Addition formulae In a nutshell Expressions of the form sin ( α + β ) \sin(\alpha + \beta) sin ( α + β ) cos ( α + β ) \cos(\alpha + \beta) cos ( α + β ) tan ( α + β ) \tan(\alpha + \beta) tan ( α + β ) sin ( α ) \sin(\alpha) sin ( α ) sin ( β ) \sin(\beta) sin ( β ) cos ( α ) \cos(\alpha) cos ( α ) cos ( β ) \cos(\beta) cos ( β ) tan ( α ) \tan(\alpha) tan ( α ) tan ( β ) \tan(\beta) tan ( β )
Addition identities The following diagram can be used to derive the identities for sin ( α + β ) \sin(\alpha + \beta) sin ( α + β ) cos ( α + β ) \cos(\alpha + \beta) cos ( α + β ) sin ( α ) \sin(\alpha) sin ( α ) sin ( β ) \sin(\beta) sin ( β ) cos ( α ) \cos(\alpha) cos ( α ) cos ( β ) \cos(\beta) cos ( β )
Observe that ∠ D A E = ∠ A D F \angle DAE = \angle ADF ∠ D A E = ∠ A D F ∠ C D F = 90 ° − ∠ A D F \angle CDF = 90\degree - \angle ADF ∠ C D F = 90° − ∠ A D F ∠ D C F = ∠ D A E = α \angle DCF = \angle DAE = \alpha ∠ D CF = ∠ D A E = α
You have that A C D ACD A C D 1 1 1 ∠ C A D = β \angle CAD = \beta ∠ C A D = β A D = cos ( β ) AD = \cos(\beta) A D = cos ( β ) C D = sin ( β ) CD = \sin(\beta) C D = sin ( β )
The equalities A E = A B + B E AE = AB + BE A E = A B + BE B C = B F + C F BC = BF + CF BC = BF + CF B E = D F BE = DF BE = D F B F = D E BF = DE BF = D E
You have that A B C ABC A BC 1 1 1 ∠ B A C = α + β \angle BAC = \alpha + \beta ∠ B A C = α + β A B = cos ( α + β ) AB = \cos(\alpha + \beta) A B = cos ( α + β ) B C = sin ( α + β ) BC = \sin(\alpha + \beta) BC = sin ( α + β )
You have that A D E ADE A D E cos ( β ) \cos(\beta) cos ( β ) ∠ D A E = α \angle DAE = \alpha ∠ D A E = α A E = cos ( β ) cos ( α ) AE = \cos(\beta)\cos(\alpha) A E = cos ( β ) cos ( α ) D E = cos ( β ) sin ( α ) DE = \cos(\beta)\sin(\alpha) D E = cos ( β ) sin ( α )
You have that C D F CDF C D F sin ( β ) \sin(\beta) sin ( β ) ∠ D C F = α \angle DCF = \alpha ∠ D CF = α C F = sin ( β ) cos ( α ) CF = \sin(\beta)\cos(\alpha) CF = sin ( β ) cos ( α ) D F = sin ( β ) sin ( α ) DF = \sin(\beta)\sin(\alpha) D F = sin ( β ) sin ( α )
Putting these together, you have that cos ( α + β ) + sin ( β ) sin ( α ) ≡ cos ( β ) cos ( α ) \cos(\alpha + \beta) + \sin(\beta)\sin(\alpha) \equiv \cos(\beta)\cos(\alpha) cos ( α + β ) + sin ( β ) sin ( α ) ≡ cos ( β ) cos ( α ) sin ( α + β ) ≡ cos ( β ) sin ( α ) + sin ( β ) cos ( α ) \sin(\alpha + \beta) \equiv \cos(\beta)\sin(\alpha) + \sin(\beta)\cos(\alpha) sin ( α + β ) ≡ cos ( β ) sin ( α ) + sin ( β ) cos ( α )
Rearranging these, you get the following identities:
cos ( α + β ) ≡ cos ( α ) cos ( β ) − sin ( α ) sin ( β ) \boxed{\cos(\alpha + \beta) \equiv \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)} cos ( α + β ) ≡ cos ( α ) cos ( β ) − sin ( α ) sin ( β )
sin ( α + β ) ≡ sin ( α ) cos ( β ) + cos ( α ) sin ( β ) \boxed{\sin(\alpha + \beta) \equiv \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)} sin ( α + β ) ≡ sin ( α ) cos ( β ) + cos ( α ) sin ( β )
Substitute these into the identity tan ( α + β ) ≡ sin ( α + β ) cos ( α + β ) \tan(\alpha + \beta) \equiv \dfrac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} tan ( α + β ) ≡ cos ( α + β ) sin ( α + β ) cos ( α ) cos ( β ) \cos(\alpha)\cos(\beta) cos ( α ) cos ( β )
