin(x))^2 = \sum_{n=0}^\infty \frac{(-1)^n 2^{2n+1} x^{2n+2}}{(2n+2)!}.$$