OxfordAQA International AS and A-level Mathematics: Inverse trigonometric functions and secant, cosecant, cotangent (9660)

This guide follows the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2, and covers the first part of content area P2.4 (Trigonometry): the inverse functions sin⁻¹, cos⁻¹ and tan⁻¹, and the reciprocal functions secant, cosecant and cotangent. The specification places this content in International A-level Unit P2 (Pure Maths), so it belongs to the A2 year and is not examined at International AS. The compound-angle, double-angle, R-form and equation-solving parts of P2.4 have their own pages and are not repeated here.

Related pages: the OxfordAQA A-level Mathematics hub, the printable checklist, the revision notes, the practice set and the exam preparation page. To find your weak spots first, try the free 10-minute diagnostics.

What this guide covers

Spec ref	What you must be able to do
P2.4	Know the functions sin⁻¹, cos⁻¹, tan⁻¹; understand their domains and graphs
P2.4	Know that −π/2 ≤ sin⁻¹x ≤ π/2, 0 ≤ cos⁻¹x ≤ π and −π/2 < tan⁻¹x < π/2
P2.4	See the inverse graphs as reflections of the relevant parts of the trig graphs in y = x
P2.4	Know secant, cosecant and cotangent, their link to cos, sin and tan, their domains and graphs
P2.4	Know and use 1 + tan²x = sec²x and 1 + cot²x = cosec²x

The specification states that addition formulae for inverse functions are not required. Angles in this guide are in radians unless a degree sign appears.

Secant, cosecant and cotangent

These are defined from the three basic functions:

sec x = 1/cos x      cosec x = 1/sin x      cot x = cos x/sin x = 1/tan x

Each is undefined wherever its denominator is zero.

Function	Undefined when	Domain (radians)	Range	Period
sec x	cos x = 0	x ≠ π/2 + nπ	outputs satisfy abs(sec x) ≥ 1	2π
cosec x	sin x = 0	x ≠ nπ	outputs satisfy abs(cosec x) ≥ 1	2π
cot x	sin x = 0	x ≠ nπ	all real numbers	π

Here n is any integer. Notice that cot x is defined at x = π/2, where it equals 0, even though tan x is not. That is why cot x = cos x/sin x is the safer definition.

Graphs

You need to sketch these from the graphs of cos, sin and tan.

y = sec x   (from y = cos x)
  vertical asymptotes at x = ±π/2, ±3π/2 ...
  U-shaped branches: minimum (0, 1), maximum (π, −1), minimum (2π, 1)
  no part of the graph lies between y = −1 and y = 1

y = cosec x   (from y = sin x)
  vertical asymptotes at x = 0, ±π, ±2π ...
  branch above the axis with minimum (π/2, 1)
  branch below the axis with maximum (3π/2, −1)

y = cot x   (from y = tan x)
  vertical asymptotes at x = 0, ±π, ±2π ...
  falls from left to right in every branch, crossing the x-axis at x = ±π/2, ±3π/2 ...

Method for sketching: draw the original curve lightly first. Wherever it crosses the x-axis, draw an asymptote. Wherever it reaches ±1, the reciprocal touches the same point. The reciprocal of a small value is large, so the curve climbs towards the asymptote.

Worked example 1. Given that sin θ = 8/17 and θ is obtuse, find the exact values of sec θ, cosec θ and cot θ.

Since θ is obtuse, cos θ is negative.

cos²θ = 1 − sin²θ = 1 − 64/289 = 225/289
cos θ = −15/17
sec θ = 1/cos θ = −17/15
cosec θ = 1/sin θ = 17/8
cot θ = cos θ/sin θ = (−15/17)/(8/17) = −15/8

Worked example 2. Give the exact value of cosec(5π/4).

sin(5π/4) = −√2/2 (the reference angle is π/4 and sine is negative in the third quadrant). So cosec(5π/4) = 1/(−√2/2) = −2/√2 = −√2.

