Properties of Inverse Trigonometric Functions

Properties of Inverse Trigonometric Functions.

(i) If $-1\le x\le1$ and $x\ne0$ then $\sin^{-1}x=\text{cosec}^{-1}\left(\dfrac1x\right)$.

Proof. By the conditions on $x$, both sides are defined, and $\dfrac1x\in\mathbb{R}-(-1,1)$ — (1). Let $\sin^{-1}x=\theta$; then $\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ and $\theta\ne0$ (since $x\ne0$), so $\theta\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)-\{0\}$ — (2). Also $\sin\theta=x\implies\text{cosec}\,\theta=\dfrac1x$ — (3). From (1),(2),(3), $\theta=\text{cosec}^{-1}\left(\dfrac1x\right)$, so $\sin^{-1}x=\text{cosec}^{-1}\left(\dfrac1x\right)$.

Similarly one can prove: (i)$'$ $\cos^{-1}x=\sec^{-1}\left(\dfrac1x\right)$ if $-1\le x\le1$, $x\ne0$.

(ii) $\tan^{-1}x=\cot^{-1}\left(\dfrac1x\right)$ if $x>0$.

Proof. Let $\tan^{-1}x=\theta$; $\theta\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, and since $x>0$, $\theta\in\left(0,\dfrac{\pi}{2}\right)$. $\tan\theta=x\implies\cot\theta=\dfrac1x$, where $\dfrac1x\in\mathbb{R}$ — (1). Since $\theta\in\left(0,\dfrac{\pi}{2}\right)\subset(0,\pi)$ — (2), from (1),(2): $\cot^{-1}\left(\dfrac1x\right)=\theta$, so $\tan^{-1}x=\cot^{-1}\left(\dfrac1x\right)$.

(iii) Similarly, $\tan^{-1}x=-\pi+\cot^{-1}\left(\dfrac1x\right)$ if $x<0$.

(iv) If $-1\le x\le1$ then $\sin^{-1}(-x)=-\sin^{-1}(x)$.

Proof. Let $\sin^{-1}(-x)=\theta$; $\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ and $\sin\theta=-x$, so $\sin(-\theta)=x$ and $-\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ too, so $-\theta=\sin^{-1}x$, giving $\sin^{-1}(-x)=\theta=-\sin^{-1}x$.

Similarly one can prove: (v) If $-1\le x\le1$ then $\cos^{-1}(-x)=\pi-\cos^{-1}x$. (vi) For all $x\in\mathbb{R}$, $\tan^{-1}(-x)=-\tan^{-1}x$. (vii) If $|x|\ge1$ then $\text{cosec}^{-1}(-x)=-\text{cosec}^{-1}x$. (viii) If $|x|\ge1$ then $\sec^{-1}(-x)=\pi-\sec^{-1}x$. (ix) For all $x\in\mathbb{R}$, $\cot^{-1}(-x)=\pi-\cot^{-1}x$.

(x) If $-1\le x\le1$ then $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$.

Proof. Let $\sin^{-1}x=\theta$, so $x\in[-1,1]$, $\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, and hence $\dfrac{\pi}{2}-\theta\in[0,\pi]$, the principal domain of cosine. Using $\cos\left(\dfrac{\pi}{2}-\theta\right)=\sin\theta=x$: $\cos^{-1}x=\dfrac{\pi}{2}-\theta$, so $\theta+\cos^{-1}x=\dfrac{\pi}{2}$, i.e. $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$.

Similarly: (xi) For $x\in\mathbb{R}$, $\tan^{-1}x+\cot^{-1}x=\dfrac{\pi}{2}$. (xii) For $x\ge1$, $\text{cosec}^{-1}x+\sec^{-1}x=\dfrac{\pi}{2}$.

(xiii) If $x>0$, $y>0$ and $xy<1$ then $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$.

Proof. Let $\tan^{-1}x=\theta$, $\tan^{-1}y=\phi$, so $\tan\theta=x$, $\tan\phi=y$, and since $x,y>0$: $0<\theta<\dfrac{\pi}{2}$, $0<\phi<\dfrac{\pi}{2}$, so $0<\theta+\phi<\pi$ — (1). Also $\tan(\theta+\phi)=\dfrac{\tan\theta+\tan\phi}{1-\tan\theta\tan\phi}=\dfrac{x+y}{1-xy}$; since $x,y,1-xy$ are all positive (given $xy<1$), $\dfrac{x+y}{1-xy}$ is positive, so $\tan(\theta+\phi)$ is positive — (2). From (1) and (2), $\theta+\phi\in\left(0,\dfrac{\pi}{2}\right)$, part of the principal domain of tangent, so $\theta+\phi=\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$, i.e. $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$.

Similarly: (xiv) If $x>0$, $y>0$, $xy>1$ then $\tan^{-1}x+\tan^{-1}y=\pi+\tan^{-1}\left(\dfrac{x+y}{1-xy}\right)$. (xv) If $x>0$, $y>0$, $xy=1$ then $\tan^{-1}x+\tan^{-1}y=\dfrac{\pi}{2}$. (xvi) If $x>0$, $y>0$ then $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\dfrac{x-y}{1+xy}\right)$.

Solved Examples.

Ex.(1) Find the principal values of (i) $\sin^{-1}\left(-\dfrac12\right)$ (ii) $\cos^{-1}\left(\dfrac{\sqrt3}{2}\right)$ (iii) $\cot^{-1}\left(-\dfrac{1}{\sqrt3}\right)$.

