<s>[INST] Trigonometric identities using $\sin x$ and $\cos x$ definition as infinite series Can someone show the way to proof that $$\cos(x+y) = \cos x\cdot\cos y - \sin x\cdot\sin y$$ and $$\cos^2x+\sin^2 x = 1$$ using the definition of $\sin x$ and $\cos x$ with infinite series.
thanks...
 [/INST] Let me do a different one.  Begin with
$$
\sin x = \sum_{k=0}^\infty \frac{(-1)^k x^{2k+1}}{(2k+1)!},
\qquad
\cos x = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!}
$$
So compute
$$\begin{align}
\sin(x+y) &= \sum_{k=0}^\infty \frac{(-1)^k(x+y)^{2k+1}}{(2k+1)!}
\\ &=
\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\sum_{j=0}^{2k+1} \binom{2k+1}{j} x^j y^{2k+1-j}
\\ &=
\sum_{k=0}^\infty (-1)^k\sum_{j=0}^{2k+1} \frac{1}{j!(2k+1-j)!} x^j y^{2k+1-j}
\\ &=
\sum_{j=0}^\infty \frac{x^j}{j!}\sum_k \frac{(-1)^k}{(2k+1-j)!} y^{2k+1-j}
\end{align}$$
where the inner sum is over all $k$ such that $2k+1 \ge j$.  Consider two cases for the inner sum, $j$ even and $j$ odd.  
If $j=2n$, then $2k+1 \ge j$ iff $2k+1 \ge 2n$ iff $k \ge n$.  So the $k$-sum is:
$$
\sum_{k=n}^\infty \frac{(-1)^k}{(2k+1-2n)!} y^{2k+1-2n}
$$
Use change of variables $i=k-n$ to get
$$
\sum_{i=0}^\infty \frac{(-1)^{i+n}}{(2i+1)!} y^{2i+1} = (-1)^n \sin y .
$$  
If $j=2n+1$, then $2k+1 \ge j$ iff $2k+1 \ge 2n+1$ iff $k \ge n$.  So the $k$-sum is
$$
\sum_{k=n}^\infty \frac{(-1)^k}{(2k+1-2n-1)!}y^{2k+1-2n-1}
$$
Again use change of variables $i=k-n$ to get
$$
\sum_{i=0}^\infty \frac{(-1)^{i+n}}{(2i)!} y^{2i} = (-1)^n \cos y.
$$
So finally we have
$$
\sin(x+y) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \sin  y
+\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\cos y
= \cos x \sin y + \sin x \cos y.
$$
</s>