The identities 1 + tan²x = sec²x and 1 + cot²x = cosec²x

Start from sin²x + cos²x = 1 and divide every term.

Divide by cos²x:   sin²x/cos²x + 1 = 1/cos²x    so   tan²x + 1 = sec²x
Divide by sin²x:   1 + cos²x/sin²x = 1/sin²x    so   1 + cot²x = cosec²x

These hold for every x where all the functions involved are defined. Being able to derive them in two lines is worth more than memorising them, because a derivation also reminds you of the restrictions.

Useful rearrangements: tan²x = sec²x − 1, cot²x = cosec²x − 1, and (sec x − 1)(sec x + 1) = tan²x.

Worked example 3. Prove that sec x − cos x ≡ sin x tan x.

LHS = 1/cos x − cos x
    = (1 − cos²x)/cos x
    = sin²x/cos x
    = sin x × (sin x/cos x)
    = sin x tan x  = RHS

Worked example 4. Given that sec²θ + tan θ = 3, find the possible values of tan θ, and the solutions for 0° ≤ θ ≤ 360°.

Replace sec²θ with 1 + tan²θ so that only tan θ remains.

1 + tan²θ + tan θ = 3
tan²θ + tan θ − 2 = 0
(tan θ − 1)(tan θ + 2) = 0
tan θ = 1  or  tan θ = −2

For tan θ = 1: θ = 45° or 225°. For tan θ = −2: the acute reference angle is 63.43…°, so θ = 180° − 63.43…° = 116.6° and θ = 360° − 63.43…° = 296.6° (1 d.p.).

Solutions: θ = 45°, 116.6°, 225°, 296.6°.

Worked example 5. Solve cosec x = 2 for 0 < x < π, then say which solution also satisfies cot x = −√3.

cosec x = 2 means sin x = 1/2, so x = π/6 or 5π/6.

Check each in cot x = cos x/sin x. At π/6: cot = (√3/2)/(1/2) = √3. At 5π/6: cot = (−√3/2)/(1/2) = −√3.

So the solutions of cosec x = 2 are x = π/6, 5π/6, and only 5π/6 also gives cot x = −√3.

Inverse trigonometric functions

Why the domain must be restricted

A function has an inverse only if each output comes from exactly one input. sin x takes the value 1/2 at x = π/6, 5π/6, 13π/6 and so on, so the full sine function has no inverse. To fix this, restrict sin x to an interval on which it is one-to-one and covers its whole range. The inverse is defined from that restricted piece.

Function	Restricted piece of the original	Domain of inverse	Range of inverse
sin⁻¹x	sin x on [−π/2, π/2]	[−1, 1]	[−π/2, π/2]
cos⁻¹x	cos x on [0, π]	[−1, 1]	[0, π]
tan⁻¹x	tan x on (−π/2, π/2)	all real x	(−π/2, π/2)

The notation sin⁻¹x means “the angle whose sine is x”. It does not mean 1/sin x. The reciprocal is cosec x.

Graphs as reflections in y = x

Reflect the restricted piece of the original graph in the line y = x and you have the inverse graph. Swap the coordinates of each key point.

y = sin⁻¹x:  (−1, −π/2), (0, 0), (1, π/2)    increasing, S-shaped, ends at x = ±1
y = cos⁻¹x:  (−1, π), (0, π/2), (1, 0)       decreasing, ends at x = ±1
y = tan⁻¹x:  (0, 0)                          increasing, horizontal asymptotes y = ±π/2

The sine and cosine inverses stop at x = −1 and x = 1 because the restricted original only has outputs between −1 and 1. The tangent inverse never stops: tan x takes every real value on its interval, so tan⁻¹x is defined for all x, and it approaches ±π/2 without reaching them.

Worked example 6. Find the domain and range of f(x) = 2cos⁻¹x + 1.

