The Double Angle Formula

 

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 \(\alpha=\beta=x, \;\;\;\; \sin(2x)=2\sin(x)\cos(x)\).

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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 \[ \sin(2x)=2x\prod_{n=1}^{\infty}\left ( 1-\frac{4 \cdot x^2}{\pi^2 n^2} \right ) =2\cdot x\prod_{n=\mathrm{even}\geq1}^{\infty}\left ( 1-\frac{4 \cdot x^2}{\pi^2 n^2} \right ) \cdot \prod_{n=\mathrm{odd}\geq1}^{\infty}\left ( 1-\frac{4 \cdot x^2}{\pi^2 n^2} \right ) \] \[ \sin(2x) =2\cdot x\prod_{n=1}^{\infty}\left ( 1-\frac{x^2}{\pi^2 n^2} \right ) \cdot \prod_{n=1}^{\infty}\left ( 1-\frac{x^2}{\pi^2 (n-1/2)^2} \right ) \] \[ \sin(2x)=2\cdot \sin(x) \cdot \cos(x) \]

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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: \( 2\sin(y)\cos(y)=\sin(2y) \)

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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 \(\cos(2x)=\cos^2(x)-\sin^2(x)\)

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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: \[ \sin\left ( \frac{y}{2} \right )=\sqrt{\frac{1-\cos(y)}{2}} \] \[ \cos\left ( \frac{y}{2} \right )=\sqrt{\frac{1+\cos(y)}{2}} \]

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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.

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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 \[ \pi/3=\underset{n\rightarrow \infty}{\lim} 2^n\sqrt{2-\underset{n\;\, \mathrm{radicals}}{\underbrace{\sqrt{2+\sqrt{2+...\sqrt{2+1}}}}}} \] \[ \pi/2=\underset{n\rightarrow \infty}{\lim} 2^n\sqrt{2-\underset{n\;\, \mathrm{radicals}}{\underbrace{\sqrt{2+\sqrt{2+...\sqrt{2}}}}}} \]

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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 \[ \frac{1}{\pi}=\sum_{k=2}^{\infty}\frac{1}{2^k}\tan \left ( \frac{\pi}{2^k} \right ) \]

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