Using the angle addition formulae In a nutshell The addition formulae for trigonometric functions, and the exact values of trigonometric functions evaluated at certain angles, can be used to compute the exact values of trigonometric functions evaluated at new angles.
Addition formulae The addition formulae for the trigonometric functions are:
sin ( α + β ) ≡ sin ( α ) cos ( β ) + cos ( α ) sin ( β ) sin ( α − β ) ≡ sin ( α ) cos ( β ) − cos ( α ) sin ( β ) cos ( α + β ) ≡ cos ( α ) cos ( β ) − sin ( α ) sin ( β ) cos ( α − β ) ≡ cos ( α ) cos ( β ) + sin ( 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 ) \begin{aligned}\sin(\alpha + \beta) &\equiv \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\\\sin(\alpha - \beta) &\equiv \sin(\alpha)\cos(\beta) - \cos(\alpha)\sin(\beta)\\\\\cos(\alpha + \beta) &\equiv \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)\\\cos(\alpha - \beta) &\equiv \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)\\\\\tan(\alpha + \beta) &\equiv \dfrac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)}\\\tan(\alpha - \beta) &\equiv \dfrac{\tan(\alpha) - \tan(\beta)}{1 + \tan(\alpha)\tan(\beta)}\end{aligned} sin ( α + β ) sin ( α − β ) cos ( α + β ) cos ( α − β ) tan ( α + β ) tan ( α − β ) ≡ sin ( α ) cos ( β ) + cos ( α ) sin ( β ) ≡ sin ( α ) cos ( β ) − cos ( α ) sin ( β ) ≡ cos ( α ) cos ( β ) − sin ( α ) sin ( β ) ≡ cos ( α ) cos ( β ) + sin ( α ) sin ( β ) ≡ 1 − tan ( α ) tan ( β ) tan ( α ) + tan ( β ) ≡ 1 + tan ( α ) tan ( β ) tan ( α ) − tan ( β )
Given the exact values of trigonometric functions evaluated at angles α \alpha α β \beta β α + β \alpha + \beta α + β α − β \alpha - \beta α − β
Example 1 Compute the exact value of tan ( 7 π 12 ) \tan\left(\dfrac{7\pi}{12}\right) tan ( 12 7 π )
You have that 7 π 12 = π 4 + π 3 \dfrac{7\pi}{12} = \dfrac{\pi}{4} + \dfrac{\pi}{3} 12 7 π = 4 π + 3 π
tan ( π 4 ) = 1 \tan\left(\dfrac{\pi}{4}\right) = 1 tan ( 4 π ) = 1 tan ( π 3 ) = 3 \tan\left(\dfrac{\pi}{3}\right) = \sqrt{3} tan ( 3 π ) = 3 tan \tan tan
tan ( 7 π 12 ) = tan ( π 4 + π 3 ) = tan ( π 4 ) + tan ( π 3 ) 1 − tan ( π 4 ) tan ( π 3 ) = 1 + 3 1 − 3 = ( 1 + 3 ) 2 ( 1 − 3 ) ( 1 + 3 ) = 4 + 2 3 − 2 = − 2 − 3 \begin{aligned}\tan\left(\dfrac{7\pi}{12}\right) &= \tan\left(\dfrac{\pi}{4} + \dfrac{\pi}{3}\right)\\&= \dfrac{\tan\left(\dfrac{\pi}{4}\right) + \tan\left(\dfrac{\pi}{3}\right)}{1 - \tan\left(\dfrac{\pi}{4}\right)\tan\left(\dfrac{\pi}{3}\right)}\\&= \dfrac{1 + \sqrt{3}}{1 - \sqrt{3}}\\&= \dfrac{(1 + \sqrt{3})^2}{(1 - \sqrt{3})(1 + \sqrt{3})}\\&= \dfrac{4 + 2\sqrt{3}}{-2}\\&= - 2 - \sqrt{3}\end{aligned} tan ( 12 7 π ) = tan ( 4 π + 3 π ) = 1 − tan ( 4 π ) tan ( 3 π ) tan ( 4 π ) + tan ( 3 π ) = 1 − 3 1 + 3 = ( 1 − 3 ) ( 1 + 3 ) ( 1 + 3 ) 2 = − 2 4 + 2 3 = − 2 − 3
Therefore, the exact value of tan ( 7 π 12 ) \tan\left(\dfrac{7\pi}{12}\right) tan ( 12 7 π ) − 2 − 3 ‾ \underline{-2 - \sqrt{3}} − 2 − 3
Example 2 Given that cos ( α ) = 20 29 \cos(\alpha) = \dfrac{20}{29} cos ( α ) = 29 20 α \alpha α sin ( β ) = 24 25 \sin(\beta) = \dfrac{24}{25} sin ( β ) = 25 24 β \beta β sin ( α + β ) \sin(\alpha + \beta) sin ( α + β )
If cos ( α ) = 20 29 \cos(\alpha) = \dfrac{20}{29} cos ( α ) = 29 20 sin ( α ) = ± 1 − cos 2 ( α ) = ± 21 29 \sin(\alpha) = \pm \sqrt{1- \cos^2(\alpha)} = \pm\dfrac{21}{29} sin ( α ) = ± 1 − cos 2 ( α ) = ± 29 21
However, α \alpha α sin ( α ) = 21 29 \sin(\alpha) = \dfrac{21}{29} sin ( α ) = 29 21
If sin ( β ) = 24 25 \sin(\beta) = \dfrac{24}{25} sin ( β ) = 25 24 cos ( β ) = ± 1 − sin 2 ( β ) = ± 7 25 \cos(\beta) = \pm\sqrt{1 - \sin^2(\beta)} = \pm \dfrac{7}{25} cos ( β ) = ± 1 − sin 2 ( β ) = ± 25 7
However, β \beta β cos ( β ) = − 7 25 \cos(\beta) = -\dfrac{7}{25} cos ( β ) = − 25 7
Put these together via the addition formula for sin \sin sin
sin ( α + β ) = sin ( α ) cos ( β ) + cos ( α ) sin ( β ) = ( 21 29 ) ( − 7 25 ) + ( 20 29 ) ( 24 25 ) = 333 725 \begin{aligned}\sin(\alpha + \beta) &= \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)\\&= \left(\dfrac{21}{29}\right)\left(-\dfrac{7}{25}\right) + \left(\dfrac{20}{29}\right)\left(\dfrac{24}{25}\right)\\&= \dfrac{333}{725}\end{aligned} sin ( α + β ) = sin ( α ) cos ( β ) + cos ( α ) sin ( β ) = ( 29 21 ) ( − 25 7 ) + ( 29 20 ) ( 25 24 ) = 725 333
Therefore, the exact value of sin ( α + β ) \sin(\alpha + \beta) sin ( α + β ) 333 725 ‾ \underline{\dfrac{333}{725}} 725 333
