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...
 A: 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
$$
\sin(x+y) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} \sin  y
+\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\cos y
= \cos x \sin y + \sin x \cos y.
$$
A: 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?
A: The functions $\sin$ and $\cos$, defined by the Taylor series have the following properties:
$$\begin{cases}
\sin'=\cos, & \sin 0=0\\
\cos'=-\sin, & \cos 0=1
\end{cases}$$
These properties are easy consequences of the definitions as power series, that can be differentiated term by term; moreover we know that the convergence radius (which is infinite, for $\sin$ and $\cos$) doesn't change when differentiating.
If we now consider the function
$$h(x)=(\sin x)^2 + (\cos x)^2$$
we have, by the chain rule,
$$
h'=2\sin\cdot\sin'+2\cos\cdot\cos'=2\sin\cdot\cos-2\cos\cdot\sin=0
$$
so $h$ is constant. Since $h(0)=0^2+1^2=1$, we have proved that, for all $x$,
$$
(\sin x)^2+(\cos x)^2=1
$$
How do you get $\cos(x+y)=\cos x\cos y-\sin x\sin y$? Just observe that when $f$ and $g$ are functions defined and differentiable over the whole real line, such that
$$f'=g,\qquad g'=-f,$$
these functions are uniquely determined as linear combinations of $\sin$ and $\cos$ by their values at $0$.
(Drawn from Lang's Introduction to mathematical analysis.)
A: I like egreg's answer, and I pick up where he left off. Fix $y \in \mathbb{R}$. Define
$$F(x) = \sin(x+y) - \sin(x) \cos(y) - \cos(x) \sin(y)$$
We want to show that $F(x) = F'(x) = 0$ for all real numbers $x$. Now
\begin{align*}
& F'(x) = \cos(x+y) - \cos(x) \cos(y) + \sin(x) \sin(y) \\
\implies & F''(x) = -\sin(x+y) + \sin(x) \cos(y) + \cos(x) \sin(y) \\
\implies &  F(x) + F''(x) = 0 \\
\implies & 2F'(x) F(x) + 2 F'(x) F''(x) = \frac{d}{d\ x} [(F(x))^2 + (F'(x))^2] = 0
\end{align*}
It is clear that $F$ is infinitely differentiable, so the Mean Value Theorem grants us that $H(x) = (F(x))^2 + (F'(x))^2$ is a constant function. Since we know from their power series definitions that $\sin(0) = 0$ and $\cos(0) = 1$, we have $F(0) = F'(0) = 0$, hence $H(x) = H(0) = 0$ for all $x \in \mathbb{R}$. Now this implies that $F(x) = F'(x) = 0$ on $\mathbb{R}$ as well, and so we now know what $\sin(x+y)$ and $\cos(x+y)$ are (just look at the expressions for $F$ and $F'$ above).
