P3.3 Reciprocal and inverse trigonometric functions
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This sub-topic is part of Pure Mathematics 3 (A2) and covers specification sections 2.1 and 2.2; P2 trigonometry is assumed.
What this page covers
7 points
- The definitions of sec, cosec and cot, with exact values in degrees and radians
- The graphs of \(y = \sec x\), \(y = \operatorname{cosec} x\) and \(y = \cot x\): asymptotes, turning points, period and range
- Simple transformations of the reciprocal graphs
- Solving equations such as \(\cot\theta = -0.4\) and \(\sec 2x = 3\) in a given interval
- arcsin, arccos and arctan: restricted domains, ranges and graphs as reflections in \(y = x\)
- The identities \(\sec^2\theta = 1 + \tan^2\theta\) and \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta\)
- Exact values, quadratic-type equations and identity proofs using these identities
In one minute
6 points
- The reciprocal functions are \(\sec\theta = \frac{1}{\cos\theta}\), \(\operatorname{cosec}\theta = \frac{1}{\sin\theta}\) and \(\cot\theta = \frac{\cos\theta}{\sin\theta}\), and the inverse functions arcsin, arccos and arctan undo sin, cos and tan on restricted domains, giving one angle only.
- Pair the reciprocals by the third letter: sec with cos, cosec with sin, cot with tan.
- Each reciprocal graph has a vertical asymptote wherever the original function is zero. sec and cosec have period \(2\pi\) and never take values strictly between \(-1\) and 1; cot has period \(\pi\).
- \(\arcsin x \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) and \(\arccos x \in [0, \pi]\) for \(-1 \le x \le 1\); \(\arctan x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) for all \(x\).
- \(\sec^2\theta = 1 + \tan^2\theta\) and \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta\) both come from \(\sin^2\theta + \cos^2\theta = 1\).
- For equations, use an identity to leave one function only, solve the quadratic, then convert back to sin, cos or tan to find every angle in the interval.
Contents
14 sections
- What are sec, cosec and cot?
- What do the graphs of sec, cosec and cot look like?
- How do you solve equations in sec, cosec and cot?
- What are arcsin, arccos and arctan?
- What do the inverse trig graphs look like?
- Where do the two new identities come from?
- How do you solve equations with these identities?
- How do you prove a trigonometric identity?
- Exam tips for P3.3
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What are sec, cosec and cot?
The three new functions are not really new at all: each one is "one over" a function you already know. Their value is that they make many expressions shorter, and they turn up naturally in calculus (the derivative of \(\tan x\) is \(\sec^2 x\)). Once you rewrite them in terms of sin, cos and tan, every technique from P2 applies.
Because a number and its reciprocal always have the same sign, the quadrant signs from the CAST diagram carry over: sec is positive wherever cos is, cosec wherever sin is, and cot wherever tan is. Calculators have no sec, cosec or cot keys, so find cos, sin or tan first and then press the reciprocal key.
(a) Find the exact values of \(\sec\frac{2\pi}{3}\), \(\operatorname{cosec} 225^\circ\) and \(\cot\frac{5\pi}{6}\). (b) Find \(\sec 1.2\) and \(\cot 40^\circ\), giving each to 3 significant figures.
(a) Rewrite each in terms of cos, sin and tan, then use exact values and the quadrant signs. \(\sec\frac{2\pi}{3} = \dfrac{1}{\cos\frac{2\pi}{3}} = \dfrac{1}{-\frac12} =\) \(-2\). \(\operatorname{cosec}225^\circ = \dfrac{1}{\sin 225^\circ} = \dfrac{1}{-\frac{\sqrt2}{2}} = -\dfrac{2}{\sqrt2} =\) \(-\sqrt2\). \(\cot\frac{5\pi}{6} = \dfrac{\cos\frac{5\pi}{6}}{\sin\frac{5\pi}{6}} = \dfrac{-\frac{\sqrt3}{2}}{\frac12} =\) \(-\sqrt3\).
