Further Trigonometry | FractionRush A-Level Pure 3

Welcome to Further Trigonometry — Pure 3

Reciprocal trigonometric functions — sec, cosec, and cot — extend your trig toolkit far beyond what Pure 1 and Pure 2 covered. They appear throughout A-Level: in integration by substitution, modelling periodic phenomena, and solving equations that would otherwise be intractable. Inverse trig functions then let you reverse these operations and arise naturally in the differentiation and integration of algebraic expressions like 1/(1+x²).

sec x = 1/cos x | cosec x = 1/sin x | cot x = cos x/sin x
1 + tan²x = sec²x | 1 + cot²x = cosec²x

Learning Objectives

Define sec x, cosec x, and cot x and evaluate them at key angles
Sketch the graphs of y = sec x, y = cosec x, and y = cot x, including asymptotes
Derive and use the identity sec²x = 1 + tan²x
Derive and use the identity cosec²x = 1 + cot²x
Prove harder trigonometric identities using reciprocal trig functions
Define arcsin, arccos, and arctan with their restricted domains and ranges
Sketch the graphs of the three inverse trigonometric functions
Solve equations involving sec, cosec, and cot by converting to sin/cos/tan
Differentiate arcsin x and arctan x using implicit differentiation
Integrate 1/(1+x²) and 1/√(1−x²) to obtain standard inverse trig results

Topic Overview

Reciprocal Functions

sec, cosec, cot: definitions, values at key angles, and their domains

Graphs

Shapes, asymptotes, and periods of the three reciprocal trig graphs

Identities

The two Pythagorean identities with sec and cosec, and how to prove harder results

Inverse Trig

arcsin, arccos, arctan: domains, ranges, and key values

Solving Equations

Convert sec/cosec/cot equations back to sin/cos/tan and solve

Differentiation & Integration

d/dx(arcsin x), d/dx(arctan x), and their integral counterparts

Learn 1 — Reciprocal Trigonometric Functions

Just as division is the reciprocal of multiplication, the three reciprocal trig functions are built by flipping sin, cos, and tan. Each is defined wherever the denominator is non-zero.

Definitions

sec x = 1 / cos x (defined when cos x ≠ 0, i.e. x ≠ 90°, 270°, ...)

cosec x = 1 / sin x (defined when sin x ≠ 0, i.e. x ≠ 0°, 180°, ...)

cot x = 1 / tan x = cos x / sin x (defined when sin x ≠ 0)

Common confusion: sec x is the reciprocal of cos x, NOT sin x. The "sec" abbreviation does NOT stand for "sine cosecant" — it stands for secant, which is 1/cos. Similarly cosec is 1/sin, not 1/cos.

Values at Key Angles

Angle θ	sin θ	cos θ	tan θ	cosec θ	sec θ	cot θ
0°	0	1	0	undef	1	undef
30°	1/2	√3/2	1/√3	2	2/√3 ≈ 1.155	√3 ≈ 1.732
45°	√2/2	√2/2	1	√2 ≈ 1.414	√2 ≈ 1.414	1
60°	√3/2	1/2	√3	2/√3 ≈ 1.155	2	1/√3 ≈ 0.577
90°	1	0	undef	1	undef	0
120°	√3/2	−1/2	−√3	2/√3	−2	−1/√3
180°	0	−1	0	undef	−1	undef

Relationship to Original Functions

Example: Evaluate sec(60°) + cosec(30°) − cot(45°)

sec(60°) = 1/cos(60°) = 1/(1/2) = 2

cosec(30°) = 1/sin(30°) = 1/(1/2) = 2

cot(45°) = cos(45°)/sin(45°) = (√2/2)/(√2/2) = 1

Total = 2 + 2 − 1 = 3

Example: If cos x = 3/5 (x acute), find sec x and cot x

sec x = 1/cos x = 5/3

By Pythagoras: sin x = 4/5, so tan x = 4/3

cot x = 1/tan x = 3/4

Learn 2 — Graphs and Properties

Graph of y = sec x

Key features:
• Vertical asymptotes at x = ±90°, ±270°, ±450°, ... (wherever cos x = 0)
• Between asymptotes, the graph forms U-shapes (opening upward where cos x > 0, downward where cos x < 0)
• Minimum value of each upward U: y = 1; maximum of each downward U: y = −1
• Period = 360° (same as cos x)
• The graph of y = sec x always sits outside the band −1 < y < 1

Graph of y = cosec x

Key features:
• Vertical asymptotes at x = 0°, ±180°, ±360°, ... (wherever sin x = 0)
• U-shaped sections between asymptotes, same qualitative shape as sec x but shifted by 90°
• Period = 360°
• Never takes values in the open interval (−1, 1)

