Sine Addition Formula

The formula for the sine of a sum 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 take two angles: a = 30° and b = 60°

$$ \sin a = \sin 30° = \frac{1}{2} $$

$$ \sin b = \sin 60° = \frac{\sqrt{3}}{2} $$

The sine of a + b is not simply the sum of the sines of the two angles

$$ \sin (a+b) \ne \sin(a) + \sin(b) $$

$$ \sin (30°+60°) \ne \sin(30°) + \sin(60°) = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} $$

Note: The sine of 30° + 60° is actually the sine of 90°, which we know equals 1. So it’s a serious mistake to calculate the sine of the sum of two angles, sin(a+b), as simply the sum of their individual sines, sin(a) + sin(b).

Now let’s compute the sine of a + b using the sine addition formula

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

Substitute a = 30° and b = 60°

$$ \sin (30°+60°) = \sin 30° \cos 60° + \sin 60° \cos 30° $$

We already know that sin(30°) = 1/2 and sin(60°) = √3/2

$$ \sin (30°+60°) = \frac{1}{2} \cos 60° + \frac{\sqrt{3}}{2} \cos 30° $$

The cosine values are cos(30°) = √3/2 and cos(60°) = 1/2

$$ \sin (30°+60°) = \frac{1}{2} \cdot \frac{1}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} $$

$$ \sin (30°+60°) = \frac{1}{4} + \frac{3}{4} $$

$$ \sin (30°+60°) = \frac{4}{4} $$

$$ \sin (30°+60°) = 1 $$

So, the sine of 30° + 60° is equal to 1.

The result is correct.

The proof

Let's prove the sine of the sum of two angles a and b:

$$ \sin(a+b) $$

We start by considering x = a + b

$$ \sin x $$

We can express the sine of x in an equivalent form using the associated angle π/2 - x

$$ \sin x = \cos \left( \frac{\pi}{2} - x \right) $$

Since x = a + b

$$ \sin (a+b) = \cos \left( \frac{\pi}{2} - (a+b) \right) $$

$$ \sin (a+b) = \cos \left( \frac{\pi}{2} - a - b \right) $$

Now let’s rearrange the cosine’s argument:

$$ \sin (a+b) = \cos \left[ \left( \frac{\pi}{2} - a \right) - b \right] $$

Next, apply the cosine subtraction formula to (π/2 - a) and b

$$ \sin (a+b) = \cos \left( \frac{\pi}{2} - a \right) \cos b + \sin b \sin \left( \frac{\pi}{2} - a \right) $$

Since cos(π/2 - a) = sin(a) (as they are associated angles)

$$ \sin (a+b) = \sin a \cos b + \sin b \sin \left( \frac{\pi}{2} - a \right) $$

And since sin(π/2 - a) = cos(a) (again, due to associated angles)

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

And that’s the sine addition formula we wanted to prove.

The proof is complete.