Double angle formulae In a nutshell The addition formulae allow you to evaluate trigonometric functions at α + β \alpha + \beta α + β α \alpha α β \beta β β = α \beta = \alpha β = α
Double angle identities The addition formulae are:
sin ( α + β ) ≡ sin ( α ) cos ( β ) + cos ( 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 ) \begin{aligned}\sin(\alpha + \beta) &\equiv \sin(\alpha)\cos(\beta) + \cos(\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)}\end{aligned} sin ( α + β ) cos ( α + β ) tan ( α + β ) ≡ sin ( α ) cos ( β ) + cos ( α ) sin ( β ) ≡ cos ( α ) cos ( β ) − sin ( α ) sin ( β ) ≡ 1 − tan ( α ) tan ( β ) tan ( α ) + tan ( β )
In each of these, let β = α \beta = \alpha β = α sin ( 2 α ) \sin(2\alpha) sin ( 2 α ) cos ( 2 α ) \cos(2\alpha) cos ( 2 α ) tan ( 2 α ) \tan(2\alpha) tan ( 2 α )
sin ( α + α ) ≡ sin ( α ) cos ( α ) + cos ( α ) sin ( α ) sin ( 2 α ) ≡ 2 sin ( α ) cos ( α ) \sin(\alpha + \alpha) \equiv \sin(\alpha)\cos(\alpha) + \cos(\alpha)\sin(\alpha)\\\boxed{\sin(2\alpha) \equiv 2\sin(\alpha)\cos(\alpha)} \quad\quad\quad\quad\quad\quad\quad sin ( α + α ) ≡ sin ( α ) cos ( α ) + cos ( α ) sin ( α ) sin ( 2 α ) ≡ 2 sin ( α ) cos ( α )
cos ( α + α ) ≡ cos ( α ) cos ( α ) − sin ( α ) sin ( α ) cos ( 2 α ) ≡ cos 2 ( α ) − sin 2 ( α ) cos ( 2 α ) ≡ 2 cos 2 ( α ) − 1 u s i n g sin 2 ( α ) = 1 − cos 2 ( α ) cos ( 2 α ) ≡ 1 − 2 sin 2 ( α ) u s i n g cos 2 ( α ) = 1 − sin 2 ( α ) \cos(\alpha + \alpha) \equiv \cos(\alpha)\cos(\alpha) - \sin(\alpha)\sin(\alpha) \quad \quad \quad \quad \quad \quad\\\boxed{\cos(2\alpha) \equiv \cos^2(\alpha) - \sin^2(\alpha)} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\\boxed{\cos(2\alpha) \equiv 2\cos^2(\alpha) - 1} \quad using \quad \sin^2(\alpha)=1-\cos^2(\alpha) \\\boxed{\cos(2\alpha) \equiv 1 -2 \sin^2(\alpha)} \quad using \quad \cos^2(\alpha)=1-\sin^2(\alpha) \\ cos ( α + α ) ≡ cos ( α ) cos ( α ) − sin ( α ) sin ( α ) cos ( 2 α ) ≡ cos 2 ( α ) − sin 2 ( α ) cos ( 2 α ) ≡ 2 cos 2 ( α ) − 1 u s in g sin 2 ( α ) = 1 − cos 2 ( α ) cos ( 2 α ) ≡ 1 − 2 sin 2 ( α ) u s in g cos 2 ( α ) = 1 − sin 2 ( α )
tan ( α + α ) ≡ tan ( α ) + tan ( α ) 1 − tan ( α ) tan ( α ) tan ( 2 α ) ≡ 2 tan ( α ) 1 − tan 2 ( α ) \tan(\alpha + \alpha) \equiv \dfrac{\tan(\alpha) + \tan(\alpha)}{1 - \tan(\alpha)\tan(\alpha)}\\\boxed{\tan(2\alpha) \equiv \dfrac{2\tan(\alpha)}{1 - \tan^2(\alpha)}} \quad\quad\quad\quad tan ( α + α ) ≡ 1 − tan ( α ) tan ( α ) tan ( α ) + tan ( α ) tan ( 2 α ) ≡ 1 − tan 2 ( α ) 2 tan ( α )
