## Deriving the formula: $\sin(2x)=2\sin(x)\cos(x)$

### Way 1: From Geometry

$RB=QA \;\;\;\;\;\;\;\;\;\; RQ=BA$ $\frac{RQ}{PQ}=\frac{QA}{OQ}=\sin(\alpha) \;\;\;\;\;\;\;\; \frac{PR}{PQ}=\frac{OA}{OQ}=\cos(\alpha)$ $\frac{PQ}{OP}=\sin(\beta) \;\;\;\;\;\;\;\; \frac{OQ}{OP}=\cos(\beta)$ $\frac{PB}{OP}=\sin(\alpha+\beta) \;\;\;\;\;\;\;\; \frac{OB}{OP}=\cos(\alpha+\beta)$ $PB=PR+RB=\frac{OA}{OQ}PQ+QA$ $\frac{PB}{OP}=\frac{OA}{OQ}\frac{PQ}{OP}+\frac{QA}{OP}=\frac{OA}{OQ}\frac{PQ}{OP}+\frac{QA}{OQ}\frac{OQ}{OP}$ $\sin(\alpha+\beta)=\cos(\alpha)\sin(\beta)+\sin(\alpha)\cos(\beta)$ Particularly, if $\alpha=\beta=x, \;\;\;\; \sin(2x)=2\sin(x)\cos(x)$.

### Way 2: From the Product Formula

Recall from this post that the product formulas for sine and cosine are, respectively: $\sin(x)=x\prod_{n=1}^{\infty}\left ( 1-\frac{x^2}{\pi^2 n^2} \right )$ And $\cos(x)=\prod_{n=1}^{\infty} \left (1-\frac{x^2}{\pi^2 (n-1/2)^2 } \right )$ Thus $\sin(2x)=2x\prod_{n=1}^{\infty}\left ( 1-\frac{4 \cdot x^2}{\pi^2 n^2} \right ) =2\cdot x\prod_{n=\mathrm{even}\geq1}^{\infty}\left ( 1-\frac{4 \cdot x^2}{\pi^2 n^2} \right ) \cdot \prod_{n=\mathrm{odd}\geq1}^{\infty}\left ( 1-\frac{4 \cdot x^2}{\pi^2 n^2} \right )$ $\sin(2x) =2\cdot x\prod_{n=1}^{\infty}\left ( 1-\frac{x^2}{\pi^2 n^2} \right ) \cdot \prod_{n=1}^{\infty}\left ( 1-\frac{x^2}{\pi^2 (n-1/2)^2} \right )$ $\sin(2x)=2\cdot \sin(x) \cdot \cos(x)$

### Way 3: From the Taylor Series

The Taylor series for sine and cosine can be construed as, respectively: $\frac{\sin(\sqrt{x})}{\sqrt{x}}=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}x^k$ $\cos(\sqrt{x})=\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k)!}x^k$ Thus $\frac{\sin(\sqrt{x})\cos(\sqrt{x})}{\sqrt{x}}=\sum_{j=0}^{\infty}\frac{(-1)^j}{(2j+1)!}x^j \sum_{k=0}^{\infty}\frac{(-1)^k}{(2k)!}x^k$ Using a Cauchy product, we find: $\frac{\sin(\sqrt{x})\cos(\sqrt{x})}{\sqrt{x}}=\sum_{j=0}^{\infty}c_j x^j$ Where $c_m=\sum_{n=0}^{m} \frac{(-1)^n}{(2n+1)!}\frac{(-1)^{m-n}}{(2(m-n))!} =\frac{(-1)^m}{(2m+1)!}\sum_{n=0}^{m} \binom{2m+1}{2n+1} =\frac{(-1)^m}{(2m+1)!}\sum_{n=0}^{m} \binom{2m}{2n+1}+\binom{2m}{2n}$ $c_m=\frac{(-1)^m}{(2m+1)!}\sum_{n=0}^{2m} \binom{2m}{n}=\frac{(-1)^m}{(2m+1)!}2^{2m}$ And thus $\frac{\sin(\sqrt{x})\cos(\sqrt{x})}{\sqrt{x}}=\sum_{m=0}^{\infty}\frac{(-1)^m}{(2m+1)!}(4x)^m=\frac{\sin(\sqrt{4x})}{\sqrt{4x}}=\frac{\sin(2\sqrt{x})}{2\sqrt{x}}$ Substituting $x=y^2$ and rearranging, we find: $2\sin(y)\cos(y)=\sin(2y)$

### Way 4: From Euler's Formula

Euler's formula is: $e^{ix}=\cos(x)+i\sin(x)$ Thus $e^{i2x}=\cos(2x)+i\sin(2x)=\left ( e^{ix} \right)^2=\left (\cos(x)+i\sin(x) \right )^2$ $e^{i2x}=\left [\cos^2(x)-\sin^2(x) \right ]+i\left [ 2\sin(x)\cos(x) \right ]$ Thus, by equating real and imaginary parts, $\sin(2x)=2\sin(x)\cos(x)$ and $\cos(2x)=\cos^2(x)-\sin^2(x)$

