- #1
Rulonegger
- 16
- 0
Homework Statement
I just have to prove the well known identities:
[tex]\cos(\alpha + \beta)=\cos(\alpha)\cos(\beta)-\sin(\alpha)\sin(\beta)[/tex]
[tex]\sin(\alpha + \beta)=\sin(\alpha)\cos(\beta)+\cos(\alpha)\sin( \beta)[/tex]
But the thing is that I've to use the Taylor power series for the sine and cosine functions to prove them.
Homework 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...
[tex](x+y)^{n}=\sum_{k=0}^{n}{{n \choose k}x^{n-k}y^{k}}[/tex]
[tex](\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}}}[/tex]
The Attempt at a Solution
I've tried to expand the [itex]\cos(x+y)[/itex] function using the taylor series, and make use of the binomial theorem, getting this:
[tex]\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}}}[/tex]
But the expresion using the same formulas for the sine and cosine fo x and y and working the product is
[tex]\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}})[/tex]
And using the Cauchy product i get
[tex]\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}}}[/tex]
So, all I've to prove is that
[tex]\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}}}[/tex]
Any ideas?