Example 1 Compute the exact value of ( sin ( π 8 ) + cos ( π 8 ) ) 2 \left(\sin\left(\dfrac{\pi}{8}\right)+ \cos\left(\dfrac{\pi}{8}\right)\right)^2 ( sin ( 8 π ) + cos ( 8 π ) ) 2
Expand the brackets, and use the identity cos 2 ( α ) + sin 2 ( α ) ≡ 1 \cos^2(\alpha) + \sin^2(\alpha) \equiv 1 cos 2 ( α ) + sin 2 ( α ) ≡ 1 sin ( 2 α ) \sin(2\alpha) sin ( 2 α )
( sin ( π 8 ) + cos ( π 8 ) ) 2 = sin 2 ( π 8 ) + cos 2 ( π 8 ) + 2 sin ( π 8 ) cos ( π 8 ) = 1 + sin ( 2 × π 8 ) = 1 + sin ( π 4 ) = 1 + 2 2 = 2 + 2 2 \begin{aligned}\left(\sin\left(\dfrac{\pi}{8}\right) + \cos\left(\dfrac{\pi}{8}\right)\right)^2 &= \sin^2\left(\dfrac{\pi}{8}\right) + \cos^2\left(\dfrac{\pi}{8}\right) + 2\sin\left(\dfrac{\pi}{8}\right)\cos\left(\dfrac{\pi}{8}\right)\\&= 1 + \sin\left(2 \times\dfrac{\pi}{8}\right)\\&= 1 + \sin\left(\dfrac{\pi}{4}\right)\\&= 1 + \dfrac{\sqrt{2}}{2}\\&= \dfrac{2 + \sqrt{2}}{2}\end{aligned} ( sin ( 8 π ) + cos ( 8 π ) ) 2 = sin 2 ( 8 π ) + cos 2 ( 8 π ) + 2 sin ( 8 π ) cos ( 8 π ) = 1 + sin ( 2 × 8 π ) = 1 + sin ( 4 π ) = 1 + 2 2 = 2 2 + 2
Therefore, the exact value of ( sin ( π 8 ) + cos ( π 8 ) ) 2 \left(\sin\left(\dfrac{\pi}{8}\right) + \cos\left(\dfrac{\pi}{8}\right)\right)^2 ( sin ( 8 π ) + cos ( 8 π ) ) 2 2 + 2 2 ‾ \underline{\dfrac{2 + \sqrt{2}}{2}} 2 2 + 2
Example 2 Show that sin ( 4 α ) ≡ 4 sin ( α ) cos 3 ( α ) − 4 sin 3 ( α ) cos ( α ) \sin(4\alpha) \equiv 4\sin(\alpha)\cos^3(\alpha) - 4\sin^3(\alpha)\cos(\alpha) sin ( 4 α ) ≡ 4 sin ( α ) cos 3 ( α ) − 4 sin 3 ( α ) cos ( α ) .
You have that 4 α = 2 × 2 α 4\alpha = 2 \times 2\alpha 4 α = 2 × 2 α , so by the double angle formula for sin \sin sin
sin ( 4 α ) ≡ 2 sin ( 2 α ) cos ( 2 α ) \sin(4\alpha) \equiv 2\sin(2\alpha)\cos(2\alpha) sin ( 4 α ) ≡ 2 sin ( 2 α ) cos ( 2 α )
Apply both the double angle formulae for sin \sin sin cos \cos cos
sin ( 4 α ) ≡ 2 sin ( 2 α ) cos ( 2 α ) ≡ 2 ( 2 sin ( α ) cos ( α ) ) ( cos 2 ( α ) − sin 2 ( α ) ) ≡ 4 sin ( α ) cos 3 ( α ) − 4 sin 3 ( α ) cos ( α ) \begin{aligned}\sin(4\alpha) &\equiv 2\sin(2\alpha)\cos(2\alpha)\\&\equiv 2(2\sin(\alpha)\cos(\alpha))(\cos^2(\alpha) - \sin^2(\alpha))\\&\equiv 4\sin(\alpha)\cos^3(\alpha) - 4\sin^3(\alpha)\cos(\alpha)\end{aligned} sin ( 4 α ) ≡ 2 sin ( 2 α ) cos ( 2 α ) ≡ 2 ( 2 sin ( α ) cos ( α )) ( cos 2 ( α ) − sin 2 ( α )) ≡ 4 sin ( α ) cos 3 ( α ) − 4 sin 3 ( α ) cos ( α )
Therefore, sin ( 4 α ) ≡ 4 sin ( α ) cos 3 ( α ) − 4 sin 3 ( α ) cos ( α ) ‾ \underline{\sin(4\alpha) \equiv 4\sin(\alpha)\cos^3(\alpha) - 4\sin^3(\alpha)\cos(\alpha)} sin ( 4 α ) ≡ 4 sin ( α ) cos 3 ( α ) − 4 sin 3 ( α ) cos ( α )
