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Mathematics · Ch 4 — Inverse Trigonometric Functions

Properties of Inverse Trigonometric Functions

4.10

Properties of Inverse Trigonometric Functions

This section develops the ten families of working properties of inverse trigonometric functions — valid strictly WITHIN each function's principal value branch, and only where each side is defined.

Property I — undoing the outer inverse (f−1(f(θ))=θf^{-1}(f(\theta))=\theta, conditional).

sin⁡−1(sin⁡θ)=θ, θ∈[−π2,π2]cos⁡−1(cos⁡θ)=θ, θ∈[0,π]tan⁡−1(tan⁡θ)=θ, θ∈(−π2,π2)\sin^{-1}(\sin\theta)=\theta,\ \theta\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right] \qquad \cos^{-1}(\cos\theta)=\theta,\ \theta\in[0,\pi] \qquad \tan^{-1}(\tan\theta)=\theta,\ \theta\in\left(-\tfrac{\pi}2,\tfrac{\pi}2\right)

cosec−1(cosec θ)=θ, θ∈[−π2,π2]∖{0}sec⁡−1(sec⁡θ)=θ, θ∈[0,π]∖{π2}cot⁡−1(cot⁡θ)=θ, θ∈(0,π)\text{cosec}^{-1}(\text{cosec}\,\theta)=\theta,\ \theta\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]\setminus\{0\} \qquad \sec^{-1}(\sec\theta)=\theta,\ \theta\in[0,\pi]\setminus\left\{\tfrac{\pi}2\right\} \qquad \cot^{-1}(\cot\theta)=\theta,\ \theta\in(0,\pi)

Proof (sine case). Let sin⁡θ=x\sin\theta=x, θ∈[−π2,π2]\theta\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]. By the very definition of inverse sine, sin⁡θ=x⇒θ=sin⁡−1x\sin\theta=x\Rightarrow\theta=\sin^{-1}x, so sin⁡−1(sin⁡θ)=θ\sin^{-1}(\sin\theta)=\theta. The other five follow identically from their own definitions.

Property II — undoing the inner inverse (f(f−1(x))=xf(f^{-1}(x))=x, unconditional on the whole domain).

sin⁡(sin⁡−1x)=x, x∈[−1,1]cos⁡(cos⁡−1x)=x, x∈[−1,1]tan⁡(tan⁡−1x)=x, x∈R\sin(\sin^{-1}x)=x,\ x\in[-1,1] \qquad \cos(\cos^{-1}x)=x,\ x\in[-1,1] \qquad \tan(\tan^{-1}x)=x,\ x\in\mathbb{R}

cosec(cosec−1x)=x, x∈R∖(−1,1)sec⁡(sec⁡−1x)=x, x∈R∖(−1,1)cot⁡(cot⁡−1x)=x, x∈R\text{cosec}(\text{cosec}^{-1}x)=x,\ x\in\mathbb{R}\setminus(-1,1) \qquad \sec(\sec^{-1}x)=x,\ x\in\mathbb{R}\setminus(-1,1) \qquad \cot(\cot^{-1}x)=x,\ x\in\mathbb{R}

Proof (sine case). For x∈[−1,1]x\in[-1,1], sin⁡−1x\sin^{-1}x is well defined; let sin⁡−1x=θ\sin^{-1}x=\theta, so by definition θ∈[−π2,π2]\theta\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right] and sin⁡θ=x\sin\theta=x. Thus sin⁡(sin⁡−1x)=x\sin(\sin^{-1}x)=x directly.

Note

For ANY trigonometric function y=f(x)y=f(x), f(f−1(x))=xf(f^{-1}(x))=x for every xx in the range of ff — this follows straight from the definition of f−1f^{-1}. Evaluating f(f−1(x))f(f^{-1}(x)): (a) if xx is already in ff's restricted (principal) domain, f(f−1(x))=xf(f^{-1}(x))=x directly; (b) if not, first find x1x_1 INSIDE ff's restricted domain with f(x)=f(x1)f(x)=f(x_1), and then f(f−1(x))=f(f−1(x1))=x1f(f^{-1}(x))=f(f^{-1}(x_1))=x_1. E.g. sin⁡−1(x)=x\sin^{-1}(x)=x if x∈[−π2,π2]x\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right] but =π−x=\pi-x if x∈[π2,3π2]x\in\left[\tfrac{\pi}2,\tfrac{3\pi}2\right]. For an expression f(g−1(x))f(g^{-1}(x)) with f≠gf\ne g (e.g. cot⁡(sin⁡−1x)\cot(\sin^{-1}x)), a reference triangle built from g−1g^{-1}'s definition is usually the way in (§4.10's composite-function properties, below).

