Edexcel IAL Maths
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The definitions of sec, cosec and cot, with exact values in degrees and radians
The graphs of y = sec x, y = cosec x and y = cot x: asymptotes, turning points, period and range
Simple transformations of the reciprocal graphs
Solving equations such as cot θ= -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 sec²θ= 1 + tan²θ and cosec²θ= 1 + cot²θ
Exact values, quadratic-type equations and identity proofs using these identities
1. The reciprocal functions are sec θ= 1/cos θ, cosecθ= 1/sin θ and cot θ= cos θ/sin θ, and the inverse functions arcsin, arccos and arctan undo sin, cos and tan on restricted domains, giving one angle only.
2. Pair the reciprocals by the third letter: sec with cos, cosec with sin, cot with tan.
3. Each reciprocal graph has a vertical asymptote wherever the original function is zero. sec and cosec have period 2π and never take values strictly between -1 and 1; cot has period π.
4. arcsin x ∈[-π/2, π/2] and arccos x ∈[0, π] for -1 ≤x ≤1; arctan x ∈(-π/2, π/2) for all x.
5. sec²θ= 1 + tan²θ and cosec²θ= 1 + cot²θ both come from sin²θ+ cos²θ= 1.
6. 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.
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 sec² x). Once you rewrite them in terms of sin, cos and tan, every technique from P2 applies.
The reciprocal trigonometric functions.
sec θ= 1/cos θ cosecθ= 1/sin θ cot θ= 1/tan θ = cos θ/sin θ sec is short for secant, cosec for cosecant (sometimes written csc) and cot for cotangent. Each is undefined where the function underneath is zero: sec where cos θ= 0, cosec and cot where sin θ= 0.
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.
WORKED EXAMPLE · QUESTION
(a) Find the exact values of sec 2π/3, cosec 225° and cot 5π/6. (b) Find sec 1.2 and cot 40°, giving each to 3 significant figures.
SOLUTION
(a) Rewrite each in terms of cos, sin and tan, then use exact values and the quadrant signs. sec 2π/3 = 1/(cos 2π/3) = 1/(-½) = -2. cosec225° = 1/sin 225° = 1/(-√2/2) = -2/√2 = -√2. cot 5π/6 = (cos 5π/6)/(sin 5π/6) = (-√3/2)/½ = -√3.
(b) Radian mode for 1.2: sec 1.2 = 1 ÷cos 1.2 = 1 ÷0.3624 = 2.76 (2.7597…). Degree mode for 40°: cot 40° = 1 ÷tan 40° = 1 ÷0.8391 = 1.19 (1.1918…).
Common error — pairing by the first letter
sec and sin both start with "s", but sec is 1/cos. Use the third letter: sec ↔ cos, cosec ↔ sin, cot ↔ tan.
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 ±1, the reciprocal is also ±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 = cosec x | y = cot x | |
|---|---|---|---|
| Asymptotes | x = π/2 + nπ (90°, 270°, …) | x = nπ (0°, 180°, 360°, …) | x = nπ |
| Period | 2π (360°) | 2π (360°) | π (180°) |
| Range | y ≤-1 or y ≥1 | y ≤-1 or y ≥1 | all real numbers |
| Turning points (0 to 2π) | min (0, 1), (2π, 1); max (π, -1) | min (π/2, 1); max (3π/2, -1) | none |
| Zeros | none | none | x = π/2, 3π/2, … |
| Symmetry | even: symmetric about the y-axis | odd: rotational symmetry about the origin | odd |
Transformations work exactly as for any graph (P3.2). y = asec 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.
WORKED EXAMPLE · QUESTION
The curve C has equation y = 2sec x - 1, 0 ≤x ≤2π. (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 2sec x - 1 = 4, giving answers to 3 significant figures. (e) Find the set of values of k for which 2sec x - 1 = k has no solutions.
SOLUTION
(a) Asymptotes where cos x = 0; the stretch and translation do not move them: x = π/2 and x = 3π/2.
