Prosthaphaeresis formulas
Prosthaphaeresis formulas allow us to transform the sum or difference of two trigonometric functions into a product of trigonometric functions.
Sine prosthaphaeresis formulas $$ \sin a + \sin b = 2 \cdot \sin \frac{a+b}{2} \cdot \cos \frac{a-b}{2} $$ $$ \sin a - \sin b = 2 \cdot \sin \frac{a-b}{2} \cdot \cos \frac{a+b}{2} $$ Cosine prosthaphaeresis formulas $$ \cos a + \cos b = 2 \cdot \cos \frac{a+b}{2} \cdot \cos \frac{a-b}{2}$$ $$ \cos a - \cos b = -2 \cdot \sin \frac{a+b}{2} \cdot \sin \frac{a-b}{2}$$
A practical example
Let’s use the sum of the sine of 60° (π/3) and the sine of 30° (π/6) as an example.
$$ \sin 60° + \sin 30° $$
Since the sine of 60° is √3/2 and the sine of 30° is 1/2, the correct result is
$$ \sin 60° + \sin 30° = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{1 + \sqrt{3}}{2} $$
Now, let’s verify if we can get the same result using the prosthaphaeresis formulas.
We’ll apply the sine addition prosthaphaeresis formula by assigning a=60° and b=30°
$$ \sin a + \sin b = 2 \cdot \sin \frac{a+b}{2} \cdot \cos \frac{a-b}{2} $$
$$ \sin 60° + \sin 30° = 2 \cdot \sin \frac{60°+30°}{2} \cdot \cos \frac{60-30°}{2} $$
$$ \sin 60° + \sin 30° = 2 \cdot \sin \frac{90°}{2} \cdot \cos \frac{30°}{2} $$
$$ \sin 60° + \sin 30° = 2 \cdot \sin 45° \cdot \cos 15° $$
The sine of 45° is √2/2.
$$ \sin 60° + \sin 30° = 2 \cdot \frac{\sqrt{2}}{2} \cdot \cos 15° $$
The cosine of 15° is (√6 + √2) / 4.
$$ \sin 60° + \sin 30° = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}+\sqrt{2}}{4} $$
$$ \sin 60° + \sin 30° = \sqrt{2} \cdot \frac{\sqrt{6}+\sqrt{2}}{4} $$
$$ \sin 60° + \sin 30° = \frac{\sqrt{2} \cdot \sqrt{6}+ \sqrt{2} \cdot\sqrt{2}}{4} $$
$$ \sin 60° + \sin 30° = \frac{ \sqrt{2 \cdot 6}+ 2}{4} $$
$$ \sin 60° + \sin 30° = \frac{ \sqrt{12}+ 2}{4} $$
Now, divide both the numerator and the denominator by two using the properties of fractions, and simplify.
$$ \sin 60° + \sin 30° = \frac{ \frac{1}{2} \cdot ( \sqrt{12}+ 2)}{\frac{1}{2} \cdot 4} $$
$$ \sin 60° + \sin 30° = \frac{ \frac{1}{2} \cdot \sqrt{12}+ \frac{1}{2} \cdot 2}{2} $$
$$ \sin 60° + \sin 30° = \frac{ \sqrt{\frac{1}{4} \cdot 12}+ 1}{2} $$
$$ \sin 60° + \sin 30° = \frac{ \sqrt{3}+ 1}{2} $$
The final result matches the initial calculation.
The proof
The sine addition prosthaphaeresis formula
To prove the sine addition prosthaphaeresis formula, I use the sine addition and subtraction formulas
$$ \begin{cases} \sin (\alpha+\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \\ \sin (\alpha-\beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \end{cases} $$
We add the two formulas together:
$$ \sin (\alpha+\beta) + \sin (\alpha-\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta + \sin \alpha \cos \beta - \cos \alpha \sin \beta $$
Then we simplify:
$$ \sin (\alpha+\beta) + \sin (\alpha-\beta) = 2 \cdot \sin \alpha \cos \beta $$
Let’s assign a=α+β to represent the sum of the angles
$$ \sin a + \sin (\alpha-\beta) = 2 \cdot \sin \alpha \cos \beta $$
and b=α-β to represent the difference
$$ \sin a + \sin b = 2 \cdot \sin \alpha \cos \beta $$
We can rewrite the angle α as (a+b)/2
$$ \sin a + \sin b = 2 \cdot \sin \frac{a+b}{2} \cos \beta $$
Explanation. If $$ a = \alpha + \beta $$ $$ b = \alpha - \beta $$ then $$ a = \alpha + \beta $$ $$ a = \alpha + (\alpha - b) $$ $$ a = \alpha + \alpha - b $$ $$ a+b = 2\alpha $$ $$ \frac{a+b}{2} = \alpha $$
The angle β can be rewritten as (a-b)/2
$$ \sin a + \sin b = 2 \cdot \sin \frac{a+b}{2} \cos \frac{a-b}{2} $$
Explanation. If $$ a = \alpha + \beta $$ $$ b = \alpha - \beta $$ then $$ \beta = a - \alpha $$ $$ \beta = a - (\beta + b) $$ $$ \beta = a - \beta - b $$ $$ \beta + \beta = a - b $$ $$ 2 \beta = a - b $$ $$ \beta = \frac{ a - b }{2} $$
This completes the proof for the sine addition prosthaphaeresis formula.