Property III — reciprocal identities.

sin⁡−1(1x)=cosec−1x, x∈R∖(−1,1)cos⁡−1(1x)=sec⁡−1x, x∈R∖(−1,1)\sin^{-1}\left(\dfrac1x\right)=\text{cosec}^{-1}x,\ x\in\mathbb{R}\setminus(-1,1) \qquad \cos^{-1}\left(\dfrac1x\right)=\sec^{-1}x,\ x\in\mathbb{R}\setminus(-1,1)

tan⁡−1(1x)={cot⁡−1x,x>0−π+cot⁡−1x,x<0\tan^{-1}\left(\dfrac1x\right)=\begin{cases}\cot^{-1}x,&x>0\\-\pi+\cot^{-1}x,&x<0\end{cases}

Proof (sine case). For x∈R∖(−1,1)x\in\mathbb{R}\setminus(-1,1), 1x∈[−1,1]\tfrac1x\in[-1,1] and x≠0x\ne0, so sin⁡−1(1x)\sin^{-1}\left(\tfrac1x\right) is well defined. Let sin⁡−1(1x)=θ\sin^{-1}\left(\tfrac1x\right)=\theta; then θ∈[−π2,π2]∖{0}\theta\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right]\setminus\{0\} and sin⁡θ=1x\sin\theta=\tfrac1x, so cosec θ=x\text{cosec}\,\theta=x, giving θ=cosec−1x\theta=\text{cosec}^{-1}x. So sin⁡−1(1x)=θ=cosec−1x\sin^{-1}\left(\tfrac1x\right)=\theta=\text{cosec}^{-1}x.

Property IV — reflection identities (negating the argument).

sin⁡−1(−x)=−sin⁡−1x, x∈[−1,1]tan⁡−1(−x)=−tan⁡−1x, x∈Rcosec−1(−x)=−cosec−1x\sin^{-1}(-x)=-\sin^{-1}x,\ x\in[-1,1] \qquad \tan^{-1}(-x)=-\tan^{-1}x,\ x\in\mathbb{R} \qquad \text{cosec}^{-1}(-x)=-\text{cosec}^{-1}x

cos⁡−1(−x)=π−cos⁡−1x, x∈[−1,1]sec⁡−1(−x)=π−sec⁡−1xcot⁡−1(−x)=π−cot⁡−1x, x∈R\cos^{-1}(-x)=\pi-\cos^{-1}x,\ x\in[-1,1] \qquad \sec^{-1}(-x)=\pi-\sec^{-1}x \qquad \cot^{-1}(-x)=\pi-\cot^{-1}x,\ x\in\mathbb{R}

Proof (cosine case). For x∈[−1,1]x\in[-1,1], −x∈[−1,1]-x\in[-1,1], so cos⁡−1(−x)\cos^{-1}(-x) is well defined. Let cos⁡−1(−x)=θ\cos^{-1}(-x)=\theta; then θ∈[0,π]\theta\in[0,\pi] and cos⁡θ=−x\cos\theta=-x. Now cos⁡θ=−x⇒cos⁡(π−θ)=x\cos\theta=-x\Rightarrow\cos(\pi-\theta)=x (since cos⁡(π−θ)=−cos⁡θ\cos(\pi-\theta)=-\cos\theta), and π−θ\pi-\theta gives x=cos⁡(π−θ)⇒cos⁡−1x=π−θ⇒θ=π−cos⁡−1xx=\cos(\pi-\theta)\Rightarrow\cos^{-1}x=\pi-\theta\Rightarrow\theta=\pi-\cos^{-1}x. So cos⁡−1(−x)=π−cos⁡−1x\cos^{-1}(-x)=\pi-\cos^{-1}x.