Graph of y = cot x

Key features:
• Vertical asymptotes at x = 0°, ±180°, ... (wherever sin x = 0 — same as cosec x)
• Decreasing curve between asymptotes (opposite direction to tan x)
• Period = 180° (same as tan x)
• Passes through zero where tan x is undefined, and is undefined where tan x = 0

Graph-sketching strategy for sec/cosec:
1. Sketch the underlying sin or cos curve lightly
2. Mark every zero of sin/cos as a vertical asymptote
3. At each maximum (value 1) of the original curve, the reciprocal also has value 1 — draw a U opening upward
4. At each minimum (value −1), the reciprocal also has value −1 — draw a U opening downward

Periodicity Summary

Function	Period	Asymptotes	Range
y = sec x	360°	x = 90° + 180°n	y ≤ −1 or y ≥ 1
y = cosec x	360°	x = 180°n	y ≤ −1 or y ≥ 1
y = cot x	180°	x = 180°n	ℝ (all reals)

Example: State the asymptotes of y = sec(2x) in [0°, 360°]

sec(2x) has asymptotes where cos(2x) = 0

cos(2x) = 0 when 2x = 90°, 270°, 450°, 630°

So x = 45°, 135°, 225°, 315°

Learn 3 — Pythagorean Identities with sec and cosec

Starting from the fundamental identity sin²x + cos²x = 1, dividing through by cos²x or sin²x yields two powerful new identities involving the reciprocal functions.

Derivation: sec²x = 1 + tan²x

Divide sin²x + cos²x = 1 by cos²x:

sin²x/cos²x + cos²x/cos²x = 1/cos²x
tan²x + 1 = sec²x
∴ sec²x = 1 + tan²x ✓

Derivation: cosec²x = 1 + cot²x

Divide sin²x + cos²x = 1 by sin²x:

sin²x/sin²x + cos²x/sin²x = 1/sin²x
1 + cot²x = cosec²x
∴ cosec²x = 1 + cot²x ✓

sec²x = 1 + tan²x cosec²x = 1 + cot²x

Useful Rearrangements

From sec²x = 1 + tan²x:
  • sec²x − 1 = tan²x
  • sec²x − tan²x = 1

From cosec²x = 1 + cot²x:
  • cosec²x − 1 = cot²x
  • cosec²x − cot²x = 1

Proving Identities

Strategy for proving identities with sec/cosec:
1. Work on the more complicated side
2. Convert everything to sin and cos if stuck
3. Use sec²x = 1+tan²x to switch between tan and sec
4. Factor where possible before simplifying

Example: Prove (sec x − tan x)(sec x + tan x) = 1

LHS = sec²x − tan²x (difference of two squares)

= (1 + tan²x) − tan²x (using sec²x = 1 + tan²x)

= 1 = RHS ✓

Example: Simplify cosec²θ − cot θ cosec θ

= cosec θ (cosec θ − cot θ) (factor out cosec θ)

= (1/sin θ)(1/sin θ − cos θ/sin θ)

= (1/sin θ) × (1 − cos θ)/sin θ

= (1 − cos θ)/sin²θ = (1 − cos θ)/(1 − cos²θ) = 1/(1 + cos θ)

Example: If tan x = 3, find sec²x and hence sec x (x acute)

sec²x = 1 + tan²x = 1 + 9 = 10

sec x = √10 (taking positive root since x is acute)

Learn 4 — Inverse Trigonometric Functions

To invert sin, cos, and tan we must restrict their domains so they become one-to-one. The resulting functions arcsin, arccos, and arctan each have a specific, fixed range.

arcsin x

y = arcsin x means sin y = x
Domain: [−1, 1] Range: [−π/2, π/2] = [−90°, 90°]

arcsin is the inverse of sin restricted to [−90°, 90°]. It is an increasing function, passing through (0, 0).

arccos x

y = arccos x means cos y = x
Domain: [−1, 1] Range: [0, π] = [0°, 180°]

arccos is the inverse of cos restricted to [0°, 180°]. It is a decreasing function, passing through (0, π/2).

arctan x

y = arctan x means tan y = x
Domain: ℝ (all reals) Range: (−π/2, π/2) = (−90°, 90°)

arctan is the inverse of tan restricted to (−90°, 90°). It has horizontal asymptotes at y = ±π/2.

