# Finding $\sin(x+y)=\sin x\cos y+\sin y\cos x$ with advanced and very advanced methods.I tried Cauchy Product, but I failed.

Cauchy Product:

Let $c_n$ be $$\left(\displaystyle\sum a_n\right)\left(\displaystyle\sum b_m\right)=\displaystyle\sum c_k\tag1$$

Then;

$$c_k=\sum_{m+n=k}a_n b_m=\displaystyle\sum_{n=0}^k a_n b_{k-n}\tag2$$

We know that;

$$\sin x=\sum_{i=0}^\infty\dfrac{(-1)^i(x)^{2i+1}}{(2i+1)!},\quad\cos y=\sum_{i=0}^\infty\dfrac{(-1)^i(y)^{2i}}{(2i)!}\tag3$$

Let $a_n,b_m$ be

$$a_n=\dfrac{(-1)^n(x)^{2n+1}}{(2n+1)!},\quad b_m=\dfrac{(-1)^m(y)^{2m}}{(2m)!}\\\text{ }\\\sum (sincos)_k=\sin x\cos y \\ \text{ }\\ \sum(cossin)_k=\cos x\sin y\tag4$$

From $(2)$ we know that we can this;

$$(sincos)_k=\displaystyle\sum_{n=0}^k a_n b_{k-n}=\sum_{n=0}^k\dfrac{(-1)^n(x)^{2n+1}}{(2n+1)!}\dfrac{(-1)^{k-n}(y)^{2(k-n)}}{(2(k-n))!}\tag5$$

And

$$(cossin)_k=\displaystyle\sum_{n=0}^k a_n b_{k-n}=\sum_{n=0}^k\dfrac{(-1)^n(y)^{2n+1}}{(2n+1)!}\dfrac{(-1)^{k-n}(x)^{2(k-n)}}{(2(k-n))!}\tag6$$

And since our definitions $\sin x\cos y+\sin y\cos x=I$ equals to ;

$$I=\displaystyle\sum_{k=0}^\infty\left[\sum_{n=0}^k(-1)^n\left(\dfrac{(x)^{2n+1}(y)^{2(k-n)}+(y)^{2n+1}(x)^{2(k-n)}}{(2n+1)!(2(k-n))!}\right)\right]\tag7$$

,

$$I=\displaystyle\sum_{k=0}^\infty\left[\sum_{n=0}^k(-1)^n\left(\dfrac{\frac{x^{2n+1}y^{2k}}{y^{2n}}+\frac{y^{2n+1}x^{2k}}{x^{2n}}}{(2n+1)!(2(k-n))!}\right)\right]\\=\displaystyle\sum_{k=0}^\infty\left[\sum_{n=0}^k(-1)^n\left(\dfrac{x^{4n+1}y^{2k}+y^{4n+1}x^{2k}}{(xy)^{2n}(2n+1)!(2(k-n))!}\right)\right]\tag8$$

I couldn't go any further,please tell whether I make mistake or not?How we can prove more quickly and are there any methods to prove this equation?How we can use this "Cauchy Product" method correctly and accurately?

• you can try to find when y is fixed, the Fourier serie for sin(x+y). – zwim Apr 3 '17 at 23:15
• Note that from (7) you can change the inner sum to $\sum_{m=0}^{2k+1}(-1)^n\frac{x^my^{2k+1-m}}{(m)!(2k+1-m)!}$ – Mosquite Apr 3 '17 at 23:19
• @Mosquite can you please gıve more ınformatıon, I couldn't sımply change into this term – user2312512851 Apr 7 '17 at 22:48
• and can you gıve me furtherreading about, how we can change index of sums ? – user2312512851 Apr 7 '17 at 22:48
• ? please, I couldn't do. – user2312512851 Apr 13 '17 at 19:30

By Euler's formula, $$e^{i(\alpha+\beta)}=\cos(\alpha+\beta)+i\sin(\alpha+\beta)$$ On the other hand, $$e^{i(\alpha+\beta)}=e^{i\alpha}e^{i\beta}=[\cos(\alpha)+i\sin(\alpha)][\cos(\beta)+i\sin(\beta)]$$$$=\cos(\alpha)\cos(\beta)+i\cos(\beta)sin(\alpha)+i\cos(\alpha)\sin(\beta)-\sin(\alpha)\sin(\beta)$$ Equate real and imaginary parts. This also yields the corresponding identity for cosine. Incidentally, many handy trig identities follow from Euler's formula.