(b) Radian mode for 1.2: \(\sec 1.2 = 1 \div \cos 1.2 = 1 \div 0.3624 =\) 2.76 (2.7597…). Degree mode for 40°: \(\cot 40^\circ = 1 \div \tan 40^\circ = 1 \div 0.8391 =\) 1.19 (1.1918…).
sec and sin both start with "s", but sec is \(\frac{1}{\cos}\). Use the third letter: sec ↔ cos, cosec ↔ sin, cot ↔ tan.
What do the graphs of sec, cosec and cot look like?
You can sketch any reciprocal graph from the original using three facts. Where the original function is zero, the reciprocal has a vertical asymptote, because dividing by a tiny number gives a huge result. Where the original is \(\pm 1\), the reciprocal is also \(\pm 1\), so the two graphs touch there; for sec and cosec these are the turning points. And where the original is small, the reciprocal is large, and the other way round.
| \(y = \sec x\) | \(y = \operatorname{cosec} x\) | \(y = \cot x\) | |
|---|---|---|---|
| Asymptotes | \(x = \frac{\pi}{2} + n\pi\) (90°, 270°, …) | \(x = n\pi\) (0°, 180°, 360°, …) | \(x = n\pi\) |
| Period | \(2\pi\) (360°) | \(2\pi\) (360°) | \(\pi\) (180°) |
| Range | \(y \le -1\) or \(y \ge 1\) | \(y \le -1\) or \(y \ge 1\) | all real numbers |
| Turning points (0 to \(2\pi\)) | min \((0, 1)\), \((2\pi, 1)\); max \((\pi, -1)\) | min \(\left(\frac{\pi}{2}, 1\right)\); max \(\left(\frac{3\pi}{2}, -1\right)\) | none |
| Zeros | none | none | \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \ldots\) |
| Symmetry | even: symmetric about the \(y\)-axis | odd: rotational symmetry about the origin | odd |
Transformations work exactly as for any graph (P3.2). \(y = a\sec x + b\) stretches by scale factor \(a\) parallel to the \(y\)-axis and then translates by \(b\). The asymptotes do not move, because no \(x\)-coordinate changes, but the turning points and the "gap" in the range do.
The curve C has equation \(y = 2\sec x - 1\), \(0 \le x \le 2\pi\). (a) Write down the equations of the asymptotes of C. (b) Find the coordinates of the turning points of C. (c) State the range of values of \(y\) on C. (d) Solve \(2\sec x - 1 = 4\), giving answers to 3 significant figures. (e) Find the set of values of \(k\) for which \(2\sec x - 1 = k\) has no solutions.
(a) Asymptotes where \(\cos x = 0\); the stretch and translation do not move them: \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\).
(b) The turning points of \(\sec x\) are \((0, 1)\), \((\pi, -1)\), \((2\pi, 1)\); apply \(y \mapsto 2y - 1\). Minimum points \((0, 1)\) and \((2\pi, 1)\); maximum point \((\pi, -3)\).
(c) \(\sec x \le -1\) or \(\sec x \ge 1\), so \(y \le -3\) or \(y \ge 1\).
(d) \(\sec x = \frac{5}{2} \Rightarrow \cos x = 0.4 \Rightarrow x = 1.1593\) or \(2\pi - 1.1593 = 5.1239\): \(x = 1.16\) or \(x = 5.12\).
(e) The line \(y = k\) misses the curve exactly when it lies in the gap: \(-3 < k < 1\).
How do you solve equations in sec, cosec and cot?
Take reciprocals first: \(\sec\theta = k \iff \cos\theta = \frac{1}{k}\), and similarly for the others. That single step turns an unfamiliar equation into a P2 equation. Then find the principal value, use the symmetry of the graph (or CAST) for the other angles, and keep only answers in the interval.
Two warnings. First, \(\sec\theta = k\) and \(\operatorname{cosec}\theta = k\) have no solutions when \(-1 < k < 1\), because the reciprocal would have to be bigger than 1 in size. Second, for an equation like \(\sec(ax + b) = k\), change the interval before solving: if \(0 \le x \le \pi\), then \(0 \le 2x \le 2\pi\).
(a) Solve \(\cot\theta = -0.4\) for \(0^\circ \le \theta < 360^\circ\), giving answers to 1 decimal place. (b) Solve \(\sec 2x = 3\) for \(0 \le x \le \pi\), giving answers to 3 significant figures.
