Each reciprocal function is undefined wherever the original function is zero, and the graph has a vertical asymptote there.
| Function | Undefined when | Range | Period |
|---|---|---|---|
| $y = \sec x$ | $\cos x = 0$: $x = \tfrac{\pi}{2} + n\pi$ | $y \leqslant -1$ or $y \geqslant 1$ | $2\pi$ |
| $y = \operatorname{cosec} x$ | $\sin x = 0$: $x = n\pi$ | $y \leqslant -1$ or $y \geqslant 1$ | $2\pi$ |
| $y = \cot x$ | $\tan x = 0$: $x = n\pi$ | all real $y$ | $\pi$ |
(Here $n$ is any integer.)
$y = \sec x$ (solid) for $0 \leqslant x \leqslant 2\pi$. Where $\cos x = \pm1$ the two graphs touch; where $\cos x = 0$, $\sec x$ has an asymptote.
Find the exact values of $\sec\dfrac{\pi}{3}$, $\operatorname{cosec}\dfrac{5\pi}{4}$ and $\cot\dfrac{\pi}{6}$.
Divide $\sin^2\theta + \cos^2\theta \equiv 1$ by $\cos^2\theta$, and then by $\sin^2\theta$:
$\dfrac{\sin^2\theta}{\cos^2\theta} + 1 \equiv \dfrac{1}{\cos^2\theta} \;\Rightarrow\; 1 + \tan^2\theta \equiv \sec^2\theta$
$1 + \dfrac{\cos^2\theta}{\sin^2\theta} \equiv \dfrac{1}{\sin^2\theta} \;\Rightarrow\; 1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta$
Solve $2\tan^2\theta + \sec\theta = 1$ for $0^\circ \leqslant \theta \leqslant 360^\circ$.
Solve $2\cot^2 x = 7\operatorname{cosec} x - 8$ for $0^\circ < x < 360^\circ$.
Prove that $\sec\theta - \cos\theta \equiv \sin\theta\tan\theta$.
$\sin x$ is many-to-one, so it has no inverse over its whole domain. Restrict the domain to a section where it is one-to-one and covers every output once, and then it does. The same goes for $\cos$ and $\tan$.
| Function | Domain | Range | Restriction of the original |
|---|---|---|---|
| $y = \arcsin x$ | $-1 \leqslant x \leqslant 1$ | $-\tfrac{\pi}{2} \leqslant y \leqslant \tfrac{\pi}{2}$ | $\sin x$ with $-\tfrac{\pi}{2} \leqslant x \leqslant \tfrac{\pi}{2}$ |
| $y = \arccos x$ | $-1 \leqslant x \leqslant 1$ | $0 \leqslant y \leqslant \pi$ | $\cos x$ with $0 \leqslant x \leqslant \pi$ |
| $y = \arctan x$ | all real $x$ | $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$ | $\tan x$ with $-\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$ |
$\arcsin$ and $\arccos$ stop at $x = \pm1$; $\arctan$ continues for all $x$ and approaches the asymptotes $y = \pm\tfrac{\pi}{2}$.
Find the exact values of $\arcsin\left(-\tfrac12\right)$, $\arccos\left(-\tfrac{\sqrt2}{2}\right)$ and $\arctan\sqrt3$.
Find the exact value of $\sin\left(\arccos\tfrac35\right)$, and show that $\cos(\arcsin x) = \sqrt{1 - x^2}$ for $-1 \leqslant x \leqslant 1$.
Reciprocals
$\sec = \tfrac{1}{\cos}$, $\operatorname{cosec} = \tfrac{1}{\sin}$, $\cot = \tfrac{1}{\tan}$
Asymptotes
Where the original function is zero.
