Deriving the Angle Addition and Subtraction Formulae

Finding the exact value of the sine, cosine, or tangent of an angle is often easier if we can rewrite the given angle in terms of two angles that have known trigonometric values. We can use the special angles, which we can review in the unit circle shown below.

Unit circle

The unit circle with the values for sine and cosine displayed for special angles.

There are formulae for the addition and subtraction of angles within each of the trigonometric functions. These allow us to find the trigonometric function of a given angle if we can break it up into the sum or difference of two of the special angles. 

To see how these formulae are derived, we can place points on a diagram of a unit circle. Suppose that the angle we want to find a trigonometric function for is the one formed by the point $A$, which measures an angle $\alpha - \beta$. The angle formed by $A$ and a point $B$ on the positive $x$-axis is the same as the angle formed between two special angles, which are denoted by $P$ and $Q$. Point $P$ is at an angle $\alpha$ from the positive $x$-axis with coordinates $(\cos \alpha, \sin \alpha)$, and point $Q$ is at an angle of $\beta$ from the positive $x$-axis with coordinates $(\cos \beta, \sin \beta)$. The angles are equal, and so the distance between points $P$ and $Q$ is the same as between points $A$ and $B$. 

Relationships between angles on the unit circle

The formulae for adding and subtracting angles are derived from the relationships between angles on the unit circle.

Using the equivalence of these distances and the distance formula:

 $d = \sqrt{\left(x_{2}-x_{1}\right)^2 + \left(y_{2}-y_{1}\right)^2}$

it is possible to derive a number of relationships between the angles. We can derive the following six formulae. 

The formulae for cosine are:

$\begin{aligned} \cos(\alpha + \beta) &= \cos \alpha \cos \beta - \sin \alpha \sin \beta \\ \cos(\alpha - \beta) &= \cos \alpha \cos \beta + \sin \alpha \sin \beta \end{aligned}$

The formulae for sine are: 

$\begin{aligned} \sin(\alpha + \beta) &= \sin \alpha \cos \beta + \cos \alpha \sin \beta \\ \sin(\alpha - \beta) &= \sin \alpha \cos \beta - \cos \alpha \sin \beta \end{aligned}$

The formulae for tangent are: 

$\displaystyle{ \begin{aligned} \tan(\alpha + \beta) &= \frac{ \tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \\ \tan(\alpha - \beta) &= \frac{ \tan \alpha - \tan \beta}{1 + \tan \alpha \tan \beta} \end{aligned} }$

$$These are useful for finding angles that can be derived by adding or subtracting special angles. For example, consider the angle $75^{\circ}$, which is the sum of two special angles: $30^{\circ}$ and $45^{\circ}$. The trigonometric functions of any such angle can be found.

Example

Using the formula for the cosine of the difference of two angles, find the exact value of $\displaystyle{\cos{\left(\frac{5\pi}{4} - \frac{\pi}{6}\right)} }$.

Apply the formula $\cos(\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$:

$\displaystyle{ \cos{\left(\frac{5\pi}{4} - \frac{\pi}{6}\right)} = \cos{\left(\frac{5\pi}{4}\right)} \cos{\left(\frac{\pi}{6}\right)} + \sin{\left(\frac{5\pi}{4}\right)} \sin{\left(\frac{\pi}{6}\right)} }$

Substitute the values of the trigonometric functions from the unit circle:

$\displaystyle{ \cos{\left(\frac{5\pi}{4} - \frac{\pi}{6}\right)} = \left(-\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right) + \left(-\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) }$

Simplify:

$\displaystyle{ \begin{aligned} \cos{\left(\frac{5\pi}{4} - \frac{\pi}{6}\right)} &= -\frac{\sqrt{6}}{4} -\frac{\sqrt{2}}{4} \\ \cos{\left(\frac{5\pi}{4} - \frac{\pi}{6}\right)} &= -\frac{\sqrt{6}-\sqrt{2}}{4} \end{aligned} }$

Example

Find the exact value of $\sin(15^{\circ})$.

First, notice that $15^{\circ}$ is the difference of two special angles: $45^{\circ} - 30^{\circ} = 15^{\circ}$. We can thus apply the formula for sine of the difference of two angles: $\sin(\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta$.

Substitute $\alpha = 45^{\circ}$ and $\beta = 30^{\circ}$ into the formula:

$\sin(45^{\circ} - 30^{\circ}) = \sin (45^{\circ}) \cos (30^{\circ}) - \cos (45^{\circ}) \sin (30^{\circ})$

Substitute the values of the trigonometric functions from the unit circle:

$\displaystyle{ \sin{\left(45^{\circ} - 30^{\circ}\right)} = \left(\frac{\sqrt{2}}{2}\right) \left(\frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{2}}{2}\right) \left(\frac{1}{2}\right) }$

Simplify:

$\displaystyle{ \begin{aligned} \sin{\left(45^{\circ} - 30^{\circ}\right)} &= \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} \\ \sin{\left(45^{\circ} - 30^{\circ}\right)} &= \frac{\sqrt{6} - \sqrt{2}}{4} \end{aligned} }$

Thus, we have: 

$\displaystyle{ \sin{\left(15^{\circ}\right)} = \frac{\sqrt{6} - \sqrt{2}}{4} }$