(b) The turning points of sec x are (0, 1), (π, -1), (2π, 1); apply y ↦2y - 1. Minimum points (0, 1) and (2π, 1); maximum point (π, -3).
(c) sec x ≤-1 or sec x ≥1, so y ≤-3 or y ≥1.
(d) sec x = (5/2) ⇒cos x = 0.4 ⇒x = 1.1593 or 2π- 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.
Take reciprocals first: sec θ= k ⇔cos θ= 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 θ= k and cosecθ= 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 ≤x ≤π, then 0 ≤2x ≤2π.
WORKED EXAMPLE · QUESTION
(a) Solve cot θ= -0.4 for 0° ≤θ< 360°, giving answers to 1 decimal place. (b) Solve sec 2x = 3 for 0 ≤x ≤π, giving answers to 3 significant figures.
SOLUTION
(a) tan θ= 1 ÷(-0.4) = -2.5, principal value tan⁻¹(-2.5) = -68.2°. tan repeats every 180°, so add 180° until the answers are in range: θ= 111.8° or 291.8°.
(b) cos 2x = ⅓ on the new interval 0 ≤2x ≤2π: 2x = 1.2310 or 2π- 1.2310 = 5.0522. Halve: x = 0.615 or x = 2.53. The next value, 2x = 2π+ 1.2310, gives x > π, so there are exactly two solutions.
Exam tip — write the reciprocal step
Put the reciprocal step on its own line ("cos 2x = ⅓"). It is the step that turns an unfamiliar equation into a familiar one, and it makes the rest of your working easy to follow.
Only a one-one function has an inverse. sin, cos and tan are many-one: sin x = ½ has infinitely many solutions, so "the angle whose sine is ½" 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 -π/2 ≤x ≤π/2 | -1 ≤x ≤1 | -π/2 ≤y ≤π/2 |
| y = arccos x | cos x on 0 ≤x ≤π | -1 ≤x ≤1 | 0 ≤y ≤π |
| y = arctan x | tan x on -π/2 < x < π/2 | all real x | -π/2 < y < π/2 |
The notation sin⁻¹x means arcsin x, the inverse function, not 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(3/5)) is to call the angle α, draw a right-angled triangle, and use the range to decide the sign.
WORKED EXAMPLE · QUESTION
Find, in radians, the exact values of (a) arcsin(-√3/2), (b) arccos(-½), (c) arctan(-1), (d) arcsin(sin 5π/6). (e) Find the exact value of cos(arcsin(3/5)).
SOLUTION
(a) The angle in [-π/2, π/2] whose sine is -√3/2: -π/3.
(b) The angle in [0, π] whose cosine is -½; it must be obtuse: π- π/3 = 2π/3.
(c) -π/4, not 3π/4, which is outside the range.
(d) sin 5π/6 = ½, and the angle in [-π/2, π/2] with sine ½ is π/6, not 5π/6, which is outside the range of arcsin.
(e) Let α= arcsin(3/5), so sin α= (3/5) and -π/2 ≤α≤π/2, where cos α≥0. Then cos α= √(1 - (9/25)) = (4/5), as the 3–4–5 triangle shows.
Common error — "arcsin undoes sin" for every angle
arcsin(sin θ) = θ only when θ is already in the range of arcsin, -π/2 ≤θ≤π/2. Similarly arccos(cos θ) = θ only for 0 ≤θ≤π.
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 (-1, -π/2) through the origin to (1, π/2); it is increasing and has rotational symmetry about the origin.
y = arccos x runs from (-1, π) through (0, π/2) 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 = ±π/2.
The graphs of arcsin and arccos are finite pieces with end points: draw them with closed dots, not arrows.
WORKED EXAMPLE · QUESTION
The function f is defined by f(x) = 2arcsin(x - 1) for its largest possible domain. (a) Find the domain of f. (b) Find the range of f. (c) Solve f(x) = π/3. (d) Find f⁻¹(x), stating its domain.
