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.
A Nested Radical 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
\[
I=\int_{0}^{\pi/2}\ln\left (\sin(x) \right )dx=-\frac{\pi}{2}\ln(2)
\]
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