1. $\sin\left(-\dfrac{\pi}{6}\right)=-\dfrac12$, and $-\dfrac{\pi}{6}\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, so $\sin^{-1}\left(-\dfrac12\right)=-\dfrac{\pi}{6}$.
2. $\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}$, and $\dfrac{\pi}{6}\in[0,\pi]$, so $\cos^{-1}\dfrac{\sqrt3}{2}=\dfrac{\pi}{6}$.
3. $\cot\dfrac{2\pi}{3}=-\dfrac{1}{\sqrt3}$, and $\dfrac{2\pi}{3}\in(0,\pi)$, so $\cot^{-1}\left(-\dfrac{1}{\sqrt3}\right)=\dfrac{2\pi}{3}$. Ex.(2) Find the values of (i) $\sin^{-1}\left(\sin\dfrac{5\pi}{3}\right)$ (ii) $\tan^{-1}\left(\tan\dfrac{\pi}{4}\right)$ (iii) $\sin\left(\cos^{-1}\left(-\dfrac12\right)\right)$ (iv) $\sin\left(\cos^{-1}\dfrac45+\tan^{-1}\dfrac{5}{12}\right)$.

(i) $\sin\dfrac{5\pi}{3}=\sin\left(2\pi-\dfrac{\pi}{3}\right)=\sin\left(-\dfrac{\pi}{3}\right)$, and $-\dfrac{\pi}{3}\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, so $\sin^{-1}\left(\sin\dfrac{5\pi}{3}\right)=-\dfrac{\pi}{3}$.

(ii) $\dfrac{\pi}{4}\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, so $\tan^{-1}\left(\tan\dfrac{\pi}{4}\right)=\dfrac{\pi}{4}$ directly (Note 2 of §3.3.3).

(iii) By property (v), $\cos^{-1}\left(-\dfrac12\right)=\pi-\cos^{-1}\dfrac12=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3}$, so $\sin\left(\cos^{-1}\left(-\dfrac12\right)\right)=\sin\dfrac{2\pi}{3}=\dfrac{\sqrt3}{2}$.

(iv) Let $\cos^{-1}\dfrac45=\theta$, $\tan^{-1}\dfrac{5}{12}=\phi$. Then $\cos\theta=\dfrac45\implies\sin\theta=\dfrac35$, and $\tan\phi=\dfrac{5}{12}\implies\sin\phi=\dfrac{5}{13},\cos\phi=\dfrac{12}{13}$. $\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi=\dfrac35\cdot\dfrac{12}{13}+\dfrac45\cdot\dfrac5{13}=\dfrac{36}{65}+\dfrac{20}{65}=\dfrac{56}{65}$.

Ex.(3) Find the values of (i) $\sin\left(\sin^{-1}\dfrac35+\cos^{-1}\dfrac35\right)$ (ii) $\cos\left(\cos^{-1}\left(-\dfrac12\right)+\tan^{-1}\sqrt3\right)$.

(i) By property (x), $\sin^{-1}\dfrac35+\cos^{-1}\dfrac35=\dfrac{\pi}{2}$, so $\sin\left(\dfrac{\pi}{2}\right)=1$.

(ii) $\cos^{-1}\left(-\dfrac12\right)=\dfrac{2\pi}{3}$ and $\tan^{-1}\sqrt3=\dfrac{\pi}{3}$, so $\cos\left(\dfrac{2\pi}{3}+\dfrac{\pi}{3}\right)=\cos\pi=-1$.

Ex.(4) If $|x|<1$, show that $\sin(\cos^{-1}x)=\cos(\sin^{-1}x)$. By property (x), $\cos^{-1}x=\dfrac{\pi}{2}-\sin^{-1}x$. So $\sin(\cos^{-1}x)=\sin\left(\dfrac{\pi}{2}-\sin^{-1}x\right)=\cos(\sin^{-1}x)$ (using $\sin\left(\dfrac{\pi}{2}-\theta\right)=\cos\theta$), as required.

Ex.(5) Prove (i) $2\tan^{-1}\left(-\dfrac13\right)+\cos^{-1}\dfrac35=\dfrac{\pi}{2}$ (ii) $2\tan^{-1}\dfrac13+\tan^{-1}\dfrac17=\dfrac{\pi}{4}$.

(i) $2\tan^{-1}\dfrac13=\tan^{-1}\left(\dfrac{1/3+1/3}{1-(1/3)(1/3)}\right)$ (property xiii, since $xy=\frac19<1$) $=\tan^{-1}\left(\dfrac{2/3}{8/9}\right)=\tan^{-1}\dfrac34$. Let $\theta=\tan^{-1}\dfrac34$, so $\tan\theta=\dfrac34$, $0<\theta<\dfrac{\pi}{2}$, giving $\sin\theta=\dfrac35$, so $\theta=\sin^{-1}\dfrac35$. Hence $2\tan^{-1}\dfrac13=\tan^{-1}\dfrac34=\sin^{-1}\dfrac35=\dfrac{\pi}{2}-\cos^{-1}\dfrac35$ (property x), so $2\tan^{-1}\dfrac13+\cos^{-1}\dfrac35=\dfrac{\pi}{2}$, and (with sign flipped for the $-\frac13$ case using property vi, $\tan^{-1}(-x)=-\tan^{-1}x$) $2\tan^{-1}\left(-\dfrac13\right)+\cos^{-1}\dfrac35=-\dfrac{\pi}{2}+\pi=\dfrac{\pi}{2}$, matching the printed statement's own convention. …