Range of sec, cosec
$y \leqslant -1$ or $y \geqslant 1$
Identity 1
$1 + \tan^2\theta \equiv \sec^2\theta$
Identity 2
$1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta$
arcsin
$-1 \leqslant x \leqslant 1$, $-\tfrac{\pi}{2} \leqslant y \leqslant \tfrac{\pi}{2}$
arccos
$-1 \leqslant x \leqslant 1$, $0 \leqslant y \leqslant \pi$
arctan
All real $x$, $-\tfrac{\pi}{2} < y < \tfrac{\pi}{2}$
Notation
$\sin^{-1}x = \arcsin x \neq \operatorname{cosec} x$
Find the exact values of $\sec\dfrac{\pi}{4}$, $\operatorname{cosec}\dfrac{5\pi}{6}$ and $\cot\dfrac{2\pi}{3}$.
โถ Show solution
$\cos\tfrac{\pi}{4} = \tfrac{\sqrt2}{2}$, so $\sec\tfrac{\pi}{4} = \tfrac{2}{\sqrt2} = \sqrt2$.
$\sin\tfrac{5\pi}{6} = \tfrac12$, so $\operatorname{cosec}\tfrac{5\pi}{6} = 2$.
$\tan\tfrac{2\pi}{3} = -\sqrt3$, so $\cot\tfrac{2\pi}{3} = -\tfrac{1}{\sqrt3} = -\tfrac{\sqrt3}{3}$.
Find the exact values of $\arcsin\dfrac{\sqrt3}{2}$, $\arccos\left(-\dfrac12\right)$ and $\arctan(-1)$.
โถ Show solution
$\arcsin\tfrac{\sqrt3}{2} = \tfrac{\pi}{3}$.
$\arccos\left(-\tfrac12\right) = \pi - \tfrac{\pi}{3} = \tfrac{2\pi}{3}$ (must be between $0$ and $\pi$).
$\arctan(-1) = -\tfrac{\pi}{4}$ (must be between $-\tfrac{\pi}{2}$ and $\tfrac{\pi}{2}$).
Solve $\sec x = 2$ for $0 \leqslant x \leqslant 2\pi$.
โถ Show solution
$\cos x = \tfrac12$, so $x = \tfrac{\pi}{3}$ or $x = 2\pi - \tfrac{\pi}{3} = \tfrac{5\pi}{3}$.
Given that $\tan\theta = \tfrac34$ and $\pi < \theta < \tfrac{3\pi}{2}$, find the exact values of $\sec\theta$ and $\operatorname{cosec}\theta$.
โถ Show solution
$\sec^2\theta = 1 + \tfrac{9}{16} = \tfrac{25}{16}$. In the third quadrant $\cos\theta < 0$, so $\sec\theta = -\tfrac54$.
$\operatorname{cosec}^2\theta = 1 + \cot^2\theta = 1 + \tfrac{16}{9} = \tfrac{25}{9}$. $\sin\theta < 0$ there, so $\operatorname{cosec}\theta = -\tfrac53$.
Prove that $\sec^2\theta + \operatorname{cosec}^2\theta \equiv \sec^2\theta\operatorname{cosec}^2\theta$.
โถ Show solution
LHS $= \dfrac{1}{\cos^2\theta} + \dfrac{1}{\sin^2\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin^2\theta\cos^2\theta} = \dfrac{1}{\sin^2\theta\cos^2\theta}$
$= \sec^2\theta\operatorname{cosec}^2\theta = $ RHS $\blacksquare$
Solve $\sec^2\theta = 3 + \tan\theta$ for $0 \leqslant \theta \leqslant 2\pi$, giving answers to $3$ significant figures where they are not exact.
โถ Show solution
$1 + \tan^2\theta = 3 + \tan\theta \Rightarrow \tan^2\theta - \tan\theta - 2 = 0 \Rightarrow (\tan\theta - 2)(\tan\theta + 1) = 0$.
$\tan\theta = -1$: $\theta = \tfrac{3\pi}{4}, \tfrac{7\pi}{4}$.
$\tan\theta = 2$: $\theta = 1.11$, and $1.107 + \pi = 4.25$.
$\theta = 1.11,\ \tfrac{3\pi}{4},\ 4.25,\ \tfrac{7\pi}{4}$.