(a) \(\tan\theta = 1 \div (-0.4) = -2.5\), principal value \(\tan^{-1}(-2.5) = -68.2^\circ\). tan repeats every 180°, so add 180° until the answers are in range: \(\theta = 111.8^\circ\) or \(291.8^\circ\).
(b) \(\cos 2x = \frac{1}{3}\) on the new interval \(0 \le 2x \le 2\pi\): \(2x = 1.2310\) or \(2\pi - 1.2310 = 5.0522\). Halve: \(x = 0.615\) or \(x = 2.53\). The next value, \(2x = 2\pi + 1.2310\), gives \(x > \pi\), so there are exactly two solutions.
Put the reciprocal step on its own line ("\(\cos 2x = \frac13\)"). It is the step that turns an unfamiliar equation into a familiar one, and it makes the rest of your working easy to follow.
What are arcsin, arccos and arctan?
Only a one-one function has an inverse. sin, cos and tan are many-one: \(\sin x = \frac12\) has infinitely many solutions, so "the angle whose sine is \(\frac12\)" is ambiguous. To fix this, each function is first restricted to an interval on which it takes every value exactly once. The inverse then returns the single angle in that interval, so its range is the restricted domain.
| Function | Restriction of the original | Domain of the inverse | Range of the inverse |
|---|---|---|---|
| \(y = \arcsin x\) | \(\sin x\) on \(-\frac{\pi}{2} \le x \le \frac{\pi}{2}\) | \(-1 \le x \le 1\) | \(-\frac{\pi}{2} \le y \le \frac{\pi}{2}\) |
| \(y = \arccos x\) | \(\cos x\) on \(0 \le x \le \pi\) | \(-1 \le x \le 1\) | \(0 \le y \le \pi\) |
| \(y = \arctan x\) | \(\tan x\) on \(-\frac{\pi}{2} < x < \frac{\pi}{2}\) | all real \(x\) | \(-\frac{\pi}{2} < y < \frac{\pi}{2}\) |
The notation \(\sin^{-1}x\) means \(\arcsin x\), the inverse function, not \(\frac{1}{\sin x}\) (that is cosec). Answers are normally given in radians; in degrees the ranges are −90° to 90°, 0° to 180° and −90° to 90°. A useful trick for mixed expressions such as \(\cos(\arcsin\frac35)\) is to call the angle \(\alpha\), draw a right-angled triangle, and use the range to decide the sign.
Find, in radians, the exact values of (a) \(\arcsin\left(-\frac{\sqrt3}{2}\right)\), (b) \(\arccos\left(-\frac12\right)\), (c) \(\arctan(-1)\), (d) \(\arcsin\left(\sin\frac{5\pi}{6}\right)\). (e) Find the exact value of \(\cos\left(\arcsin\frac35\right)\).
(a) The angle in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) whose sine is \(-\frac{\sqrt3}{2}\): \(-\frac{\pi}{3}\).
(b) The angle in \([0, \pi]\) whose cosine is \(-\frac12\); it must be obtuse: \(\pi - \frac{\pi}{3} =\) \(\frac{2\pi}{3}\).
(c) \(-\frac{\pi}{4}\), not \(\frac{3\pi}{4}\), which is outside the range.
(d) \(\sin\frac{5\pi}{6} = \frac12\), and the angle in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) with sine \(\frac12\) is \(\frac{\pi}{6}\), not \(\frac{5\pi}{6}\), which is outside the range of arcsin.
(e) Let \(\alpha = \arcsin\frac35\), so \(\sin\alpha = \frac35\) and \(-\frac{\pi}{2} \le \alpha \le \frac{\pi}{2}\), where \(\cos\alpha \ge 0\). Then \(\cos\alpha = \sqrt{1 - \frac{9}{25}} =\) \(\frac45\), as the 3–4–5 triangle shows.
\(\arcsin(\sin\theta) = \theta\) only when \(\theta\) is already in the range of arcsin, \(-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}\). Similarly \(\arccos(\cos\theta) = \theta\) only for \(0 \le \theta \le \pi\).
What do the inverse trig graphs look like?
The graph of an inverse function is the reflection of the (restricted) original in the line \(y = x\), so the domain and range swap over. To see the reflection true to shape, the axes need equal scales.
