# Showing that $\sin(2x)=2\sin x\cos x$ by multiplying power series

okay so

$$\sin(x)= \sum_0^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}\\ \cos(x)= 1+\sum_0^\infty (-1)^n \frac{x^{2n}}{(2n)!}$$

I am asked to show directly by multiplying power series. Tried finding the Cauchy product by setting $a_n=(-1)^n\frac{x^{2n+1}}{(2n+1)!}$ and $b_n=(-1)^n \frac{x^{2n}}{(2n)!}$

$$\sum_0^\infty c_n ~\text{ where }~ c_n=a_0b_n+a_1b_{n-1}+ \cdots +a_nb_0$$

But couldnt work it out any help?

• What expression do you get ? – Yves Daoust Mar 1 '17 at 20:43
• It helps if you write down the series for $\sin (2x)$ to see where you want to end up. – Daniel Fischer Mar 1 '17 at 20:44
• Be careful about indexing. You're using $a_n$ and $b_n$ to indicate every other coefficient in the power series, which might make it more confusing to perform power series products. Instead, treat the coefficient at zero for every other $n$. – Christopher A. Wong Mar 1 '17 at 20:44
• The second series is wrong: either you don't have $1+{}$ or start the summation at $1$ (and not $0$). – egreg Mar 1 '17 at 21:05

## 3 Answers

$$2\sin x \cos x =2 \left(\sum_{k=0}^\infty (-1)^k\frac{x^{2k+1}}{(2k+1)!}\right) \left(\sum_{k=0}^\infty (-1)^k\frac{x^{2k}}{(2k)!}\right)$$

$$= 2\sum_{k=0}^\infty \sum_{i=0}^k (-1)^i \frac{x^{2i+1}}{(2i+1)!} (-1)^{k-i} \frac{x^{2(k-i)}}{(2k-2i)!} = \sum_{k=0}^\infty (-1)^k x^{2k+1} \sum_{i=0}^k\frac{2}{(2i+1)! (2k-2i)!}$$

$$= \sum_{k=0}^\infty (-1)^k \frac{(2x)^{2k+1}}{(2k+1)!} = \sin(2x),$$

since

$$\sum_{i=0}^k \frac{2}{(2i+1)!(2k-2i)!} = \frac{2^{2k+1}}{(2k+1)!}.$$

• Couldn't understand the last row – user346936 Mar 1 '17 at 21:28
• It follows from the formula $$\sum_{i=0}^k \frac{(2k+1)!}{(2i+1)!(2k-2i)!} = \sum_{i=0}^k \begin{pmatrix} 2k+1 \\ 2i+1 \end{pmatrix} = 2^{2k},$$ which can be proven by induction. – Stefano Mar 1 '17 at 21:29
• It had a typo. Corrected now! – Stefano Mar 1 '17 at 21:34

To see the details: $$2\sin x\cos x=2\left(\frac x{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}-\cdots \right)\left(\frac1{0!}-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\cdots \right)=\\ 2\left(\frac1{0!1!}\right)x-2\left(\frac{1}{0!3!}+\frac{1}{1!2!}\right)x^3+2\left(\frac1{0!5!}+\frac1{1!4!}+\frac1{2!3!}\right)x^5-\cdots= \\ \left(\frac1{0!1!}+\frac1{1!0!}\right)x-\left(\frac{1}{0!3!}+\frac{1}{1!2!}+\frac1{2!1!}+\frac{1}{3!0!}\right)x^3+\left(\frac1{0!5!}+\frac1{1!4!}+\frac1{2!3!}+\frac1{3!2!}+\frac1{4!1!}+\frac1{5!0!}\right)x^5-\cdots= \\ \sum_{i=0}^{\infty}\sum_{j=0}^{2i+1}\frac{(-1)^i}{j!(2i+1-j)!}x^{2i+1}=\\ \sum_{i=0}^{\infty}(-1)^ix^{2i+1}\sum_{j=0}^{2i+1}\frac{1}{j!(2i+1-j)!}=\\ \sum_{i=0}^{\infty}(-1)^ix^{2i+1}\sum_{j=0}^{2i+1}\frac{{2i+1\choose j}}{(2i+1)!}=\\ \sum_{i=0}^{\infty}\frac{(-1)^ix^{2i+1}}{(2i+1)!}\cdot\sum_{j=0}^{2i+1}{2i+1\choose j}=\\ \sum_{i=0}^{\infty}\frac{(-1)^ix^{2i+1}}{(2i+1)!}\cdot 2^{2i+1}=\\ \sum_{i=0}^{\infty}\frac{(-1)^i(2x)^{2i+1}}{(2i+1)!}=\\ \sin (2x).$$

$$=\dfrac{1}{(2k+1)!}\displaystyle\sum_{i=0}^k \frac{(2k+1)!}{(2i+1)!(2k)(2i)!}$$

$$=\dfrac{1}{(2k+1)!} \displaystyle\sum_{i=0}^k \begin{pmatrix} 2k+1 \\ 2i+1 \end{pmatrix}$$

$$=\dfrac{1}{(2k+1)!}\begin{pmatrix} 2k+1 \\ 1 \end{pmatrix} + \begin{pmatrix} 2k+1 \\ 3 \end{pmatrix} +....$$

$$=\dfrac{1}{(2k+1)!}(2^{2k+1-1})$$

$$=\dfrac{1}{(2k+1)!}2^{2k}$$

$$=2\displaystyle\sum_{k=0}^\infty (-1)^k x^{2k+1}\dfrac{1}{(2k+1)!}2^{2k}$$

$$=\displaystyle\sum_{k=0}^\infty (-1)^k\dfrac{(2x)^{2k+1}}{(2k+1)!}$$

$$=\sin2x$$