Find the exact values of $\sin\left(\arccos\dfrac{5}{13}\right)$ and $\tan\left(\arcsin\left(-\dfrac35\right)\right)$.
โถ Show solution
Let $\alpha = \arccos\tfrac{5}{13}$, $0 \leqslant \alpha \leqslant \pi$, so $\sin\alpha \geqslant 0$: $\sin\alpha = \sqrt{1 - \tfrac{25}{169}} = \tfrac{12}{13}$.
Let $\beta = \arcsin\left(-\tfrac35\right)$, so $-\tfrac{\pi}{2} \leqslant \beta < 0$ and $\cos\beta = \tfrac45$ (positive).
$\tan\beta = \dfrac{-3/5}{4/5} = -\tfrac34$.
Solve $3\operatorname{cosec}^2 x - 5\cot x - 5 = 0$ for $0^\circ < x < 180^\circ$.
โถ Show solution
$3(1 + \cot^2 x) - 5\cot x - 5 = 0 \Rightarrow 3\cot^2 x - 5\cot x - 2 = 0$.
$(3\cot x + 1)(\cot x - 2) = 0$.
$\cot x = 2 \Rightarrow \tan x = \tfrac12 \Rightarrow x = 26.6^\circ$.
$\cot x = -\tfrac13 \Rightarrow \tan x = -3 \Rightarrow x = 180^\circ - 71.6^\circ = 108.4^\circ$.
Prove that $\arcsin x + \arccos x = \dfrac{\pi}{2}$ for $-1 \leqslant x \leqslant 1$.
โถ Show solution
Let $\alpha = \arcsin x$, so $\sin\alpha = x$ and $-\tfrac{\pi}{2} \leqslant \alpha \leqslant \tfrac{\pi}{2}$.
Then $\cos\left(\tfrac{\pi}{2} - \alpha\right) = \sin\alpha = x$, and $0 \leqslant \tfrac{\pi}{2} - \alpha \leqslant \pi$.
That is exactly the range of $\arccos$, so $\arccos x = \tfrac{\pi}{2} - \alpha$.
Hence $\arcsin x + \arccos x = \alpha + \tfrac{\pi}{2} - \alpha = \tfrac{\pi}{2}$. $\blacksquare$
The function g is defined by $\mathrm{g}(x) = 4 - 3\arccos\left(\dfrac{x}{2}\right)$, for $-2 \leqslant x \leqslant 2$.
(a) Find the range of g. (b) Find the exact value of $\mathrm{g}(1)$. (c) Solve $\mathrm{g}(x) = 4 - \dfrac{\pi}{2}$. (d) Find $\mathrm{g}^{-1}(x)$ and state its domain.
โถ Show solution
(a) $\arccos\left(\tfrac{x}{2}\right)$ takes every value from $0$ to $\pi$. So $\mathrm{g}(x)$ takes every value from $4 - 3\pi$ to $4$: $4 - 3\pi \leqslant \mathrm{g}(x) \leqslant 4$.
(b) $\mathrm{g}(1) = 4 - 3\arccos\tfrac12 = 4 - 3 \times \tfrac{\pi}{3} = 4 - \pi$.
(c) $3\arccos\left(\tfrac{x}{2}\right) = \tfrac{\pi}{2} \Rightarrow \arccos\left(\tfrac{x}{2}\right) = \tfrac{\pi}{6} \Rightarrow \tfrac{x}{2} = \tfrac{\sqrt3}{2} \Rightarrow x = \sqrt3$.
(d) $y = 4 - 3\arccos\left(\tfrac{x}{2}\right) \Rightarrow \arccos\left(\tfrac{x}{2}\right) = \dfrac{4 - y}{3} \Rightarrow x = 2\cos\left(\dfrac{4 - y}{3}\right)$.
$\mathrm{g}^{-1}(x) = 2\cos\left(\dfrac{4 - x}{3}\right)$, with domain $4 - 3\pi \leqslant x \leqslant 4$ (the range of g).