Proof the identities of the sine and cosine sum of angles

1. Feb 21, 2013

Rulonegger

1. The problem statement, all variables and given/known data
I just have to prove the well known identities:
$$\cos(\alpha + \beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)$$
$$\sin(\alpha + \beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin( \beta)$$
But the thing is that i've to use the Taylor power series for the sine and cosine functions to prove them.

2. Relevant equations
They say that the binomial theorem and the Cauchy product would help, but i can't do it, maybe cause i'm a rookie working with infinite sums...
$$(x+y)^{n}=\sum_{k=0}^{n}{{n \choose k}x^{n-k}y^{k}}$$
$$(\sum_{n=0}^{\infty}{a_{n}})(\sum_{n=0}^{\infty}{b_{n}})=\sum_{n=0}^{\infty}{\sum_{k=0}^{n}{a_{n}b_{n-k}}}$$

3. The attempt at a solution
I've tried to expand the $\cos(x+y)$ function using the taylor series, and make use of the binomial theorem, getting this:
$$\cos(x+y)=\sum_{n=0}^{\infty}{\frac{(-1)^n}{(2n)!}(x+y)^{2n}}=\sum_{n=0}^{\infty}{\frac{(-1)^n}{(2n)!}\sum_{k=0}^{n}{{2n \choose k}x^{k}y^{2n-k}}}$$
But the expresion using the same formulas for the sine and cosine fo x and y and working the product is
$$\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)=(\sum_{n=0}^{\infty}{\frac{(-1)^n}{(2n)!}x^{2n}})(\sum_{n=0}^{\infty}{\frac{(-1)^n}{(2n)!}y^{2n}})-(\sum_{n=0}^{\infty}{\frac{(-1)^n}{(2n+1)!}x^{2n+1}})(\sum_{n=0}^{\infty}{\frac{(-1)^n}{(2n+1)!}y^{2n+1}})$$
And using the Cauchy product i get
$$\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)=\sum_{n=0}^{\infty}{\sum_{k=0}^{n}{\frac{(-1)^n}{(2n)!}\frac{(-1)^{n-k}}{(2(n-k))!}x^{2n}y^{2(n-k)}}}-\sum_{n=0}^{\infty}{\sum_{k=0}^{n}{\frac{(-1)^n}{(2n+1)!}\frac{(-1)^{n-k}}{(2(n-k)+1)!}x^{2n+1}y^{2(n-k)+1}}}$$
So, all i've to prove is that
$$\sum_{n=0}^{\infty}{\frac{(-1)^n}{(2n)!}\sum_{k=0}^{n}{{2n \choose k}x^{k}y^{2n-k}}}=\sum_{n=0}^{\infty}{\sum_{k=0}^{n}{\frac{(-1)^n}{(2n)!}\frac{(-1)^{n-k}}{(2(n-k))!}x^{2n}y^{2(n-k)}}}-\sum_{n=0}^{\infty}{\sum_{k=0}^{n}{\frac{(-1)^n}{(2n+1)!}\frac{(-1)^{n-k}}{(2(n-k)+1)!}x^{2n+1}y^{2(n-k)+1}}}$$
Any ideas?

2. Feb 21, 2013

jbunniii

I haven't read through the whole thing yet, but the first thing I noticed is that the rightmost sum should be $\sum_{k=0}^{2n}$, not $\sum_{k=0}^{n}$.

3. Feb 21, 2013

jbunniii

Another thing is that your Cauchy product expression is wrong:
The subscript on the $a$ should be $k$, not $n$:
$$\sum_{n=0}^{\infty}{\sum_{k=0}^{n}{a_{k}b_{ n-k}}}$$
Both of these errors show up in all of your expressions. After you make the corrections, if you are still stuck, post your new equations and we'll see where to go from there.

4. Feb 22, 2013

HallsofIvy

Staff Emeritus
If it were me, I would use a trick. Those power series are convergent for any x and so "uniformly convergent" on any closed and bounded interval. That means, in particular, that they are "term by term" differentiable at any x.