The domain is the same as for cos⁻¹x: −1 ≤ x ≤ 1. For the range, 0 ≤ cos⁻¹x ≤ π gives 0 ≤ 2cos⁻¹x ≤ 2π, then adding 1: 1 ≤ f(x) ≤ 1 + 2π.

Worked example 7. Find the domain of g(x) = sin⁻¹(2x − 1).

The input to sin⁻¹ must lie in [−1, 1]: −1 ≤ 2x − 1 ≤ 1, so 0 ≤ 2x ≤ 2, giving 0 ≤ x ≤ 1.

Exact values

Work backwards from a known sine, cosine or tangent value, then check the answer lies in the principal range.

sin⁻¹(√3/2) = π/3        cos⁻¹(−1/2) = 2π/3        tan⁻¹(−√3) = −π/3

For cos⁻¹(−1/2), the angles with cosine −1/2 include 2π/3 and 4π/3, but only 2π/3 lies in [0, π].

Compositions

Because the functions are inverses on the restricted pieces:

sin(sin⁻¹x) = x       for −1 ≤ x ≤ 1
sin⁻¹(sin x) = x      holds just for x in [−π/2, π/2]

The same pattern holds for cos⁻¹ (restricted piece [0, π]) and tan⁻¹ (restricted piece (−π/2, π/2)). Outside the restricted interval, find the angle inside it that has the same sine (or cosine, or tangent).

Worked example 8. Evaluate sin⁻¹(sin 5π/6) and tan⁻¹(tan 3π/4).

5π/6 is outside [−π/2, π/2]. sin(5π/6) = sin(π/6), and π/6 is in the range, so sin⁻¹(sin 5π/6) = π/6.

3π/4 is outside (−π/2, π/2). tan(3π/4) = −1 = tan(−π/4), so tan⁻¹(tan 3π/4) = −π/4.

Worked example 9. Find the exact value of sin(cos⁻¹(−7/25)).

Let α = cos⁻¹(−7/25). Then cos α = −7/25 and 0 ≤ α ≤ π, so α is obtuse and sin α is positive.

sin²α = 1 − 49/625 = 576/625
sin α = 24/25   (positive, since α is in [0, π])

So the answer is 24/25. A sketch of a right-angled triangle with sides 7, 24 and 25 confirms the lengths; the range of cos⁻¹ settles the sign.

Worked example 10. Find the exact value of tan(sin⁻¹(−4/5)).

Let β = sin⁻¹(−4/5), so sin β = −4/5 and −π/2 ≤ β ≤ π/2. Since sin β is negative, β lies in [−π/2, 0], where cos β is positive.

cos²β = 1 − 16/25 = 9/25,   cos β = 3/5
tan β = sin β/cos β = (−4/5)/(3/5) = −4/3

Answer: −4/3.

Common errors

• Writing sin⁻¹x = 1/sin x. The reciprocal of sine is cosec x.
• Giving sin⁻¹(sin x) = x for any x. It only works inside the principal interval.
• Giving cos⁻¹(−1/2) as −2π/3 or 4π/3. The range of cos⁻¹ is [0, π], so negative angles never appear.
• Sketching sec x with branches that dip into the strip −1 < y < 1.
• Drawing cot x rising like tan x. In each branch cot x falls.
• Forgetting that tan⁻¹x has asymptotes but no end points, while sin⁻¹x and cos⁻¹x stop at x = ±1.
• Using 1 + tan²x = sec²x in an equation and then stopping at the quadratic. Solve for tan x, then go back to find the angles, including every solution in the interval.
• Taking the wrong sign for the root when finding sin(cos⁻¹x) or cos(sin⁻¹x). Use the range to decide.

Next steps

Try the practice questions, then condense with the revision notes. Check your progress against the checklist and the free diagnostics.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, published by OxfordAQA: content area P2.4 Trigonometry (inverse trigonometric functions; secant, cosecant and cotangent).