Addition formulae

​​In a nutshell

Expressions of the form $$\sin(\alpha + \beta)$$​, $$\cos(\alpha + \beta)$$​, or $$\tan(\alpha + \beta)$$ can be expressed in terms of $$\sin(\alpha)$$, $$\sin(\beta)$$​, $$\cos(\alpha)$$​, $$\cos(\beta)$$​, $$\tan(\alpha)$$​ and $$\tan(\beta)$$​.

Addition identities

The following diagram can be used to derive the identities for $$\sin(\alpha + \beta)$$​ and $$\cos(\alpha + \beta)$$​ in terms of $$\sin(\alpha)$$​, $$\sin(\beta)$$​, $$\cos(\alpha)$$​ and $$\cos(\beta)$$​.

Observe that $$\angle DAE = \angle ADF$$, and therefore $$\angle CDF = 90\degree - \angle ADF$$. It follows that $$\angle DCF = \angle DAE = \alpha$$​.

You have that $$ACD$$ is a right angled triangle with a hypotenuse of length $$1$$​, and $$\angle CAD = \beta$$​.  Therefore the lengths $$AD = \cos(\beta)$$, and $$CD = \sin(\beta)$$​.

The equalities $$AE = AB + BE$$​ and $$BC = BF + CF$$ hold. Moreover, $$BE = DF$$​ and $$BF = DE$$​.

You have that $$ABC$$ is a right angled triangle with a hypotenuse of length $$1$$​, and $$\angle BAC = \alpha + \beta$$. Therefore the lengths $$AB = \cos(\alpha + \beta)$$​, and $$BC = \sin(\alpha + \beta)$$​.

You have that $$ADE$$​ is a right angled triangle with a hypotenuse of length $$\cos(\beta)$$, and $$\angle DAE = \alpha$$​. Therefore the lengths $$AE = \cos(\beta)\cos(\alpha)$$​, and $$DE = \cos(\beta)\sin(\alpha)$$​.

You have that $$CDF$$​ is a right angled triangle with a hypotenuse of length $$\sin(\beta)$$, and $$\angle DCF = \alpha$$​. Therefore the lengths $$CF = \sin(\beta)\cos(\alpha)$$, and $$DF = \sin(\beta)\sin(\alpha)$$​.

Putting these together, you have that $$\cos(\alpha + \beta) + \sin(\beta)\sin(\alpha) \equiv \cos(\beta)\cos(\alpha)$$​, and $$\sin(\alpha + \beta) \equiv \cos(\beta)\sin(\alpha) + \sin(\beta)\cos(\alpha)$$​.

Rearranging these, you get the following identities:

​$$\boxed{\cos(\alpha + \beta) \equiv \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)}$$​​

​$$\boxed{\sin(\alpha + \beta) \equiv \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)}$$​​

Substitute these into the identity $$\tan(\alpha + \beta) \equiv \dfrac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}$$, and divide the numerator and denominator by $$\cos(\alpha)\cos(\beta)$$​:

​$$\begin{aligned}\tan(\alpha + \beta) &\equiv \dfrac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)}\\& \equiv \dfrac{\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)}\\&\equiv \dfrac{\left(\dfrac{\sin(\alpha)\cos(\beta)}{\cos(\alpha)\cos(\beta)}\right) + \left(\dfrac{\cos(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}\right)}{\left(\dfrac{\cos(\alpha)\cos(\beta)}{\cos(\alpha)\cos(\beta)}\right) - \left(\dfrac{\sin(\alpha)\sin(\beta)}{\cos(\alpha)\cos(\beta)}\right)}\\&\equiv \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}\end{aligned}$$​​

Therefore, you get the following addition identity:

​$$\boxed{\tan(\alpha + \beta) \equiv \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}}$$​​

Difference identitites

Using the facts that $$\sin(-\alpha) = -\sin(\alpha)$$​, $$\cos(-\alpha) = \cos(\alpha)$$​, and $$\tan(-\alpha) = -\tan(\alpha)$$​, you can deduce the following identities:

​$$\cos(\alpha - \beta) \equiv \cos(\alpha)\cos(-\beta) - \sin(\alpha)(\sin(-\beta)\\\boxed{\cos(\alpha - \beta) \equiv \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)}$$​​

​$$\sin(\alpha - \beta) \equiv \sin(\alpha)\cos(-\beta) + \cos(\alpha)\sin(-\beta)\\\boxed{\sin(\alpha - \beta) \equiv \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)}$$​​

​

​$$\tan(\alpha - \beta) \equiv \dfrac{\tan(\alpha) + \tan(-\beta)}{1 - \tan(\alpha)\tan(-\beta)}\\\boxed{\tan(\alpha - \beta) \equiv \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}}$$​​

Example 1

Given that $$\cos(\alpha) = \dfrac35$$​ and $$\cos(\beta) = \dfrac{5}{13}$$​, find all possible values of $$\cos(\alpha + \beta)$$​.

​

​$$\cos(\alpha) = \dfrac35$$​ implies $$\sin(\alpha) = \pm \sqrt{1 - \cos^2(\alpha)} = \pm \sqrt{1 - \dfrac{9}{25}}= \pm \dfrac45$$​.

​$$\cos(\beta) = \dfrac{5}{13}$$ implies $$\sin(\beta) = \pm\sqrt{1- \cos^2(\beta)} = \pm\sqrt{1 - \dfrac{25}{169}} = \pm\dfrac{12}{13}$$.

Using the identity $$\cos(\alpha + \beta) \equiv \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$$, you have:

$$\begin{aligned}\cos(\alpha + \beta) &= \left(\dfrac35\right)\left(\dfrac{5}{13}\right) \pm \left(\dfrac45\right)\left(\dfrac{12}{13}\right)\\&=\dfrac{63}{65}, -\dfrac{33}{65}\end{aligned}$$​​

Therefore, the possible values of $$\cos(\alpha + \beta)$$ are $$\underline{\dfrac{63}{65}}$$ and $$\underline{-\dfrac{33}{65}}$$.