Reciprocal and inverse trigonometric functionsEdexcel A-Level Maths: Revision notes

The three reciprocal functions are defined by $\sec\theta=\frac{1}{\cos\theta},\qquad\mathrm{cosec}\,\theta=\frac{1}{\sin\theta},\qquad\cot\theta=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}.$ $\sec\theta$ is undefined where $\cos\theta=0$ ($\theta=90^\circ+180^\circ n$); $\mathrm{cosec}\,\theta$ and $\cot\theta$ are undefined where $\sin\theta=0$ ($\theta=180^\circ n$). Both $\sec\theta$ and $\mathrm{cosec}\,\theta$ have range $y\le-1$ or $y\ge1$, and $\cot\theta$ takes every real value. Their graphs have the same periods as the originals ($360^\circ$ for $\sec$ and $\mathrm{cosec}$, $180^\circ$ for $\cot$) with vertical asymptotes where the denominator is zero. The graph of $\sec\theta$ touches $\pm1$ where $\cos\theta=\pm1$.

$\arcsin$, $\arccos$ and $\arctan$ (also written $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$) reverse $\sin$, $\cos$ and $\tan$ on restricted domains so that each is a function: $\arcsin x:\ \text{domain }[-1,1],\ \text{range }\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ $\arccos x:\ \text{domain }[-1,1],\ \text{range }[0,\pi]$ $\arctan x:\ \text{domain }\mathbb{R},\ \text{range }\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$ Their graphs are reflections of the restricted $\sin$, $\cos$ and $\tan$ graphs in $y=x$. $\arctan x$ has horizontal asymptotes at $y=\pm\frac{\pi}{2}$. In degrees the ranges are $[-90^\circ,90^\circ]$, $[0^\circ,180^\circ]$ and $(-90^\circ,90^\circ)$.

$\sin(\arcsin x)=x$ for $-1\le x\le1$, but $\arcsin(\sin x)=x$ only when $x$ is in the principal range. Example: $\arcsin\left(\sin\frac{5\pi}{6}\right)=\arcsin\frac12=\frac{\pi}{6}$, not $\frac{5\pi}{6}$. Likewise $\arccos\left(\cos\left(-\frac{\pi}{3}\right)\right)=\frac{\pi}{3}$. Exact values: $\arctan1=\frac{\pi}{4}$, $\arctan(-1)=-\frac{\pi}{4}$, $\arccos\left(-\frac12\right)=\frac{2\pi}{3}$.

Divide $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ and by $\sin^2\theta$: $\sec^2\theta=1+\tan^2\theta,\qquad\mathrm{cosec}^2\theta=1+\cot^2\theta.$ Example: if $\tan\theta=\frac{5}{12}$ and $\theta$ is acute then $\sec^2\theta=\frac{169}{144}$, so $\sec\theta=\frac{13}{12}$, and $\cot\theta=\frac{12}{5}$, giving $\mathrm{cosec}^2\theta=\frac{169}{25}$ and $\mathrm{cosec}\,\theta=\frac{13}{5}$.

Rewrite $\sec$, $\mathrm{cosec}$ and $\cot$ as $\sin$ and $\cos$, combine over a common denominator and use $\sin^2\theta+\cos^2\theta=1$. Example: $\cot\theta+\tan\theta=\frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta}=\frac{1}{\sin\theta\cos\theta}=\sec\theta\,\mathrm{cosec}\,\theta$, and $\frac{1}{1-\sin\theta}+\frac{1}{1+\sin\theta}=\frac{2}{1-\sin^2\theta}=\frac{2}{\cos^2\theta}=2\sec^2\theta$. Work on one side only, and write each line.

Replace the reciprocal functions or use the squared identities to reach an equation in one function, then solve in the interval. Half angles: for $\mathrm{cosec}\frac{\theta}{2}=2$ with $0\le\theta<360^\circ$, $\sin\frac{\theta}{2}=\frac12$ with $0\le\frac{\theta}{2}<180^\circ$, so $\frac{\theta}{2}=30^\circ,150^\circ$ and $\theta=60^\circ,300^\circ$. Quadratics: $\tan^2\theta+\sec\theta=1$ becomes $\sec^2\theta+\sec\theta-2=0$, so $\sec\theta=1$ or $-2$, giving $\theta=0^\circ,120^\circ,240^\circ$. For $\cot^2x+\mathrm{cosec}\,x=5$ use $\cot^2x=\mathrm{cosec}^2x-1$, leading to $\mathrm{cosec}\,x=2$ or $-3$. Work in radians when the interval is in radians.