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

Condensed for the final weeks. For full explanations and worked examples, use the study guide. These notes follow content area P2.4 (Trigonometry, first part) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2. The specification places this content in International A-level Unit P2, so it is A2 content and not examined at International AS. Angles are in radians unless a degree sign appears.

Also useful: the hub, the checklist, the practice set and the free diagnostics.

Definitions

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

Function	Undefined at	Range	Period
sec x	x = π/2 + nπ	abs(sec x) ≥ 1	2π
cosec x	x = nπ	abs(cosec x) ≥ 1	2π
cot x	x = nπ	all reals	π

cot(π/2) = 0 even though tan(π/2) does not exist.

Exact values to know

Angle	sec	cosec	cot
π/6	2/√3	2	√3
π/4	√2	√2	1
π/3	2	2/√3	1/√3
π/2	undefined	1	0
0	1	undefined	undefined

Get any other angle by finding the reference angle, then fixing the sign with the quadrant: sec has the sign of cos, cosec the sign of sin, and cot the sign of tan.

Graph features

Graph	Asymptotes	Turning points / crossings	Shape
y = sec x	x = ±π/2, ±3π/2 …	minimum (0, 1), maximum (π, −1)	U-shaped branches, never inside −1 < y < 1
y = cosec x	x = 0, ±π, ±2π …	minimum (π/2, 1), maximum (3π/2, −1)	same shape as sec, shifted by π/2
y = cot x	x = 0, ±π, ±2π …	crosses the x-axis at ±π/2, ±3π/2 …	every branch falls left to right

Method: sketching a reciprocal graph

1. Sketch the original (cos, sin or tan) lightly.
2. Draw an asymptote at each x-intercept of the original.
3. Where the original equals 1 or −1, the reciprocal passes through the same point.
4. Draw each branch curving away from that point towards the asymptotes.
5. Label asymptote equations and the coordinates of the turning points.

The two identities

sin²x + cos²x = 1
÷ cos²x:   tan²x + 1 = sec²x
÷ sin²x:   1 + cot²x = cosec²x

Rearranged: tan²x = sec²x − 1, cot²x = cosec²x − 1.

Method: equations with sec or cosec squared

1. Spot a squared reciprocal alongside tan (or cot) in the same equation.
2. Replace sec²x by 1 + tan²x (or cosec²x by 1 + cot²x).
3. Collect terms to get a quadratic in tan x (or cot x).
4. Factorise or use the formula, then solve each linear piece.
5. Find every angle in the stated interval for each value.

Worked reminder: equation with cosec²

Solve 3cosec²x = 7 − cot x for 0° < x < 180°.

3(1 + cot²x) = 7 − cot x
3cot²x + cot x − 4 = 0
(3cot x + 4)(cot x − 1) = 0
cot x = 1   or   cot x = −4/3
tan x = 1   or   tan x = −3/4
x = 45°     or   x = 180° − 36.87...° = 143.1° (1 d.p.)

Only one solution comes from each value because the interval is a single period of tan x.

Worked reminder: proof

Prove that cosec x − sin x ≡ cos x cot x.

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

Method: proving an identity

1. Pick the more complicated side and work only on that side.
2. Rewrite sec, cosec and cot in terms of sin x and cos x.
3. Combine fractions over a common denominator.
4. Use sin²x + cos²x = 1, or one of the two identities above, to simplify.
5. Stop when the expression matches the other side exactly, and write the final line as the target.

A second route is to substitute the identities directly. To prove sec²x − cosec²x ≡ tan²x − cot²x:

LHS = (1 + tan²x) − (1 + cot²x) = tan²x − cot²x = RHS

Inverse functions

Function	Restricted piece used	Domain	Range
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)

Graph points

sin⁻¹x:  (−1, −π/2), (0, 0), (1, π/2)     increasing
cos⁻¹x:  (−1, π), (0, π/2), (1, 0)        decreasing
tan⁻¹x:  (0, 0), asymptotes y = ±π/2      increasing, defined for all x

Each graph is the reflection of the restricted piece of the original in y = x. Swap x and y at each key point.

Compositions

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

Outside the principal interval, replace the angle by the one inside it with the same sine, cosine or tangent.

Method: exact value of sin(cos⁻¹k) and similar

1. Let α = cos⁻¹k, so cos α = k.
2. Use the range of cos⁻¹ to decide the sign of sin α.
3. Find sin α from sin²α = 1 − k².
4. Check with a right-angled triangle if needed.