SOLUTION
(a) arcsin needs an input between −1 and 1: -1 ≤x - 1 ≤1 ⇒ 0 ≤x ≤2.
(b) arcsin(x - 1) runs from -π/2 to π/2; doubling gives -π≤f(x) ≤π.
(c) arcsin(x - 1) = π/6 ⇒x - 1 = sin π/6 = ½ ⇒ x = (3/2).
(d) Swap and rearrange: x = 2arcsin(y - 1) ⇒y - 1 = sin x/2. So f⁻¹(x) = 1 + sin x/2, domain -π≤x ≤π (the range of f).
Start from sin²θ+ cos²θ≡1 (P2) and divide every term by cos²θ, or by sin²θ. Each division produces one of the new functions, and the identities appear in one line.
The Pythagorean identities.
sin²θ/cos²θ + cos²θ/cos²θ ≡1/cos²θ ⇒ tan²θ+ 1 ≡sec²θ sin²θ/sin²θ + cos²θ/sin²θ ≡1/sin²θ ⇒ 1 + cot²θ≡cosec²θ The rearranged forms tan²θ= sec²θ- 1 and cot²θ= cosec²θ- 1 are used most. They hold for every θ where both sides are defined.
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.
WORKED EXAMPLE · QUESTION
(a) Given that θ is obtuse and tan θ= -2√2, find the exact values of sec θ, cos θ and cosecθ. (b) Find the exact value of sec(arctan √5/2).
SOLUTION
(a) sec²θ= 1 + tan²θ= 1 + 8 = 9, so sec θ= ±3. θ is obtuse, so cos θ< 0 and therefore sec θ= -3, cos θ= -⅓. Next, cot θ= -1/2√2 and cosec²θ= 1 + ⅛ = (9/8). sin θ> 0 for an obtuse angle, so cosecθ= 3/2√2 = 3√2/4. Check: sin θ= tan θcos θ= (-2√2)(-⅓) = 2√2/3, and 1/sin θ = 3/2√2.
(b) Let β= arctan √5/2, so tan β= √5/2 and -π/2 < β< π/2, where cos β> 0. Then sec²β= 1 + (5/4) = (9/4), so sec(arctan √5/2) = (3/2).
Look at what the equation contains. tan² with sec (or sec² with tan) calls for sec²θ= 1 + tan²θ; cot² with cosec (or cosec² with cot) calls for cosec²θ= 1 + cot²θ. 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 θ= ½), 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.
WORKED EXAMPLE · QUESTION
(a) Solve 2tan²θ+ sec θ= 1 for 0° ≤θ< 360°, giving answers to 1 decimal place where appropriate. (b) Solve 3cosec² x - 5cot x = 5 for 0 < x < 2π, giving answers to 3 significant figures.
SOLUTION
(a) tan²θ= sec²θ- 1: 2(sec²θ- 1) + sec θ= 1 ⇒2sec²θ+ sec θ- 3 = 0 ⇒(2sec θ+ 3)(sec θ- 1) = 0. So sec θ= 1 or sec θ= -(3/2). cos θ= 1 gives θ= 0°; cos θ= -⅔ gives θ= 131.8° or 360° - 131.8° = 228.2°. θ= 0°, 131.8°, 228.2°.
(b) cosec² x = 1 + cot² x: 3 + 3cot² x - 5cot x = 5 ⇒3cot² x - 5cot x - 2 = 0 ⇒(3cot x + 1)(cot x - 2) = 0. So tan x = ½ or tan x = -3. tan x = ½: x = 0.4636 or π+ 0.4636. tan x = -3: x = π- 1.2490 or 2π- 1.2490. x = 0.464, 1.89, 3.61, 5.03.
Common error — the wrong sign in the identity
Writing tan²θ= 1 - sec²θ, or sec²θ= tan²θ- 1. If you are unsure, re-derive the identity by dividing sin² + cos² = 1 by cos², or test it with θ= 0: sec² 0 = 1 and tan² 0 = 0.
