More trig identities (HL)

That word from English A strikes back where it hurts.

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Double angle identity for tan⁡\bm\tan

tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}

Compound angle identities

Note, ±\pm and ∓\mp in one equation means all use top or all use bottom

sin⁡(α±β)=sin⁡αcos⁡β±cos⁡αsin⁡β\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \cos\alpha\sin\beta
cos⁡(α±β)=cos⁡αcos⁡β∓sin⁡αsin⁡β\cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta
tan⁡(α±β)=tan⁡α±tan⁡β1∓tan⁡αtan⁡β\tan(\alpha \pm \beta) = \frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\tan\beta}

Note: α=β\alpha = \beta simplifies to the double angle identities for ++, and the values for sin⁡0,cos⁡0,tan⁡0\sin0, \cos0,\tan0 for −-.

Supplementary angle identities

This relates to odd/even symmetry across π2 rad\frac{\pi}{2} \text{ rad}

sin⁡(π−θ)=sin⁡(θ)\sin(\pi - \theta) = \sin(\theta)
cos⁡(π−θ)=−cos⁡(θ)\cos(\pi - \theta) = -\cos(\theta)
tan⁡(π−θ)=−tan⁡(θ)\tan(\pi - \theta) = -\tan(\theta)

Example: Evaluate arctan⁡52+arctan⁡73\displaystyle\arctan \frac52 + \arctan\frac73.

Consider the double angle formula for tan⁡\tan

tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β\tan(\alpha + \beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}

Let α=arctan⁡52\alpha = \arctan\frac52 and β=arctan⁡73\beta = \arctan\frac73. And use tan⁡arctan⁡x=x, −π2<x<π2{\tan \arctan x = x,\,-\frac\pi2 < x < \frac\pi2} to simplify the right hand side.

tan⁡(α+β)=52+731−52⋅73=2961−356=296−296=−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*}

Given arctan⁡\arctan is a strictly increasing function, we note that arctan⁡1=π4\arctan 1 = \frac{\pi}{4} and π2>arctan⁡52>arctan⁡73>π4{\frac{\pi}{2} > \arctan\frac52 > \arctan\frac73 > \frac\pi4}, this also means that π>arctan⁡52+arctan⁡73>π2{\pi > \arctan\frac52 + \arctan\frac73 > \frac\pi2}.

This means arctan⁡52+arctan⁡73\arctan \frac52 + \arctan\frac73 is the angle θ\theta in quadrant II such that tan⁡θ=−1{\tan \theta = -1}.

arctan⁡52+arctan⁡73=π−π4=3π4■\arctan \frac52 + \arctan\frac73 = \pi - \frac\pi4 = \frac{3\pi}{4}\qed

Reciprocal trig functions

See also properties of reciprocal functions

Not to be confused with inverse trig functions

csc⁡θ=cosec⁡θ=1sin⁡θ\csc\theta = \cosec\theta = \frac{1}{\sin\theta}
sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}
cot⁡θ=1tan⁡θ\cot\theta = \frac{1}{\tan\theta}

Pythagorean identities II

OG:

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

Dividing all terms by sin⁡2θ\sin^2\theta, or by cos⁡2θ\cos^2\theta, result in

1+cot⁡2θ=cosec⁡2θ1 + \cot^2\theta = \cosec^2\theta
tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta
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