Worked reminder: cos(tan⁻¹(−3/4)). Let β = tan⁻¹(−3/4), with β in (−π/2, 0). Then cos β is positive, and sec²β = 1 + 9/16 = 25/16, so sec β = 5/4 and cos β = 4/5.

Domain and range after a transformation

Write the restriction on the inside of the inverse function, then solve it. For the range, apply the transformation to the end points of the principal range.

Function	Domain	Range
3sin⁻¹x − π/2	−1 ≤ x ≤ 1	−2π ≤ y ≤ π
cos⁻¹(x/2)	−2 ≤ x ≤ 2	0 ≤ y ≤ π
tan⁻¹x + π/4	all real x	−π/4 < y < 3π/4

The last graph keeps its shape and moves up by π/4, so its asymptotes are y = −π/4 and y = 3π/4. A stretch or translation changes the range, but the domain of tan⁻¹ stays all real numbers until the input is changed.

Must-know distinctions

• sin⁻¹x is the inverse function. It is not 1/sin x, which is cosec x.
• sec, cosec and cot are reciprocals or ratios. sin⁻¹, cos⁻¹ and tan⁻¹ are inverses. Their graphs have different origins.
• sin⁻¹ and cos⁻¹ end at x = ±1. tan⁻¹ has asymptotes and is defined for every x.
• cos⁻¹ never returns a negative angle. sin⁻¹ and tan⁻¹ do.
• cot x falls in each branch, and tan x rises.
• Domain of the inverse is the range of the original piece. Range of the inverse is the restricted domain.

Checking an answer

• Substitute a solution back into the original equation, using the reciprocal definitions.
• For an inverse value, confirm the angle lies in the stated range of the function.
• For a sketch, check that each asymptote sits where the original function crosses the x-axis.
• For a range, test one value from inside the domain and see that the output lies within your stated interval.

Quick self-test

1. Find the exact value of cosec(5π/3).
2. Find the exact value of cot(π/3).
3. State the values of x that must be excluded from the domain of cosec x.
4. Find sin⁻¹(−1/2) in radians, exactly.
5. Find tan⁻¹(√3) in radians, exactly.
6. Find cos⁻¹(cos(−π/4)).
7. Simplify (1 + tan²θ)cos²θ.
8. State the range of tan⁻¹x.
9. Find the exact value of cos(sin⁻¹(20/29)).
10. Given that tan x = 3/4 and x is acute, find sec x.
11. State the domain of y = sin⁻¹(x/3).

Answers

1. sin(5π/3) = −√3/2, so cosec(5π/3) = −2/√3.
2. tan(π/3) = √3, so cot(π/3) = 1/√3.
3. sin x = 0, so exclude x = nπ for any integer n.
4. −π/6.
5. π/3.
6. cos(−π/4) = cos(π/4), and π/4 is in [0, π], so the answer is π/4.
7. (1 + tan²θ) = sec²θ, and sec²θ cos²θ = 1.
8. −π/2 < tan⁻¹x < π/2.
9. Let α = sin⁻¹(20/29). Then cos α is positive and cos²α = 1 − 400/841 = 441/841, so 21/29.
10. sec²x = 1 + 9/16 = 25/16, and sec x is positive for an acute angle: 5/4.
11. −1 ≤ x/3 ≤ 1, so −3 ≤ x ≤ 3.

Where marks are usually lost

• Writing sin⁻¹x for 1/sin x, or the other way round.
• Leaving out the asymptotes on a sketch of sec, cosec or cot, or drawing the curves through them.
• Sketching sec x or cosec x with a branch that enters the strip −1 < y < 1.
• Giving an angle outside the range of the inverse function, for example cos⁻¹(−1/2) = −2π/3.
• Evaluating sin⁻¹(sin x) as x when x lies outside [−π/2, π/2].
• Choosing the wrong sign for the square root when finding sin(cos⁻¹k) or cos(sin⁻¹k).
• Proving an identity by working on both sides at once. Start from one side and reach the other.
• Solving the quadratic in tan x but then listing only one angle per value.
• Mixing degrees and radians in the final answers.
• Stating the domain of sin⁻¹(2x − 1) as −1 ≤ x ≤ 1 instead of solving for the inside expression.

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).