Note

The inverse of a one-to-one, ODD function is again odd (as with sin⁡−1,tan⁡−1,cosec−1\sin^{-1},\tan^{-1},\text{cosec}^{-1}). Asking whether the inverse of an EVEN function is even does not even make sense — an even function cannot be one-to-one anywhere except at 00, so it never has a genuine inverse to test; cos⁡−1x,sec⁡−1x\cos^{-1}x,\sec^{-1}x are simply neither even nor odd.

Property V — cofunction inverse identities.

sin⁡−1x+cos⁡−1x=π2, x∈[−1,1]tan⁡−1x+cot⁡−1x=π2, x∈Rcosec−1x+sec⁡−1x=π2, x∈R∖(−1,1) or x≥1\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}2,\ x\in[-1,1] \qquad \tan^{-1}x+\cot^{-1}x=\dfrac{\pi}2,\ x\in\mathbb{R} \qquad \text{cosec}^{-1}x+\sec^{-1}x=\dfrac{\pi}2,\ x\in\mathbb{R}\setminus(-1,1)\text{ or }x\ge1

Proof (sine case). Let sin⁡−1x=θ\sin^{-1}x=\theta, so θ∈[−π2,π2]\theta\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right] and sin⁡θ=x\sin\theta=x. Since −π2≤θ≤π2  ⟺  0≤π2−θ≤π-\tfrac{\pi}2\le\theta\le\tfrac{\pi}2\iff0\le\tfrac{\pi}2-\theta\le\pi, and cos⁡(π2−θ)=sin⁡θ=x\cos\left(\tfrac{\pi}2-\theta\right)=\sin\theta=x, we get cos⁡−1x=π2−θ\cos^{-1}x=\tfrac{\pi}2-\theta, i.e. cos⁡−1x=π2−sin⁡−1x\cos^{-1}x=\tfrac{\pi}2-\sin^{-1}x.

Property VI — addition and subtraction formulas.

sin⁡−1x+sin⁡−1y=sin⁡−1(x1−y2+y1−x2),if x2+y2≤1 or xy<0\sin^{-1}x+\sin^{-1}y=\sin^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right),\quad\text{if } x^2+y^2\le1 \text{ or } xy<0

sin⁡−1x−sin⁡−1y=sin⁡−1(x1−y2−y1−x2),if x2+y2≤1 or xy>0\sin^{-1}x-\sin^{-1}y=\sin^{-1}\left(x\sqrt{1-y^2}-y\sqrt{1-x^2}\right),\quad\text{if } x^2+y^2\le1 \text{ or } xy>0

cos⁡−1x+cos⁡−1y=cos⁡−1(xy−1−x21−y2),if x+y≥0\cos^{-1}x+\cos^{-1}y=\cos^{-1}\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right),\quad\text{if } x+y\ge0

cos⁡−1x−cos⁡−1y=cos⁡−1(xy+1−x21−y2),if x≤y\cos^{-1}x-\cos^{-1}y=\cos^{-1}\left(xy+\sqrt{1-x^2}\sqrt{1-y^2}\right),\quad\text{if } x\le y

tan⁡−1x+tan⁡−1y=tan⁡−1(x+y1−xy), xy<1tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy), xy>−1\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\dfrac{x+y}{1-xy}\right),\ xy<1 \qquad \tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\dfrac{x-y}{1+xy}\right),\ xy>-1

Proof (sine addition case). Let A=sin⁡−1xA=\sin^{-1}x, B=sin⁡−1yB=\sin^{-1}y; so A,B∈[−π2,π2]A,B\in\left[-\tfrac{\pi}2,\tfrac{\pi}2\right], x=sin⁡Ax=\sin A, y=sin⁡By=\sin B, and cos⁡A,cos⁡B≥0\cos A,\cos B\ge0, giving cos⁡A=1−x2\cos A=\sqrt{1-x^2}, cos⁡B=1−y2\cos B=\sqrt{1-y^2}. Then sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B=x1−y2+y1−x2\sin(A+B)=\sin A\cos B+\cos A\sin B=x\sqrt{1-y^2}+y\sqrt{1-x^2}, with x,y≤1x,y\le1 so x2+y2≤1x^2+y^2\le1 (or, with mixed signs, xy<0xy<0); hence A+B=sin⁡−1(x1−y2+y1−x2)A+B=\sin^{-1}\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right).

Property VII — double-angle-style formulas (setting y=xy=x in Property VI).