The sine difference prosthaphaeresis formula
To prove the sine difference prosthaphaeresis formula, we use the sine addition and subtraction formulas
$$ \begin{cases} \sin (\ alpha+\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \\ \sin (\alpha-\beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \end{cases} $$
We subtract the two formulas:
$$ \sin (\alpha+\beta) - \sin (\alpha-\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta - ( \sin \alpha \cos \beta - \cos \alpha \sin \beta ) $$
And simplify:
$$ \sin (\alpha+\beta) - \sin (\alpha-\beta) = 2 \cos \alpha \sin \beta $$
Let’s assign a=α+β for the sum of the angles
$$ \sin a - \sin (\alpha-\beta) = 2 \cos \alpha \sin \beta $$
and b=α-β for the difference
$$ \sin a - \sin b = 2 \cos \alpha \sin \beta $$
The angle α can be rewritten as (a+b)/2
$$ \sin a - \sin b = 2 \cdot \cos \frac{a+b}{2} \sin \beta $$
Explanation. If $$ a = \alpha + \beta $$ $$ b = \alpha - \beta $$ then $$ a = \alpha + \beta $$ $$ a = \alpha + (\alpha - b) $$ $$ a = \alpha + \alpha - b $$ $$ a+b = 2\alpha $$ $$ \frac{a+b}{2} = \alpha $$
The angle β can be rewritten as (a-b)/2
$$ \sin a - \sin b = 2 \cdot \cos \frac{a+b}{2} \sin \frac{a-b}{2} $$
Explanation. If $$ a = \alpha + \beta $$ $$ b = \alpha - \beta $$ then $$ \beta = a - \alpha $$ $$ \beta = a - (\beta + b) $$ $$ \beta = a - \beta - b $$ $$ \beta + \beta = a - b $$ $$ 2 \beta = a - b $$ $$ \beta = \frac{ a - b }{2} $$
This completes the proof for the sine difference prosthaphaeresis formula.
The cosine addition prosthaphaeresis formula
To prove the cosine addition prosthaphaeresis formula, we use the cosine addition and subtraction formulas
$$ \begin{cases} \cos (\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \\ \cos (\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{cases} $$
We add the two formulas:
$$ \cos (\alpha+\beta) + \cos (\alpha-\beta) = 2 \cdot \cos \alpha \cos \beta $$
Let’s assign a=α+β to represent the sum of the angles
$$ \cos a + \cos (\alpha-\beta) = 2 \cdot \cos \alpha \cos \beta $$
and b=α-β for the difference
$$ \cos a + \cos b = 2 \cdot \cos \alpha \cos \beta $$
The angle α can be rewritten as (a+b)/2
$$ \cos a + \cos b = 2 \cdot \cos \frac{a+b}{2} \cos \beta $$
Explanation. If $$ a = \alpha + \beta $$ $$ b = \alpha - \beta $$ then $$ a = \alpha + \beta $$ $$ a = \alpha + (\alpha - b) $$ $$ a = \alpha + \alpha - b $$ $$ a+b = 2\alpha $$ $$ \frac{a+b}{2} = \alpha $$
The angle β can be rewritten as (a-b)/2
$$ \cos a + \cos b = 2 \cdot \cos \frac{a+b}{2} \cos \frac{a-b}{2} $$
Explanation. If $$ a = \alpha + \beta $$ $$ b = \alpha - \beta $$ then $$ \beta = a - \alpha $$ $$ \beta = a - (\beta + b) $$ $$ \beta = a - \beta - b $$ $$ \beta + \beta = a - b $$ $$ 2 \beta = a - b $$ $$ \beta = \frac{ a - b }{2} $$
This completes the proof for the cosine addition prosthaphaeresis formula.
The cosine difference prosthaphaeresis formula
To prove the cosine difference prosthaphaeresis formula, we use the cosine addition and subtraction formulas
$$ \begin{cases} \cos (\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \\ \cos (\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{cases}$$
We subtract the two formulas:
$$ \cos (\alpha+\beta) - \cos (\alpha-\beta) = - 2 \cdot \sin \alpha \sin \beta $$
Let’s assign a=α+β to represent the sum of the angles
$$ \cos a - \cos (\alpha-\beta) = - 2 \cdot \sin \alpha \sin \beta $$
and b=α-β for the difference
$$ \cos a - \cos b = - 2 \cdot \sin \alpha \sin \beta $$
We can rewrite the angle α as (a+b)/2
$$ \cos a - \cos b = - 2 \cdot \sin \frac{a+b}{2} \sin \beta $$
Explanation. If $$ a = \alpha + \beta $$ $$ b = \alpha - \beta $$ then $$ \beta = a - \alpha $$ $$ \beta = a - (\beta + b) $$ $$ \beta = a - \beta - b $$ $$ \beta + \beta = a - b $$ $$ 2 \beta = a - b $$ $$ \beta = \frac{ a - b }{2} $$
This completes the proof for the cosine difference prosthaphaeresis formula.
And so on.