- \(y = \arcsin x\) runs from \(\left(-1, -\frac{\pi}{2}\right)\) through the origin to \(\left(1, \frac{\pi}{2}\right)\); it is increasing and has rotational symmetry about the origin.
- \(y = \arccos x\) runs from \((-1, \pi)\) through \(\left(0, \frac{\pi}{2}\right)\) to \((1, 0)\); it is decreasing and is not symmetric about the \(y\)-axis.
- \(y = \arctan x\) is defined for every \(x\) and has horizontal asymptotes \(y = \pm\frac{\pi}{2}\).
- The graphs of arcsin and arccos are finite pieces with end points: draw them with closed dots, not arrows.
The function f is defined by \(f(x) = 2\arcsin(x - 1)\) for its largest possible domain. (a) Find the domain of f. (b) Find the range of f. (c) Solve \(f(x) = \frac{\pi}{3}\). (d) Find \(f^{-1}(x)\), stating its domain.
(a) arcsin needs an input between −1 and 1: \(-1 \le x - 1 \le 1 \Rightarrow\) \(0 \le x \le 2\).
(b) \(\arcsin(x - 1)\) runs from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\); doubling gives \(-\pi \le f(x) \le \pi\).
(c) \(\arcsin(x - 1) = \frac{\pi}{6} \Rightarrow x - 1 = \sin\frac{\pi}{6} = \frac12 \Rightarrow\) \(x = \frac32\).
(d) Swap and rearrange: \(x = 2\arcsin(y - 1) \Rightarrow y - 1 = \sin\frac{x}{2}\). So \(f^{-1}(x) = 1 + \sin\frac{x}{2}\), domain \(-\pi \le x \le \pi\) (the range of f).
Where do the two new identities come from?
Start from \(\sin^2\theta + \cos^2\theta \equiv 1\) (P2) and divide every term by \(\cos^2\theta\), or by \(\sin^2\theta\). Each division produces one of the new functions, and the identities appear in one line.
The identities only tell you the square of a ratio. To get the ratio itself you must choose a sign, using the quadrant of the angle or the range of an inverse function.
(a) Given that \(\theta\) is obtuse and \(\tan\theta = -2\sqrt2\), find the exact values of \(\sec\theta\), \(\cos\theta\) and \(\operatorname{cosec}\theta\). (b) Find the exact value of \(\sec\left(\arctan\frac{\sqrt5}{2}\right)\).
(a) \(\sec^2\theta = 1 + \tan^2\theta = 1 + 8 = 9\), so \(\sec\theta = \pm 3\). \(\theta\) is obtuse, so \(\cos\theta < 0\) and therefore \(\sec\theta = -3\), \(\cos\theta = -\frac13\). Next, \(\cot\theta = -\frac{1}{2\sqrt2}\) and \(\operatorname{cosec}^2\theta = 1 + \frac18 = \frac98\). \(\sin\theta > 0\) for an obtuse angle, so \(\operatorname{cosec}\theta = \frac{3}{2\sqrt2} =\) \(\frac{3\sqrt2}{4}\). Check: \(\sin\theta = \tan\theta\cos\theta = (-2\sqrt2)\left(-\frac13\right) = \frac{2\sqrt2}{3}\), and \(\frac{1}{\sin\theta} = \frac{3}{2\sqrt2}\).
(b) Let \(\beta = \arctan\frac{\sqrt5}{2}\), so \(\tan\beta = \frac{\sqrt5}{2}\) and \(-\frac{\pi}{2} < \beta < \frac{\pi}{2}\), where \(\cos\beta > 0\). Then \(\sec^2\beta = 1 + \frac54 = \frac94\), so \(\sec\left(\arctan\frac{\sqrt5}{2}\right) =\) \(\frac32\).
How do you solve equations with these identities?
Look at what the equation contains. tan² with sec (or sec² with tan) calls for \(\sec^2\theta = 1 + \tan^2\theta\); cot² with cosec (or cosec² with cot) calls for \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta\). Replace the squared term so that only one function is left, collect everything on one side, and solve the quadratic. Reject impossible values (such as \(\sec\theta = \frac12\)), then convert each remaining value to cos, sin or tan and find every angle in the interval.
