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...

In both cases, you'll want to use the Cauchy product, and the binomial theorem will be useful (for at least the first one), too. I leave the second one to you.
For the first one, $$\begin{align}\sin x\cdot\sin y &= \left(\sum_{n=0}^\infty\frac{(-1)^nx^{2n+1}}{(2n+1)!}\right)\cdot\left(\sum_{n=0}^\infty\frac{(-1)^ny^{2n+1}}{(2n+1)!}\right)\\ &= \sum_{n=0}^\infty\left(\sum_{k=0}^n \frac{(-1)^kx^{2k+1}}{(2k+1)!}\cdot\frac{(-1)^{n-k}y^{2(n-k)+1}}{(2(n-k)+1)!}\right)\\ &= \sum_{n=0}^\infty\left(\sum_{k=0}^n \frac{(-1)^nx^{2k+1}y^{2(n+1)-(2k+1)}}{(2k+1)!(2(n+1)-(2k+1))!}\right)\\ &= \sum_{n=0}^\infty\frac{(-1)^n}{(2(n+1))!}\left(\sum_{k=0}^n\frac{(2(n+1))!x^{2k+1}y^{2(n+1)-(2k+1)}}{(2k+1)!(2(n+1)-(2k+1))!}\right)\\ &= \sum_{n=0}^\infty\frac{(-1)^n}{(2(n+1))!}\left(\sum_{k=0}^n\binom{2(n+1)}{2k+1}x^{2k+1}y^{2(n+1)-(2k+1)}\right)\\ &= -\sum_{n=0}^\infty\frac{(-1)^{n+1}}{(2(n+1))!}\left(\sum_{k=0}^n\binom{2(n+1)}{2k+1}x^{2k+1}y^{2(n+1)-(2k+1)}\right)\\ &= -\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left(\sum_{k=0}^{n-1}\binom{2n}{2k+1}x^{2k+1}y^{2n-(2k+1)}\right),\end{align}$$ whence $$-\sin x\cdot\sin y=\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left(\sum_{k=0}^{n-1}\binom{2n}{2k+1}x^{2k+1}y^{2n-(2k+1)}\right).$$
You can do some of the same sorts of manipulations to see that $$\cos x\cdot\cos y = 1+\sum_{n=1}^\infty\frac{(-1)^n}{(2n)!}\left(\sum_{k=0}^n\binom{2n}{2k}x^{2k}y^{2n-2k}\right).$$ On the other hand, the binomial theorem shows us that for $n\ge 1$ we have $$(x+y)^{2n} = \sum_{j=0}^{2n}\binom{2n}jx^jy^{2n-j},$$ and splitting the right-hand side into two sums--one for even $j$ and one for odd $j$--gives us $$(x+y)^{2n} = \left(\sum_{k=0}^{n}\binom{2n}{2k}x^{2k}y^{2n-2k}\right)+\left(\sum_{k=0}^{n-1}\binom{2n}{2k+1}x^{2k+1}y^{2n-(2k+1)}\right).$$ Can you put the pieces together and fill in the omitted steps/justifications?
