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[parent] Prosthaphaeresis formulas (Proof)

The Prosthaphaeresis formulas convert sums of sines or cosines to products of them:


$\displaystyle \sin A + \sin B$ $\displaystyle =$ $\displaystyle 2 \sin \left( \frac{A+B}{2} \right) \cos \left (\frac{A-B}{2} \right)$  
$\displaystyle \sin A - \sin B$ $\displaystyle =$ $\displaystyle 2 \sin \left( \frac{A-B}{2} \right) \cos \left (\frac{A+B}{2} \right)$  
$\displaystyle \cos A + \cos B$ $\displaystyle =$ $\displaystyle 2 \cos \left( \frac{A+B}{2} \right) \cos \left (\frac{A-B}{2} \right)$  
$\displaystyle \cos A - \cos B$ $\displaystyle =$ $\displaystyle -2 \sin \left( \frac{A+B}{2} \right) \sin \left (\frac{A-B}{2} \right)$  

We prove the first two using the sine of a sum and sine of a difference formulas:


$\displaystyle \sin (X+Y)$ $\displaystyle =$ $\displaystyle \sin X \cos Y + \cos X \sin Y$  
$\displaystyle \sin (X-Y)$ $\displaystyle =$ $\displaystyle \sin X \cos Y - \cos X \sin Y$  

Adding or subtracting the two equations yields

$\displaystyle \sin (X+Y) + \sin (X-Y)$ $\displaystyle =$ $\displaystyle 2 \sin X \cos Y$  
$\displaystyle \sin (X+Y) - \sin (X-Y)$ $\displaystyle =$ $\displaystyle 2 \sin Y \cos X$  

If we let $ X = \frac{A+B}{2}$ and $ Y = \frac{A-B}{2}$, then $ X+Y = \frac{2A}{2} = A$ and $ X-Y = \frac{2B}{2} = B$, and the last two equations become


$\displaystyle \sin A + \sin B$ $\displaystyle =$ $\displaystyle 2 \sin \left( \frac{A+B}{2} \right) \cos \left (\frac{A-B}{2} \right)$  
$\displaystyle \sin A - \sin B$ $\displaystyle =$ $\displaystyle 2 \sin \left( \frac{A-B}{2} \right) \cos \left (\frac{A+B}{2} \right)$  

as desired.

The last two can be proven similarly, this time using the cosine of a sum and cosine of a difference formulas:


$\displaystyle \cos (X+Y)$ $\displaystyle =$ $\displaystyle \cos X \cos Y - \sin X \sin Y$  
$\displaystyle \cos (X-Y)$ $\displaystyle =$ $\displaystyle \cos X \cos Y + \sin X \sin Y$  

Adding or subtracting the two equations yields

$\displaystyle \cos (X+Y) + \cos(X-Y)$ $\displaystyle =$ $\displaystyle 2 \cos X \cos Y$  
$\displaystyle \cos (X+Y) - \cos(X-Y)$ $\displaystyle =$ $\displaystyle - 2 \sin Y \sin X$  

Again, if we let $ X = \frac{A+B}{2}$ and $ Y = \frac{A-B}{2}$, then $ X+Y = \frac{2A}{2} = A$ and $ X-Y = \frac{2B}{2} = B$, and the last two equations become


$\displaystyle \cos A + \cos B$ $\displaystyle =$ $\displaystyle 2 \cos \left( \frac{A+B}{2} \right) \cos \left (\frac{A-B}{2} \right)$  
$\displaystyle \cos A - \cos B$ $\displaystyle =$ $\displaystyle - 2 \sin \left( \frac{A-B}{2} \right) \sin \left (\frac{A+B}{2} \right)$  

as desired.

Notes

'Prosthaphaeresis' comes from the Greek: “prosthesi” = addition + “afairo” = subtraction.

The Prosthaphaeresis formula $ \cos x \cos y = \frac{\cos (x+y) + \cos (x-y)}{2}$ was used by scientists to transform multiplication into addition. For example, to calculate the product $ ab$, where $ 0 < a, b < 1$ (for $ a$ and $ b$ outside of this range, it is a simple matter to multiply or divide by a factor of 10 and divide or multiply this back in later), one would let $ \cos x = a$ and $ \cos y = b$. Using a table of cosines, one could then find an approximate value for $ x$ and $ y$, then find $ x+y$ and $ x-y$, and look up the cosines of the resulting two quantities (that is, $ \cos (x+y)$ and $ \cos (x-y)$). The average of these numbers is the desired product $ ab$. This technique was used by Tycho Brahe to perform astronomical calculations.



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Other names:  Simpson's formulas

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derivatives of sine and cosine (Derivation) by Wkbj79
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Cross-references: numbers, average, factor, divide, simple, range, calculate, multiplication, Transform, subtraction, addition, equations, difference, products, cosines, sines, sums
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This is version 4 of Prosthaphaeresis formulas, born on 2004-09-01, modified 2004-10-12.
Object id is 6121, canonical name is ProsthaphaeresisFormulas.
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AMS MSC26A09 (Real functions :: Functions of one variable :: Elementary functions)

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