Never divide both sides by a trig function that could be zero; factorise instead, or you lose solutions.
(a) Solve \(2\tan^2\theta + \sec\theta = 1\) for \(0^\circ \le \theta < 360^\circ\), giving answers to 1 decimal place where appropriate. (b) Solve \(3\operatorname{cosec}^2 x - 5\cot x = 5\) for \(0 < x < 2\pi\), giving answers to 3 significant figures.
(a) \(\tan^2\theta = \sec^2\theta - 1\): \(2(\sec^2\theta - 1) + \sec\theta = 1 \Rightarrow 2\sec^2\theta + \sec\theta - 3 = 0 \Rightarrow (2\sec\theta + 3)(\sec\theta - 1) = 0\). So \(\sec\theta = 1\) or \(\sec\theta = -\frac32\). \(\cos\theta = 1\) gives \(\theta = 0^\circ\); \(\cos\theta = -\frac23\) gives \(\theta = 131.8^\circ\) or \(360^\circ - 131.8^\circ = 228.2^\circ\). \(\theta = 0^\circ,\ 131.8^\circ,\ 228.2^\circ\).
(b) \(\operatorname{cosec}^2 x = 1 + \cot^2 x\): \(3 + 3\cot^2 x - 5\cot x = 5 \Rightarrow 3\cot^2 x - 5\cot x - 2 = 0 \Rightarrow (3\cot x + 1)(\cot x - 2) = 0\). So \(\tan x = \frac12\) or \(\tan x = -3\). \(\tan x = \frac12\): \(x = 0.4636\) or \(\pi + 0.4636\). \(\tan x = -3\): \(x = \pi - 1.2490\) or \(2\pi - 1.2490\). \(x = 0.464,\ 1.89,\ 3.61,\ 5.03\).
Writing \(\tan^2\theta = 1 - \sec^2\theta\), or \(\sec^2\theta = \tan^2\theta - 1\). If you are unsure, re-derive the identity by dividing \(\sin^2 + \cos^2 = 1\) by \(\cos^2\), or test it with \(\theta = 0\): \(\sec^2 0 = 1\) and \(\tan^2 0 = 0\).
How do you prove a trigonometric identity?
Start from one side, usually the more complicated one, and work until you reach the other side. Do not work on both sides at once, and never start by assuming the result. Useful moves are: rewrite sec, cosec and cot in terms of sin and cos; combine fractions over a common denominator; factorise a difference of two squares; and use the Pythagorean identities in whichever direction helps. Finish with a clear statement such as "= RHS, as required".
(a) Prove that \(\sec^4\theta - \tan^4\theta \equiv \sec^2\theta + \tan^2\theta\). (b) Hence solve \(\sec^4\theta - \tan^4\theta = 4 - \tan\theta\) for \(0 \le \theta < 2\pi\), giving exact answers where possible and other answers to 3 significant figures.
(a) Difference of two squares: LHS \(= (\sec^2\theta - \tan^2\theta)(\sec^2\theta + \tan^2\theta)\). By the identity, \(\sec^2\theta - \tan^2\theta = 1\), so LHS \(= \sec^2\theta + \tan^2\theta =\) RHS.
(b) Using (a), then \(\sec^2\theta = 1 + \tan^2\theta\): \(1 + 2\tan^2\theta = 4 - \tan\theta \Rightarrow 2\tan^2\theta + \tan\theta - 3 = 0 \Rightarrow (2\tan\theta + 3)(\tan\theta - 1) = 0\). \(\tan\theta = 1\) gives \(\theta = \frac{\pi}{4}\) or \(\frac{5\pi}{4}\). \(\tan\theta = -\frac32\) has principal value \(-0.9828\), so \(\theta = \pi - 0.9828\) or \(2\pi - 0.9828\). \(\theta = \frac{\pi}{4},\ \frac{5\pi}{4},\ 2.16,\ 5.30\).
"Hence" means part (a) must be used: replacing \(\sec^4 - \tan^4\) by \(\sec^2 + \tan^2\) is what makes the quadratic appear.