## The Half-Angle Formulas

We find from the last demonstration $\cos(2x)=\cos^2(x)-\sin^2(x)=2\cos^2(x)-1=1-2\sin^2(x)$ Substituting $2x=y$ and solving, we find: $\sin\left ( \frac{y}{2} \right )=\sqrt{\frac{1-\cos(y)}{2}}$ $\cos\left ( \frac{y}{2} \right )=\sqrt{\frac{1+\cos(y)}{2}}$

## An Infinite Product Formula

We can write the double-angle formula as $\sin(x)=2\sin\left ( \frac{x}{2} \right )\cos\left ( \frac{x}{2} \right )$ Iterating this, we then have $\sin(x)=2^n\sin\left ( \frac{x}{2^n} \right ) \prod_{k=1}^{n}\cos\left ( \frac{x}{2^k} \right )$ However, in the limit as n gets large, $2^n\sin\left ( \frac{x}{2^n} \right )\rightarrow x$. Thus, letting n go to infinity, we have $\sin(x)=x \prod_{k=1}^{n}\cos\left ( \frac{x}{2^k} \right )$ A simple theorem of this general result is $\frac{\pi}{2}=\frac{1}{\cos(\tfrac{\pi}{4})\cos(\tfrac{\pi}{8})\cos(\tfrac{\pi}{16})\cdots } =\frac{1}{\sqrt{\tfrac{1}{2}}\sqrt{\tfrac{1}{2}+\tfrac{1}{2}\sqrt{\tfrac{1}{2}}}\sqrt{\tfrac{1}{2}+\tfrac{1}{2}\sqrt{\tfrac{1}{2}+\tfrac{1}{2}\sqrt{\tfrac{1}{2}}}}\cdots }=\frac{2}{\sqrt{2}}\frac{2}{\sqrt{2+\sqrt{2}}}\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}\cdots$ This is known as Viète's formula.

We note that $2\cos(x/2)=\sqrt{2+2\cos(x)}$ Thus, by iterating, we find $2\cos(x/2^n)=\underset{n\;\, \mathrm{radicals}}{\underbrace{\sqrt{2+\sqrt{2+...\sqrt{2+2\cos(x)}}}}}$ Thus $2\sin(x/2^{n+1})=\sqrt{2-\underset{n\;\, \mathrm{radicals}}{\underbrace{\sqrt{2+\sqrt{2+...\sqrt{2+2\cos(x)}}}}}}$ And we can thus conclude that $x=\underset{n\rightarrow \infty}{\lim} 2^n\sqrt{2-\underset{n\;\, \mathrm{radicals}}{\underbrace{\sqrt{2+\sqrt{2+...\sqrt{2+2\cos(x)}}}}}}$ For example $\pi/3=\underset{n\rightarrow \infty}{\lim} 2^n\sqrt{2-\underset{n\;\, \mathrm{radicals}}{\underbrace{\sqrt{2+\sqrt{2+...\sqrt{2+1}}}}}}$ $\pi/2=\underset{n\rightarrow \infty}{\lim} 2^n\sqrt{2-\underset{n\;\, \mathrm{radicals}}{\underbrace{\sqrt{2+\sqrt{2+...\sqrt{2}}}}}}$

## An Infinite Series

Above, we derived $\sin(x)=x \prod_{k=1}^{n}\cos\left ( \frac{x}{2^k} \right )$ Taking the log of both sides and differentiating $\frac{\mathrm{d} }{\mathrm{d} x}\ln\left (\sin(x) \right )=\frac{\mathrm{d} }{\mathrm{d} x}\ln\left (x \prod_{k=1}^{n}\cos\left ( \frac{x}{2^k} \right ) \right )$ $\cot(x)=\frac{1}{x}-\sum_{k=1}^{\infty}\frac{1}{2^k}\tan \left ( \frac{x}{2^k} \right )$ $\\ \frac{1}{x}-\cot(x)=\sum_{k=1}^{\infty}\frac{1}{2^k}\tan \left ( \frac{x}{2^k} \right )$ From this we can easily derive $\frac{1}{\pi}=\sum_{k=2}^{\infty}\frac{1}{2^k}\tan \left ( \frac{\pi}{2^k} \right )$

## A Definite Integral

Let $I=\int_{0}^{\pi/2}\ln\left ( \sin(x) \right )dx =\int_{\pi/2}^{\pi}\ln\left ( \sin(x) \right )dx =\int_{0}^{\pi/2}\ln\left ( \cos(x) \right )dx$ Then $2I=\int_{0}^{\pi}\ln\left ( \sin(x) \right )dx =2\int_{0}^{\pi/2}\ln\left ( \sin(x) \right )dx =\int_{0}^{\pi/2}\ln\left ( \sin(x) \right )+\ln\left ( \cos(x) \right )dx$ $2I=\int_{0}^{\pi/2}\ln\left ( \sin(x) \cos(x) \right )dx=\int_{0}^{\pi/2}\ln\left (\tfrac{1}{2} \sin(2x) \right )dx=-\frac{\pi}{2}\ln(2)+\int_{0}^{\pi/2}\ln\left (\sin(2x) \right )dx$ By the substitution $u=2x$, we then have $2I=-\frac{\pi}{2}\ln(2)+\tfrac{1}{2}\int_{0}^{\pi}\ln\left (\sin(u) \right )du=-\frac{\pi}{2}\ln(2)+I$ Therefore $I=\int_{0}^{\pi/2}\ln\left (\sin(x) \right )dx=-\frac{\pi}{2}\ln(2)$