# Addition and Subtraction Formulas for Sine and Cosine

## Cosine Subtraction Formula

Let α, β be two angles such that α > β. Take the following points on a unit circle: A(0), M(α), N(β), P(αβ).

The coordinates of these points are

$A = A\left( {1,0} \right),\;\;M = M\left( {\cos \alpha ,\sin \alpha } \right),\;\;N = N\left( {\cos \beta ,\sin \beta } \right),\;\;P = P\left( {\cos \left( {\alpha - \beta } \right),\sin \left( {\alpha - \beta } \right)} \right).$

Since $$\angle MON = \angle POA = \alpha - \beta,$$ the line segments $$\color{#cc00ff}{MN}$$ and $$\color{#0099ff}{AP}$$ have the same length:

$\left| \color{#cc00ff}{MN} \right| = \left| \color{#0099ff}{AP} \right|.$

The distance between two points on a plane is given by the formula

$d = \sqrt {{{\left( {{x_1} - {x_2}} \right)}^2} + {{\left( {{y_1} - {y_2}} \right)}^2}} .$

Therefore,

${\left| \color{#cc00ff}{MN} \right|^2} = {\left( {{x_M} - {x_N}} \right)^2} + {\left( {{y_M} - {y_N}} \right)^2} = {\left( {\cos \alpha - \cos \beta } \right)^2} + {\left( {\sin \alpha - \sin \beta } \right)^2} = {\cos ^2}\alpha - 2\cos \alpha \cos \beta + {\cos ^2}\beta + {\sin ^2}\alpha - 2\sin \alpha \sin \beta + {\sin ^2}\beta = \underbrace {{{\cos }^2}\alpha + {{\sin }^2}\alpha }_1 + \underbrace {{{\cos }^2}\beta + {{\sin }^2}\beta }_1 - 2\left( {\cos \alpha \cos \beta + \sin \alpha \sin \beta } \right) = 2 - 2\left( {\cos \alpha \cos \beta + \sin \alpha \sin \beta } \right);$

Similarly we find the distance $$\left| \color{#0099ff}{AP} \right|$$ squared:

${\left| \color{#0099ff}{AP} \right|^2} = {\left( {{x_A} - {x_P}} \right)^2} + {\left( {{y_A} - {y_P}} \right)^2} = {\left( {1 - \cos \left( {\alpha - \beta } \right)} \right)^2} + {\left( {0 - \sin \left( {\alpha - \beta } \right)} \right)^2} = 1 - 2\cos \left( {\alpha - \beta } \right) + \underbrace {{{\cos }^2}\left( {\alpha - \beta } \right) + {{\sin }^2}\left( {\alpha - \beta } \right)}_1 = 2 - 2\cos \left( {\alpha - \beta } \right).$

The cosine subtraction formula follows from the equality $${\left| \color{#cc00ff}{MN} \right|^2} = {\left| \color{#0099ff}{AP} \right|^2}:$$

$\cos\left({\alpha - \beta}\right) = \cos\alpha \cos\beta + \sin\alpha \sin\beta$

Consider two points $$N\left( \beta \right)$$ and $$L\left( { - \beta } \right)$$ lying on the terminal sides of the angles $$\beta$$ and $$-\beta,$$ respectively.

These points are symmetric with respect to the $$x-$$axis. Therefore, they have the same $$x-$$coordinates. Their $$y-$$coordinates are equal in magnitude but opposite in sign. In other words, the cosine function is even and the sine function is odd:

$\cos\left({- \beta}\right) = \cos\beta, \;\;\sin\left({- \beta}\right) = -\sin\beta$

Now, let's go back to the cosine subtraction formula and replace $$\beta \to -\beta:$$

$\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \left( { - \beta } \right) + \sin \alpha \sin \left( { - \beta } \right).$

Since cosine is even and sine is odd, we get the cosine addition identity in the form

$\cos\left({\alpha + \beta}\right) = \cos\alpha \cos\beta - \sin\alpha \sin\beta$

## Special Cases

Substitute $$\alpha = \frac{\pi }{2}$$ in the cosine subtraction formula:

$\cos \left( {\frac{\pi }{2} - \beta } \right) = \cos \frac{\pi }{2}\cos \beta + \sin \frac{\pi }{2}\sin \beta = 0 \cdot \cos \beta + 1 \cdot \sin \beta = \sin \beta ,$

that is,

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

Replace $$\beta \to \frac{\pi }{2} - \beta$$ in the last expression:

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

or

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

Similarly, take the cosine addition formula and substitute $$\alpha = \frac{\pi }{2}:$$

$\cos \left( {\frac{\pi }{2} + \beta } \right) = \cos \frac{\pi }{2}\cos \beta - \sin \frac{\pi }{2}\sin \beta = 0 \cdot \cos \beta - 1 \cdot \sin \beta = - \sin \beta .$

We got the following identity:

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

By changing $$\beta \to -\beta$$ in the identity $$\sin \left( {\frac{\pi }{2} - \beta } \right) = \cos \beta ,$$ we obtain

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

that is,

$\sin\left( {\frac{\pi }{2} + \beta } \right) = \cos\beta$

## Sine Subtraction Formula

Using cofunction identities from the previous section, we derive the sine subtraction formula:

$\sin \left( {\alpha - \beta } \right) = \cos \left( {\frac{\pi }{2} - \left( {\alpha - \beta } \right)} \right) = \cos \left( {\left( {\frac{\pi }{2} - \alpha } \right) + \beta } \right) = \cos \left( {\frac{\pi }{2} - \alpha } \right)\cos \beta - \sin \left( {\frac{\pi }{2} - \alpha } \right)\sin \beta = \sin \alpha \cos \beta - \cos \alpha \sin \beta .$

Thus, we have

$\sin\left({\alpha - \beta}\right) = \sin\alpha \cos\beta - \cos\alpha \sin\beta$

If we replace $$\beta \to -\beta$$ in this formula, we get the sine addition identity:
$\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \left( { - \beta } \right) - \cos \alpha \sin \left( { - \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .$
$\sin\left({\alpha + \beta}\right) = \sin\alpha \cos\beta + \cos\alpha \sin\beta$