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

The three reciprocal functions are $\sec\theta=\frac{1}{\cos\theta},\qquad\operatorname{cosec}\theta=\frac{1}{\sin\theta},\qquad\cot\theta=\frac{1}{\tan\theta}=\frac{\cos\theta}{\sin\theta}.$ Each is undefined where its denominator is zero: $\sec\theta$ when $\theta=\frac{\pi}{2}+n\pi$; $\operatorname{cosec}\theta$ and $\cot\theta$ when $\theta=n\pi$ (that is $0^\circ,180^\circ,\ldots$). Example: if $\sin\theta=\frac35$ with $\theta$ acute then $\cos\theta=\frac45$, $\operatorname{cosec}\theta=\frac53$, $\sec\theta=\frac54$ and $\cot\theta=\frac43$.

The graphs of $y=\sec x$ and $y=\operatorname{cosec}x$ have vertical asymptotes where $\cos x=0$ or $\sin x=0$. Both have period $2\pi$ and, because $|\sin x|\le1$ and $|\cos x|\le1$, their values satisfy $y\le-1$ or $y\ge1$ (they never lie between $-1$ and $1$). Wherever $\sin x$ or $\cos x$ equals $\pm1$, the reciprocal equals the same value and touches the original graph. $y=\cot x$ has period $\pi$, asymptotes at $x=n\pi$ and takes every real value, so its range is all real numbers. $y=\sec x$: domain $x\ne\frac{\pi}{2}+n\pi$. $y=\operatorname{cosec}x$ and $y=\cot x$: domain $x\ne n\pi$.

Divide $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ to get $\sec^2\theta=1+\tan^2\theta.$ Divide by $\sin^2\theta$ to get $\operatorname{cosec}^2\theta=1+\cot^2\theta.$ Use them to convert an equation involving $\sec^2$ or $\operatorname{cosec}^2$ and a single other function into a quadratic. Example: $2\sec^2\theta=5-\tan\theta$ becomes $2\tan^2\theta+\tan\theta-3=0$, so $\tan\theta=1$ or $-\frac32$.

The inverse functions return an angle from a ratio, using these principal value ranges:

• $y=\arcsin x$: domain $-1\le x\le1$, range $-\frac{\pi}{2}\le y\le\frac{\pi}{2}$.
• $y=\arccos x$: domain $-1\le x\le1$, range $0\le y\le\pi$.
• $y=\arctan x$: domain all real $x$, range $-\frac{\pi}{2}<y<\frac{\pi}{2}$. Each graph is the reflection of the restricted original in the line $y=x$. Examples: $\arccos\left(-\frac12\right)=\frac{2\pi}{3}$, $\arctan(-\sqrt3)=-\frac{\pi}{3}$, $\arcsin\left(-\frac12\right)=-\frac{\pi}{6}$. Also $\sin(\arcsin x)=x$ for $-1\le x\le1$, but $\arcsin(\sin x)=x$ only when $x$ is in the range: $\arcsin\left(\sin\frac{5\pi}{6}\right)=\frac{\pi}{6}$.

To solve, rewrite with a single function, factorise and find all solutions in the interval. Example: $\operatorname{cosec}^2\theta-3\cot\theta-1=0$ becomes $\cot^2\theta-3\cot\theta=0$, so $\cot\theta=0$ ($\theta=\frac{\pi}{2},\frac{3\pi}{2}$) or $\cot\theta=3$ ($\tan\theta=\frac13$, $\theta=0.322,\ 3.46$). To prove an identity, work on one side only. Write everything in $\sin$ and $\cos$, use a common denominator, and finish with $\sin^2\theta+\cos^2\theta=1$: $\tan\theta+\cot\theta=\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}=\sec\theta\operatorname{cosec}\theta.$