tan ( α + β ) ≡ sin ( α + β ) cos ( α + β ) ≡ sin ( α ) cos ( β ) + cos ( α ) sin ( β ) cos ( α ) cos ( β ) − sin ( α ) sin ( β ) ≡ ( sin ( α ) cos ( β ) cos ( α ) cos ( β ) ) + ( cos ( α ) sin ( β ) cos ( α ) cos ( β ) ) ( cos ( α ) cos ( β ) cos ( α ) cos ( β ) ) − ( sin ( α ) sin ( β ) cos ( α ) cos ( β ) ) ≡ tan ( α ) + tan ( β ) 1 − tan ( α ) tan ( β ) \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} tan ( α + β ) ≡ cos ( α + β ) sin ( α + β ) ≡ cos ( α ) cos ( β ) − sin ( α ) sin ( β ) sin ( α ) cos ( β ) + cos ( α ) sin ( β ) ≡ ( cos ( α ) cos ( β ) cos ( α ) cos ( β ) ) − ( cos ( α ) cos ( β ) sin ( α ) sin ( β ) ) ( cos ( α ) cos ( β ) sin ( α ) cos ( β ) ) + ( cos ( α ) cos ( β ) cos ( α ) sin ( β ) ) ≡ 1 − tan ( α ) tan ( β ) tan ( α ) + tan ( β )
Therefore, you get the following addition identity:
tan ( α + β ) ≡ tan ( α ) + tan ( β ) 1 − tan ( α ) tan ( β ) \boxed{\tan(\alpha + \beta) \equiv \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}} tan ( α + β ) ≡ 1 − tan ( α ) tan ( β ) tan ( α ) + tan ( β )
Difference identitites Using the facts that sin ( − α ) = − sin ( α ) \sin(-\alpha) = -\sin(\alpha) sin ( − α ) = − sin ( α ) cos ( − α ) = cos ( α ) \cos(-\alpha) = \cos(\alpha) cos ( − α ) = cos ( α ) tan ( − α ) = − tan ( α ) \tan(-\alpha) = -\tan(\alpha) tan ( − α ) = − tan ( α )
cos ( α − β ) ≡ cos ( α ) cos ( − β ) − sin ( α ) ( sin ( − β ) cos ( α − β ) ≡ cos ( α ) cos ( β ) + sin ( α ) sin ( β ) \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)} cos ( α − β ) ≡ cos ( α ) cos ( − β ) − sin ( α ) ( sin ( − β ) cos ( α − β ) ≡ cos ( α ) cos ( β ) + sin ( α ) sin ( β )
sin ( α − β ) ≡ sin ( α ) cos ( − β ) + cos ( α ) sin ( − β ) sin ( α − β ) ≡ sin ( α ) cos ( β ) − cos ( α ) sin ( β ) \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)} sin ( α − β ) ≡ sin ( α ) cos ( − β ) + cos ( α ) sin ( − β ) sin ( α − β ) ≡ sin ( α ) cos ( β ) − cos ( α ) sin ( β )
tan ( α − β ) ≡ tan ( α ) + tan ( − β ) 1 − tan ( α ) tan ( − β ) tan ( α − β ) ≡ tan ( α ) − tan ( β ) 1 + tan ( α ) tan ( β ) \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)}} tan ( α − β ) ≡ 1 − tan ( α ) tan ( − β ) tan ( α ) + tan ( − β ) tan ( α − β ) ≡ 1 + tan ( α ) tan ( β ) tan ( α ) − tan ( β )
Example 1 Given that cos ( α ) = 3 5 \cos(\alpha) = \dfrac35 cos ( α ) = 5 3 and cos ( β ) = 5 13 \cos(\beta) = \dfrac{5}{13} cos ( β ) = 13 5 , find all possible values of cos ( α + β ) \cos(\alpha + \beta) cos ( α + β ) .
cos ( α ) = 3 5 \cos(\alpha) = \dfrac35 cos ( α ) = 5 3 implies sin ( α ) = ± 1 − cos 2 ( α ) = ± 1 − 9 25 = ± 4 5 \sin(\alpha) = \pm \sqrt{1 - \cos^2(\alpha)} = \pm \sqrt{1 - \dfrac{9}{25}}= \pm \dfrac45 sin ( α ) = ± 1 − cos 2 ( α ) = ± 1 − 25 9 = ± 5 4 .
cos ( β ) = 5 13 \cos(\beta) = \dfrac{5}{13} cos ( β ) = 13 5 sin ( β ) = ± 1 − cos 2 ( β ) = ± 1 − 25 169 = ± 12 13 \sin(\beta) = \pm\sqrt{1- \cos^2(\beta)} = \pm\sqrt{1 - \dfrac{25}{169}} = \pm\dfrac{12}{13} sin ( β ) = ± 1 − cos 2 ( β ) = ± 1 − 169 25 = ± 13 12
Using the identity cos ( α + β ) ≡ cos ( α ) cos ( β ) − sin ( α ) sin ( β ) \cos(\alpha + \beta) \equiv \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta) cos ( α + β ) ≡ cos ( α ) cos ( β ) − sin ( α ) sin ( β )
cos ( α + β ) = ( 3 5 ) ( 5 13 ) ± ( 4 5 ) ( 12 13 ) = 63 65 , − 33 65 \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} cos ( α + β ) = ( 5 3 ) ( 13 5 ) ± ( 5 4 ) ( 13 12 ) = 65 63 , − 65 33
Therefore, the possible values of cos ( α + β ) \cos(\alpha + \beta) cos ( α + β ) 63 65 ‾ \underline{\dfrac{63}{65}} 65 63 − 33 65 ‾ \underline{-\dfrac{33}{65}} − 65 33