Exam tips for P3.3
Whenever you are stuck, rewrite sec, cosec and cot as \(\frac{1}{\cos}\), \(\frac{1}{\sin}\) and \(\frac{\cos}{\sin}\). Equations become familiar, and proofs often simplify to a single fraction.
tan and sec together: \(\sec^2 = 1 + \tan^2\). cot and cosec together: \(\operatorname{cosec}^2 = 1 + \cot^2\). Replace the squared term so that only one function remains, and you will usually see a quadratic.
When an identity gives \(\sec^2\theta = 9\), the answer is \(\pm 3\) until you decide. Use the quadrant (obtuse, reflex, "\(\theta\) in the third quadrant") or the range of an inverse function to choose the sign, and say why.
arcsin, arccos and arctan each return a single angle in their own range. Unless the question uses degrees, give the answer in radians, exactly where possible (\(-\frac{\pi}{3}\), \(\frac{2\pi}{3}\), \(-\frac{\pi}{4}\)).
Before solving \(\sec 2x = 3\) on \(0 \le x \le \pi\), rewrite the interval as \(0 \le 2x \le 2\pi\) and find every value of \(2x\) in it. Then divide. In a proof, write every line as a full expression joined by "=" and finish with "= RHS".
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(\sec x = \frac{1}{\sin x}\). | The pairs are matched by the third letter, not the first. | sec ↔ cos, cosec ↔ sin, cot ↔ tan. |
| \(\sin^{-1}x\) means \(\frac{1}{\sin x}\). | The −1 here is function-inverse notation. | \(\sin^{-1}x = \arcsin x\); \(\frac{1}{\sin x} = (\sin x)^{-1} = \operatorname{cosec}x\). |
| \(\arcsin(\sin\theta) = \theta\) always. | arcsin only returns angles in \(\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\). | \(\arcsin\left(\sin\frac{5\pi}{6}\right) = \frac{\pi}{6}\). The rule holds only when \(\theta\) is in the range. |
| \(\sec x = 0.5\) has two solutions in \([0, 2\pi]\). | It needs \(\cos x = 2\), which is impossible. | sec and cosec never lie strictly between −1 and 1. |
| \(\tan^2\theta = 1 - \sec^2\theta\). | The sign is wrong: at \(\theta = \frac{\pi}{4}\) this gives \(1 = -1\). | \(\tan^2\theta = \sec^2\theta - 1\). |
| Dividing \(\sec\theta\tan\theta = 2\tan\theta\) by \(\tan\theta\) is fine. | It throws away every solution of \(\tan\theta = 0\). | Factorise: \(\tan\theta(\sec\theta - 2) = 0\), then solve both factors. |
How this topic connects
P2 trigonometry and P3.1 functions
The identities here come straight from \(\sin^2\theta + \cos^2\theta = 1\), and every equation finishes with the P2 method of principal value plus symmetry. arcsin, arccos and arctan are the clearest example of restricting a domain to make a one-one function, with graphs reflected in \(y = x\).
P3.4 Compound angles, double angles and R-form
The compound-angle and double-angle formulae are combined with sec, cosec and cot in harder identities and equations.
P3.6 Differentiation and P4 integration
The derivative of \(\tan x\) is \(\sec^2 x\), and the derivatives of \(\sec x\), \(\operatorname{cosec} x\) and \(\cot x\) are written in terms of these functions. In P4, \(\int\tan^2 x\,dx\) is found by rewriting \(\tan^2 x = \sec^2 x - 1\), and substitutions such as \(x = \sin\theta\) lead to arcsin in the answer.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(\sec\theta = \frac{1}{\cos\theta}\) | definition of secant | No — learn it |
| \(\operatorname{cosec}\theta = \frac{1}{\sin\theta}\) | definition of cosecant | No — learn it |
| \(\cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}\) | definition of cotangent | No — learn it |
| \(\sec^2\theta \equiv 1 + \tan^2\theta\) | equations and proofs with tan and sec | No — learn it |
| \(\operatorname{cosec}^2\theta \equiv 1 + \cot^2\theta\) | equations and proofs with cot and cosec | No — learn it |
| \(-\frac{\pi}{2} \le \arcsin x \le \frac{\pi}{2}\), \(0 \le \arccos x \le \pi\), \(-\frac{\pi}{2} < \arctan x < \frac{\pi}{2}\) | ranges of the inverse functions (domains \(-1 \le x \le 1\) for arcsin and arccos) | No — learn it |
| \(\cos(\arcsin x) = \sqrt{1 - x^2}\) | exact values of mixed expressions | No — derive it |
