Werner Formulas
What Are the Werner Formulas?
In trigonometry, the Werner formulas transform a product of trigonometric functions into a sum of trigonometric functions. $$ \sin \alpha \cos \beta = \frac{1}{2} \cdot [ \sin (\alpha+\beta) + \sin(\alpha-\beta) ] $$ $$ \cos \alpha \cos \beta = \frac{1}{2} \cdot [ \cos (\alpha+\beta) + \cos(\alpha-\beta) ] $$ $$ \sin \alpha \sin \beta = \frac{1}{2} \cdot [ \cos (\alpha-\beta) - \cos(\alpha+\beta) ] $$
What Are They Used For?
The Werner formulas are essentially the reverse of the prosthaphaeresis formulas.
While the prosthaphaeresis formulas turn a sum into a product of trigonometric functions, the Werner formulas convert a product into a sum of trigonometric functions.
Note: Interestingly, the mathematician Johann Werner, who developed the Werner formulas, is also credited with the prosthaphaeresis formulas. This shows a strong connection between these two sets of formulas, especially in their derivations.
A Practical Example
Let’s walk through a simple example, using the sine of 30° (π/6) and the cosine of 60° (π/3):
$$ \sin 30° = \frac{1}{2} $$
$$ \cos 60° = \frac{1}{2} $$
The product of these two trigonometric functions is:
$$ \sin 30° \cdot \cos 60° = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} $$
Now, let's check if the Werner formula yields the same result.
$$ \sin \alpha \cos \beta = \frac{1}{2} \cdot [ \sin (\alpha+\beta) + \sin(\alpha-\beta) ] $$
Here, α=30° and β=60°.
$$ \sin 30° \cos 60° = \frac{1}{2} \cdot [ \sin (30°+60°) + \sin(30°-60°) ] $$
$$ \sin 30° \cos 60° = \frac{1}{2} \cdot [ \sin (90°) + \sin(-30°) ] $$
We know the sine of 90° is 1:
$$ \sin 30° \cos 60° = \frac{1}{2} \cdot [ 1 + \sin(-30°) ] $$
The sine of -30° is -1/2:
$$ \sin 30° \cos 60° = \frac{1}{2} \cdot [ 1 - \frac{1}{2} ] $$
Now, simplify the expression:
$$ \sin 30° \cos 60° = \frac{1}{2} \cdot \frac{1}{2} $$
$$ \sin 30° \cos 60° = \frac{1}{4} $$
As expected, the result matches.
The Proof
The First Werner Formula
$$ \sin \alpha \cos \beta = \frac{1}{2} \cdot [ \sin (\alpha+\beta) + \sin(\alpha-\beta) ] $$
To prove the first Werner formula, we’ll 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’ll now add both equations:
$$ \sin (\alpha+\beta) + \sin (\alpha-\beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta + \sin \alpha \cos \beta - \cos \alpha \sin \beta $$
Simplifying, we get:
$$ \sin (\alpha+\beta) + \sin (\alpha-\beta) = 2 \cdot \sin \alpha \cos \beta $$
Now, divide both sides by 2:
$$ \frac{1}{2} \cdot [ \sin (\alpha+\beta) + \sin (\alpha-\beta) ] = \sin \alpha \cos \beta $$
And that’s the proof.
The Second Werner Formula
$$ \cos \alpha \cos \beta = \frac{1}{2} \cdot [ \cos (\alpha+\beta) + \cos (\alpha-\beta) ] $$
To prove the second Werner formula, we’ll use the cosine addition and subtraction formulas:
$$ \begin{cases} \cos (\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \\ \cos (\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{cases} $$
Adding both equations gives us:
$$ \cos (\alpha+\beta) + \cos (\alpha-\beta) = 2 \cdot \cos \alpha \cos \beta $$
Again, divide both sides by 2:
$$ \frac{1}{2} \cdot [ \cos (\alpha+\beta) + \cos (\alpha-\beta) ] = \cos \alpha \cos \beta $$
And that completes the proof.
The Third Werner Formula
$$ \sin \alpha \sin \beta = \frac{1}{2} \cdot [ \cos (\alpha-\beta) - \cos(\alpha+\beta) ] $$
For the third Werner formula, we use the cosine addition and subtraction formulas again:
$$ \begin{cases} \cos (\alpha+\beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \\ \cos (\alpha-\beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{cases} $$
Subtracting the two equations gives us:
$$ \cos (\alpha+\beta) - \cos (\alpha-\beta) = -2 \cdot \sin \alpha \sin \beta $$
Now, divide both sides by 2:
$$ \frac{1}{2} \cdot [ \cos (\alpha+\beta) - \cos (\alpha-\beta) ] = - \sin \alpha \sin \beta $$
Multiplying both sides by -1, we get:
$$ \frac{1}{2} \cdot [ \cos (\alpha-\beta) - \cos (\alpha+\beta) ] = \sin \alpha \sin \beta $$
And that’s the proof.