Then, by differentiating the power series, term by term, you arrive at (sin(x))'= cos(x) and (cos(x))'= -sin(x). Differentiating again, (sin(x))''= -sin(x) and (cos(x))''= -cos(x). That is, both y= sin(x) and y= cos(x) satisfy the linear differential equation y''= -y. For initial conditions, again from the power series, for y= cos(x), y(0)= 1, y'(0)= 0 while for y= sin(x), y(0)= 0, y'(0)= 1.

From that you can show that "The solution to the initial value problem, y''= -y, y(0)= A, y'(0)= B, is y(x)= Acos(x)+ Bsin(x)."

Now note that y(x)= sin(x+ a) satisfies y''= -y with y(0)= sin(a) and y'(0)= cos(a) while y(x)= cos(x+a) satisfies y''= -y with y(0)= cos(a) and y'(0)= -sin(a).

5. Feb 22, 2013

Rulonegger

Thanks jbunniii and HallsofIvy. I'm sorry, but I've just made a typing mistake, and i've just copied and pasted the first expression.
So, the correct expression to prove is
$$\sum_{n=0}^{\infty}{\frac{(-1)^{n}}{(2n)!}\sum_{k=0}^{2n}{{2n \choose k}x^{k}y^{2n-k}}}=\sum_{n=0}^{\infty}{\sum_{k=0}^{n}{\frac{(-1)^k}{(2k)!}\frac{(-1)^{n-k}}{(2(n-k))!}x^{2k}y^{2(n-k)}}}-\sum_{n=0}^{\infty}{\sum_{k=0}^{n}{\frac{(-1)^k}{(2k+1)!}\frac{(-1)^{n-k}}{(2(n-k)+1)!}x^{2k+1}y^{2(n-k)+1}}}$$
But i don't know how to combine the two terms in the right side of the equation, cause i'm not sure if it is correct to take both sums as a whole, because the two sums to be computed from n=0 to ∞. If not, what should i do?

And HallsofIvy, i think your way is good, but the professor says that we have to prove them with taylor series. Thanks a lot!

6. Feb 22, 2013

jbunniii

OK, let's work with the right hand side. The first term is
\begin{align} & \sum_{n=0}^{\infty}(-1)^n \sum_{k=0}^{n} \frac{1}{(2k)!}\frac{1}{(2(n-k))!}x^{2k}y^{2(n-k)} \\ &= \sum_{n=0}^{\infty}(-1)^n \sum_{j=0,\textrm{ } j\textrm{ even }}^{2n} \frac{1}{j!} \frac{1}{(2n-j)!} x^j y^{2n-j} \end{align}
and similarly, the second term becomes
\begin{align} & -\sum_{n=0}^{\infty}{\sum_{k=0}^{n}{\frac{(-1)^k}{(2k+1)!}\frac{(-1)^{n-k}}{(2(n-k)+1)!}x^{2k+1}y^{2(n-k)+1}}} \\ &= -\sum_{n=0}^{\infty}(-1)^n \sum_{j=1,\textrm{ } j\textrm{ odd}}^{2n+1} \frac{1}{j!} \frac{1}{(2(n+1)-j)!} x^{j} y^{2(n+1)-j} \\ &= \sum_{n=1}^{\infty}(-1)^{n}\sum_{j=1,\textrm{ } j\textrm{ odd}}^{2n-1} \frac{1}{j!} \frac{1}{(2n-j)!} x^{j} y^{2n-j} \\ \end{align}
(Hopefully I didn't make any index errors - please check my work.) I think you should be able to simplify this to match the left hand side.

7. Feb 23, 2013

Rulonegger

Thanks jbunniii, you didn't make any error, and you left the problem almost done.

8. Feb 23, 2013

joshmccraney

why not use taylor series of sine/cosine to prove $$e^{ix}=cosx+isinx$$ then let $x=\alpha + \beta$ giving us $$e^{i(\alpha + \beta)}=cos(\alpha + \beta)+isin(\alpha + \beta)$$ knowing $e^{i(\alpha + \beta)}=e^{\alpha}e^{\beta}$ allows us to write: $$cos(\alpha + \beta)+isin(\alpha + \beta)=(cos{\alpha}+isin{\alpha})(cos{\beta}+isin{\beta})$$ now simply perform the multiplication and equate each side of the equation, letting the real and imaginary sides equal. for the subtraction let $\beta=-\beta$ and use properties of sine and cosine as odd/even functions, which you can do since you know the series expansion!