Check your understanding
1. Find the exact values of \(\operatorname{cosec}\left(-\frac{\pi}{6}\right)\) and \(\sec 300^\circ\).
\(\sin\left(-\frac{\pi}{6}\right) = -\frac12\), so \(\operatorname{cosec}\left(-\frac{\pi}{6}\right) = -2\). \(\cos 300^\circ = \frac12\), so \(\sec 300^\circ = 2\).
2. State the domain and range of \(y = \arccos x\), and the range of \(y = \arctan x\).
arccos: domain \(-1 \le x \le 1\), range \(0 \le y \le \pi\). arctan: range \(-\frac{\pi}{2} < y < \frac{\pi}{2}\) (its domain is all real numbers).
3. Given that \(\cot\theta = \frac34\) and \(180^\circ < \theta < 270^\circ\), find the exact value of \(\operatorname{cosec}\theta\).
\(\operatorname{cosec}^2\theta = 1 + \frac{9}{16} = \frac{25}{16}\). In the third quadrant \(\sin\theta < 0\), so \(\operatorname{cosec}\theta = -\frac54\).
4. Explain why the equation \(\operatorname{cosec}x = 0.6\) has no solutions.
It would need \(\sin x = \frac{1}{0.6} = \frac53 > 1\), which is impossible. The range of cosec is \(y \le -1\) or \(y \ge 1\).
5. Solve \(\tan^2\theta = 2\sec\theta + 2\) for \(0^\circ \le \theta < 360^\circ\).
\(\sec^2\theta - 1 = 2\sec\theta + 2 \Rightarrow \sec^2\theta - 2\sec\theta - 3 = 0 \Rightarrow (\sec\theta - 3)(\sec\theta + 1) = 0\). \(\cos\theta = \frac13\) gives \(\theta = 70.5^\circ\) or \(289.5^\circ\); \(\cos\theta = -1\) gives \(\theta = 180^\circ\).
6. Find the exact value of \(\tan\left(\arccos\left(-\frac{5}{13}\right)\right)\).
Let \(\gamma = \arccos\left(-\frac{5}{13}\right)\), so \(\cos\gamma = -\frac{5}{13}\) and \(0 \le \gamma \le \pi\); here \(\gamma\) is obtuse, so \(\sin\gamma = +\sqrt{1 - \frac{25}{169}} = \frac{12}{13}\). Then \(\tan\gamma = \frac{12/13}{-5/13} = -\frac{12}{5}\).
Specification checklist
- (P3 2.1) I know the definitions of sec, cosec and cot and can find their exact values, and their values to a given accuracy, in degrees and in radians.
- (P3 2.1) I can sketch \(y = \sec x\), \(y = \operatorname{cosec}x\) and \(y = \cot x\), showing asymptotes, turning points, period and range, and sketch simple transformations of them.
- (P3 2.1) I can solve equations such as \(\sec 2x = 3\) or \(\cot\theta = k\) in a given interval, and recognise when \(\sec x = k\) or \(\operatorname{cosec}x = k\) has no solutions.
- (P3 2.1) I can explain why sin, cos and tan must be restricted before they have inverses, and state the domain and range of arcsin, arccos and arctan.
- (P3 2.1) I can sketch \(y = \arcsin x\), \(y = \arccos x\) and \(y = \arctan x\) as reflections in \(y = x\), and find exact values such as \(\arccos\left(-\frac12\right)\).
- (P3 2.2) I can derive \(\sec^2\theta = 1 + \tan^2\theta\) and \(\operatorname{cosec}^2\theta = 1 + \cot^2\theta\) from \(\sin^2\theta + \cos^2\theta = 1\), and use them to find exact values with the correct sign.
- (P3 2.2) I can use these identities to turn an equation into a quadratic in one function and find all its solutions in an interval.
- (P3 2.2) I can prove identities involving sec, cosec and cot, working from one side to the other.