More trig identities (HL) That word from English A strikes back where it hurts.
Contents Double angle identity for tan \bm\tan t a n tan 2 θ = 2 tan θ 1 − tan 2 θ \tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} tan 2 θ = 1 − tan 2 θ 2 tan θ Compound angle identities Note, ± \pm ± ∓ \mp ∓
sin ( α ± β ) = sin α cos β ± cos α sin β \sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta sin ( α ± β ) = sin α cos β ± cos α sin β cos ( α ± β ) = cos α cos β ∓ sin α sin β \cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta cos ( α ± β ) = cos α cos β ∓ sin α sin β tan ( α ± β ) = tan α ± tan β 1 ∓ tan α tan β \tan(\alpha \pm \beta) = \frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta} tan ( α ± β ) = 1 ∓ tan α tan β tan α ± tan β Note: α = β \alpha = \beta α = β + + + sin 0 , cos 0 , tan 0 \sin0, \cos0,\tan0 sin 0 , cos 0 , tan 0 − - −
Supplementary angle identities This relates to odd/even symmetry across π 2 rad \frac{\pi}{2} \text{ rad} 2 π rad
sin ( π − θ ) = sin ( θ ) \sin(\pi - \theta) = \sin(\theta) sin ( π − θ ) = sin ( θ ) cos ( π − θ ) = − cos ( θ ) \cos(\pi - \theta) = -\cos(\theta) cos ( π − θ ) = − cos ( θ ) tan ( π − θ ) = − tan ( θ ) \tan(\pi - \theta) = -\tan(\theta) tan ( π − θ ) = − tan ( θ ) Example: Evaluate arctan 5 2 + arctan 7 3 \displaystyle\arctan \frac52 + \arctan\frac73 arctan 2 5 + arctan 3 7
Consider the double angle formula for tan \tan tan
tan ( α + β ) = tan α + tan β 1 − tan α tan β \tan(\alpha + \beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} tan ( α + β ) = 1 − tan α tan β tan α + tan β Let α = arctan 5 2 \alpha = \arctan\frac52 α = arctan 2 5 β = arctan 7 3 \beta = \arctan\frac73 β = arctan 3 7 tan arctan x = x , − π 2 < x < π 2 {\tan \arctan x = x,\,-\frac\pi2 < x < \frac\pi2} tan arctan x = x , − 2 π < x < 2 π
tan ( α + β ) = 5 2 + 7 3 1 − 5 2 ⋅ 7 3 = 29 6 1 − 35 6 = 29 6 − 29 6 = − 1 \begin{align*}
\tan(\alpha + \beta) &= \frac{\frac52+\frac73}{1-\frac52\cdot\frac73} \\
&= \frac{\frac{29}{6}}{1 - \frac{35}{6}} \\
&= \frac{\frac{29}{6}}{\frac{-29}{6}} \\
&= -1
\end{align*} tan ( α + β ) = 1 − 2 5 ⋅ 3 7 2 5 + 3 7 = 1 − 6 35 6 29 = 6 − 29 6 29 = − 1 Given arctan \arctan arctan strictly increasing function , we note that arctan 1 = π 4 \arctan 1 = \frac{\pi}{4} arctan 1 = 4 π π 2 > arctan 5 2 > arctan 7 3 > π 4 {\frac{\pi}{2} > \arctan\frac52 > \arctan\frac73 > \frac\pi4} 2 π > arctan 2 5 > arctan 3 7 > 4 π π > arctan 5 2 + arctan 7 3 > π 2 {\pi > \arctan\frac52 + \arctan\frac73 > \frac\pi2} π > arctan 2 5 + arctan 3 7 > 2 π
This means arctan 5 2 + arctan 7 3 \arctan \frac52 + \arctan\frac73 arctan 2 5 + arctan 3 7 θ \theta θ tan θ = − 1 {\tan \theta = -1} tan θ = − 1
arctan 5 2 + arctan 7 3 = π − π 4 = 3 π 4 ■ \arctan \frac52 + \arctan\frac73 = \pi - \frac\pi4 = \frac{3\pi}{4}\qed arctan 2 5 + arctan 3 7 = π − 4 π = 4 3 π ■ Reciprocal trig functions See also properties of reciprocal functions
Not to be confused with inverse trig functions
csc θ = cosec θ = 1 sin θ \csc\theta = \cosec\theta = \frac{1}{\sin\theta} csc θ = cosec θ = sin θ 1 sec θ = 1 cos θ \sec\theta = \frac{1}{\cos\theta} sec θ = cos θ 1 cot θ = 1 tan θ \cot\theta = \frac{1}{\tan\theta} cot θ = tan θ 1 Pythagorean identities II OG:
sin 2 θ + cos 2 θ = 1 \sin^2\theta + \cos^2\theta = 1 sin 2 θ + cos 2 θ = 1 Dividing all terms by sin 2 θ \sin^2\theta sin 2 θ cos 2 θ \cos^2\theta cos 2 θ
1 + cot 2 θ = cosec 2 θ 1 + \cot^2\theta = \cosec^2\theta 1 + cot 2 θ = cosec 2 θ tan 2 θ + 1 = sec 2 θ \tan^2\theta + 1 = \sec^2\theta tan 2 θ + 1 = sec 2 θ