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 model proof.
Prove that (cosecx - cot x)² ≡(1 - cos x)/(1 + cos x). LHS = (1/sin x - cos x/sin x)² = ((1 - cos x)²)/(sin² x) = ((1 - cos x)²)/(1 - cos² x) = ((1 - cos x)²)/((1 - cos x)(1 + cos x)) = (1 - cos x)/(1 + cos x) = RHS The key steps: everything in sin and cos, sin² x = 1 - cos² x, then the difference of two squares lets a factor cancel.
WORKED EXAMPLE · QUESTION
(a) Prove that sec⁴θ- tan⁴θ≡sec²θ+ tan²θ. (b) Hence solve sec⁴θ- tan⁴θ= 4 - tan θ for 0 ≤θ< 2π, giving exact answers where possible and other answers to 3 significant figures.
SOLUTION
(a) Difference of two squares: LHS = (sec²θ- tan²θ)(sec²θ+ tan²θ). By the identity, sec²θ- tan²θ= 1, so LHS = sec²θ+ tan²θ= RHS.
(b) Using (a), then sec²θ= 1 + tan²θ: 1 + 2tan²θ= 4 - tan θ⇒2tan²θ+ tan θ- 3 = 0 ⇒(2tan θ+ 3)(tan θ- 1) = 0. tan θ= 1 gives θ= π/4 or 5π/4. tan θ= -(3/2) has principal value -0.9828, so θ= π- 0.9828 or 2π- 0.9828. θ= π/4, 5π/4, 2.16, 5.30.
"Hence" means part (a) must be used: replacing sec⁴ - tan⁴ by sec² + tan² is what makes the quadratic appear.
Exam tip 1 — get back to sin, cos and tan
Whenever you are stuck, rewrite sec, cosec and cot as 1/cos, 1/sin and cos/sin. Equations become familiar, and proofs often simplify to a single fraction.
Exam tip 2 — let the equation choose the identity
tan and sec together: sec² = 1 + tan². cot and cosec together: cosec² = 1 + cot². Replace the squared term so that only one function remains, and you will usually see a quadratic.
Exam tip 3 — use signs, not just squares
When an identity gives sec²θ= 9, the answer is ±3 until you decide. Use the quadrant (obtuse, reflex, "θ in the third quadrant") or the range of an inverse function to choose the sign, and say why.
Exam tip 4 — inverse functions give one value, in radians
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 (-π/3, 2π/3, -π/4).
Exam tip 5 — adjust the interval for 2x, x + a and so on
Before solving sec 2x = 3 on 0 ≤x ≤π, rewrite the interval as 0 ≤2x ≤2π 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".
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| sec x = 1/sin x. | The pairs are matched by the third letter, not the first. | sec ↔ cos, cosec ↔ sin, cot ↔ tan. |
| sin⁻¹x means 1/sin x. | The −1 here is function-inverse notation. | sin⁻¹x = arcsin x; 1/sin x = (sin x)⁻¹ = cosecx. |
| arcsin(sin θ) = θ always. | arcsin only returns angles in [-π/2, π/2]. | arcsin(sin 5π/6) = π/6. The rule holds only when θ is in the range. |
| sec x = 0.5 has two solutions in [0, 2π]. | It needs cos x = 2, which is impossible. | sec and cosec never lie strictly between −1 and 1. |
| tan²θ= 1 - sec²θ. | The sign is wrong: at θ= π/4 this gives 1 = -1. | tan²θ= sec²θ- 1. |
| Dividing sec θtan θ= 2tan θ by tan θ is fine. | It throws away every solution of tan θ= 0. | Factorise: tan θ(sec θ- 2) = 0, then solve both factors. |
P2 trigonometry and P3.1 functions
The identities here come straight from sin²θ+ cos²θ= 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 sec² x, and the derivatives of sec x, cosec x and cot x are written in terms of these functions. In P4, ∫tan² x dx is found by rewriting tan² x = sec² x - 1, and substitutions such as x = sin θ lead to arcsin in the answer.