2tan⁡−1x=tan⁡−1(2x1−x2), ∣x∣<12tan⁡−1x=cos⁡−1(1−x21+x2), x≥02tan⁡−1x=sin⁡−1(2x1+x2), ∣x∣≤12\tan^{-1}x=\tan^{-1}\left(\dfrac{2x}{1-x^2}\right),\ |x|<1 \qquad 2\tan^{-1}x=\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right),\ x\ge0 \qquad 2\tan^{-1}x=\sin^{-1}\left(\dfrac{2x}{1+x^2}\right),\ |x|\le1

Proof (cosine case). Let θ=2tan⁡−1x\theta=2\tan^{-1}x, so tan⁡θ2=x\tan\tfrac{\theta}2=x. The identity cos⁡θ=1−tan⁡2(θ/2)1+tan⁡2(θ/2)=1−x21+x2\cos\theta=\dfrac{1-\tan^2(\theta/2)}{1+\tan^2(\theta/2)}=\dfrac{1-x^2}{1+x^2} gives θ=cos⁡−1(1−x21+x2)\theta=\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right), i.e. 2tan⁡−1x=cos⁡−1(1−x21+x2)2\tan^{-1}x=\cos^{-1}\left(\dfrac{1-x^2}{1+x^2}\right), x≥0x\ge0.

Property VIII — a same-left-side, two-different-right-sides pair.

sin⁡−1(2x1−x2)=2sin⁡−1x,∣x∣≤12(equivalently −12≤x≤12)\sin^{-1}\left(2x\sqrt{1-x^2}\right)=2\sin^{-1}x,\quad |x|\le\dfrac1{\sqrt2}\quad\text{(equivalently }-\tfrac1{\sqrt2}\le x\le\tfrac1{\sqrt2}\text{)}

sin⁡−1(2x1−x2)=2cos⁡−1x,12≤x≤1\sin^{-1}\left(2x\sqrt{1-x^2}\right)=2\cos^{-1}x,\quad \dfrac1{\sqrt2}\le x\le1

Proof. Let x=sin⁡θx=\sin\theta; then 2x1−x2=2sin⁡θcos⁡θ=sin⁡2θ2x\sqrt{1-x^2}=2\sin\theta\cos\theta=\sin2\theta, so sin⁡−1(2x1−x2)=sin⁡−1(sin⁡2θ)=2θ=2sin⁡−1x\sin^{-1}\left(2x\sqrt{1-x^2}\right)=\sin^{-1}(\sin2\theta)=2\theta=2\sin^{-1}x (when 2θ2\theta stays in range). The cos⁡−1\cos^{-1} form on the upper piece of the domain is proved the same way with x=cos⁡θx=\cos\theta.

Property IX — mixed cofunction/Pythagorean substitutions.

sin⁡−1x=cos⁡−11−x2, 0≤x≤1sin⁡−1x=−cos⁡−11−x2, −1≤x<0\sin^{-1}x=\cos^{-1}\sqrt{1-x^2},\ 0\le x\le1 \qquad \sin^{-1}x=-\cos^{-1}\sqrt{1-x^2},\ -1\le x<0

cos⁡−1x=sin⁡−11−x2, 0≤x≤1cos⁡−1x=π−sin⁡−11−x2, −1≤x<0\cos^{-1}x=\sin^{-1}\sqrt{1-x^2},\ 0\le x\le1 \qquad \cos^{-1}x=\pi-\sin^{-1}\sqrt{1-x^2},\ -1\le x<0

tan⁡−1x=sin⁡−1(x1+x2)=cos⁡−1(11+x2),x>0\tan^{-1}x=\sin^{-1}\left(\dfrac{x}{\sqrt{1+x^2}}\right)=\cos^{-1}\left(\dfrac1{\sqrt{1+x^2}}\right),\quad x>0

Proof (first case). Let sin⁡−1x=θ\sin^{-1}x=\theta; since 0≤x≤10\le x\le1, θ∈[0,π2]\theta\in\left[0,\tfrac{\pi}2\right]. Then cos⁡θ=1−sin⁡2θ=1−x2\cos\theta=\sqrt{1-\sin^2\theta}=\sqrt{1-x^2}, i.e. cos⁡−11−x2=θ=sin⁡−1x\cos^{-1}\sqrt{1-x^2}=\theta=\sin^{-1}x. …

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