Sine subtraction formula

The formula for the subtraction of sine is: $$ \sin(\alpha - \beta) = \sin \alpha \cos \beta - \sin \beta \cos \alpha $$

It’s incorrect to write:

$$ \sin(\alpha - \beta) = \sin \alpha - \sin \beta $$

A practical example

Let’s consider two angles, a = 90° and b = 30°

$$ \sin a = \sin 90° = 1 $$

$$ \sin b = \sin 30° = \frac{1}{2} $$

Note: The sine of a - b is not simply the difference between the sines of the two angles. $$ \sin(a - b) \ne \sin a - \sin b $$ $$ \sin(90° - 30°) \ne \sin 90° - \sin 30° = 1 - \frac{1}{2} = \frac{1}{2} $$ In fact, the sine of 90° - 30° is the sine of 60°, which equals the square root of three over two. $$ \sin(90° - 30°) = \sin(60°) = \frac{\sqrt{3}}{2} $$

Now, let’s calculate the sine of a - b using the sine subtraction formula:

$$ \sin(a - b) = \sin a \cos b - \sin b \cos a $$

Substituting a = 90° and b = 30°, we get:

$$ \sin(90° - 30°) = \sin 90° \cos 30° - \sin 30° \cos 90° $$

Since we know that sin(90°) = 1 and cos(90°) = 0, we can simplify:

$$ \sin(90° - 30°) = 1 \cdot \cos 30° - \sin 30° \cdot 0 $$

$$ \sin(90° - 30°) = \cos 30° $$

And since cos(30°) = √3/2, we find:

$$ \sin(90° - 30°) = \frac{\sqrt{3}}{2} $$

Therefore, the sine of 90° - 30° is √3/2.

$$ \sin(90° - 30°) = \sin(60°) = \frac{\sqrt{3}}{2} $$

The result is correct.

Proof of the formula

We start with the sine of the difference of two angles:

$$ \sin(a - b) $$

We can rewrite this as:

$$ \sin(a - b) = \sin[a + (-b)] $$

This allows us to apply the sine addition formula:

$$ \sin(a - b) = \sin[a + (-b)] = \sin a \cos(-b) + \sin(-b) \cos a $$

Sine is an odd function, so sin(-b) = -sin(b).

$$ \sin(a - b) = \sin[a + (-b)] = \sin a \cos(-b) + [-\sin(b)] \cos a $$

$$ \sin(a - b) = \sin[a + (-b)] = \sin a \cos(-b) - \sin b \cos a $$

Cosine, on the other hand, is an even function, so cos(-b) = cos(b).

$$ \sin(a - b) = \sin[a + (-b)] = \sin a \cos(b) - \sin b \cos a $$

This gives us the desired formula.

And that’s the proof.