| Formula | Use | In the formulae list? |
|---|---|---|
| sec θ= 1/cos θ | definition of secant | No — learn it |
| cosecθ= 1/sin θ | definition of cosecant | No — learn it |
| cot θ= 1/tan θ = cos θ/sin θ | definition of cotangent | No — learn it |
| sec²θ≡1 + tan²θ | equations and proofs with tan and sec | No — learn it |
| cosec²θ≡1 + cot²θ | equations and proofs with cot and cosec | No — learn it |
| -π/2 ≤arcsin x ≤π/2, 0 ≤arccos x ≤π, -π/2 < arctan x < π/2 | ranges of the inverse functions (domains -1 ≤x ≤1 for arcsin and arccos) | No — learn it |
| cos(arcsin x) = √(1 - x²) | exact values of mixed expressions | No — derive it |
1. Find the exact values of cosec(-π/6) and sec 300°.
2. State the domain and range of y = arccos x, and the range of y = arctan x.
3. Given that cot θ= ¾ and 180° < θ< 270°, find the exact value of cosecθ.
4. Explain why the equation cosecx = 0.6 has no solutions.
5. Solve tan²θ= 2sec θ+ 2 for 0° ≤θ< 360°.
6. Find the exact value of tan(arccos(-(5/13))).
1. sin(-π/6) = -½, so cosec(-π/6) = -2. cos 300° = ½, so sec 300° = 2.
2. arccos: domain -1 ≤x ≤1, range 0 ≤y ≤π. arctan: range -π/2 < y < π/2 (its domain is all real numbers).
3. cosec²θ= 1 + (9/16) = (25/16). In the third quadrant sin θ< 0, so cosecθ= -(5/4).
4. It would need sin x = (1/0.6) = (5/3) > 1, which is impossible. The range of cosec is y ≤-1 or y ≥1.
5. sec²θ- 1 = 2sec θ+ 2 ⇒sec²θ- 2sec θ- 3 = 0 ⇒(sec θ- 3)(sec θ+ 1) = 0. cos θ= ⅓ gives θ= 70.5° or 289.5°; cos θ= -1 gives θ= 180°.
6. Let γ= arccos(-(5/13)), so cos γ= -(5/13) and 0 ≤γ≤π; here γ is obtuse, so sin γ= +√(1 - (25/169)) = (12/13). Then tan γ= (12/13)/(-5/13) = -(12/5).
Check understanding
The answer and each step appear one click at a time.
This question is about the values of the reciprocal trigonometric functions. Answer parts (a) to (c) without a calculator, showing your working.
(a) Find the exact value of cosec(4π/3), giving your answer with a rational denominator. [2]
(b) Find the exact value of sec 135°. [1]
(c) Find the exact value of cot(7π/4). [1]
(d) Find the value of cosec 2.5, giving your answer to 3 significant figures. [1]
Mark scheme click to reveal each step
cosec(4π/3) = 1/sin(4π/3) = 1/(−√3/2)
= −2/√3 = −2√3/3
1/cos 135° = 1/(−√2/2) = −√2
1/tan(7π/4) = 1/(−1) = −1
1/sin 2.5 = 1/0.5985 = 1.67 (radian mode)
Give each answer to this question as an exact value in radians or as an exact number. The inverse trigonometric functions arcsin, arccos and arctan take their usual ranges.
(a) Write down the value of arcsin(½). [1]
(b) Write down the value of arccos(−√3/2). [1]
(c) Write down the value of arctan(−√3). [1]
(d) Find the exact value of sin(arccos(−5/13)). [2]
(e) Find the value of arccos(cos(4π/3)). [1]
Mark scheme click to reveal each step
π/6
5π/6
−π/3
let α = arccos(−5/13); 0 ≤ α ≤ π so sin α ≥ 0 and sin α = √(1 − 25/169)
= 12/13
cos(4π/3) = −½, so arccos(−½) = 2π/3
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