Key Values

Expression	Value (radians)	Value (degrees)
arcsin(0)	0	0°
arcsin(1/2)	π/6	30°
arcsin(√2/2)	π/4	45°
arcsin(1)	π/2	90°
arccos(1)	0	0°
arccos(1/2)	π/3	60°
arccos(0)	π/2	90°
arccos(−1)	π	180°
arctan(0)	0	0°
arctan(1)	π/4	45°
arctan(√3)	π/3	60°
arctan(−1)	−π/4	−45°

Derivatives

d/dx(arcsin x) = 1/√(1 − x²) for |x| < 1

d/dx(arctan x) = 1/(1 + x²) for all x

Chain rule extension: d/dx(arctan(f(x))) = f'(x)/(1 + [f(x)]²)
e.g. d/dx(arctan(3x)) = 3/(1 + 9x²)

Important: arccos has no standard derivative formula required at A-Level P3, though it is d/dx(arccos x) = −1/√(1−x²). Focus on arcsin and arctan.

Learn 5 — Solving Equations and Integration

Solving Equations with sec, cosec, cot

The key technique is always to convert to sin, cos, or tan first, then use standard methods.

Example: Solve sec x = 2 for 0° ≤ x ≤ 360°

sec x = 2 ⟹ 1/cos x = 2 ⟹ cos x = 1/2

cos x = 1/2 ⟹ x = 60° or x = 300°

Example: Solve cosec x = −2 for 0° ≤ x ≤ 360°

cosec x = −2 ⟹ sin x = −1/2

sin is negative in 3rd and 4th quadrants

Reference angle: arcsin(1/2) = 30°

x = 180° + 30° = 210° or x = 360° − 30° = 330°

Solving Harder Equations using Identities

Example: Solve 2sec²x + tan x = 3 for 0° ≤ x ≤ 360°

Replace sec²x with 1 + tan²x:

2(1 + tan²x) + tan x = 3

2 + 2tan²x + tan x = 3

2tan²x + tan x − 1 = 0

(2tan x − 1)(tan x + 1) = 0

tan x = 1/2: x = 26.6°, 206.6° tan x = −1: x = 135°, 315°

Integration — Standard Forms

∫ 1/(1 + x²) dx = arctan x + C

∫ 1/√(1 − x²) dx = arcsin x + C

Scaled Versions (by substitution or recognition)

∫ 1/(a² + x²) dx = (1/a) arctan(x/a) + C

Example: Find ∫ 1/(9 + x²) dx

Write as ∫ 1/(3² + x²) dx where a = 3

= (1/3) arctan(x/3) + C

Example: Evaluate ∫₀¹ 1/(1 + x²) dx

= [arctan x]₀¹ = arctan(1) − arctan(0) = π/4 − 0 = π/4

Examination tip: For ∫ 1/(a²+x²) dx, always write a² explicitly. If the integral is ∫ 1/(4+x²) dx, then a = 2 (not a = 4), so the answer is (1/2)arctan(x/2) + C.

Worked Examples

Example 1: Find the exact value of cosec(30°) + sec(60°) + cot(45°)

cosec(30°) = 1/sin(30°) = 1/(1/2) = 2

sec(60°) = 1/cos(60°) = 1/(1/2) = 2

cot(45°) = cos(45°)/sin(45°) = (√2/2)/(√2/2) = 1

Total = 2 + 2 + 1 = 5

Example 2: Prove sec²θ − cosec²θ = tan²θ − cot²θ

LHS = sec²θ − cosec²θ = (1 + tan²θ) − (1 + cot²θ)

= 1 + tan²θ − 1 − cot²θ

= tan²θ − cot²θ = RHS ✓ M1 A1

Example 3: Solve 2sec²x − tan x − 3 = 0 for 0° ≤ x ≤ 360°

Replace sec²x = 1 + tan²x: 2(1 + tan²x) − tan x − 3 = 0 M1

2tan²x − tan x − 1 = 0

(2tan x + 1)(tan x − 1) = 0 M1

tan x = −1/2: x = 180° − 26.6° = 153.4° or 360° − 26.6° = 333.4° A1

tan x = 1: x = 45° or 225° A1

Solutions: 45°, 153.4°, 225°, 333.4°

Example 4: Sketch y = sec x for −180° ≤ x ≤ 360°, identifying asymptotes

Asymptotes occur where cos x = 0: x = −90°, 90°, 270°

Between x = −90° and x = 90°: U-shape with minimum at (0°, 1)

Between x = 90° and x = 270°: inverted U with maximum at (180°, −1)

Between x = 270° and x = 360°: partial U-shape, minimum at (360°, 1)

Mark asymptotes as dashed vertical lines. Range: y ≤ −1 or y ≥ 1

Example 5: Differentiate y = arctan(3x)

Using d/dx(arctan u) = u'/(1 + u²) with u = 3x, u' = 3: M1

dy/dx = 3/(1 + (3x)²) = 3/(1 + 9x²) A1

Example 6: Integrate ∫ 1/(9 + x²) dx

Write as ∫ 1/(3² + x²) dx, using standard form ∫ 1/(a²+x²) dx = (1/a)arctan(x/a) + C M1

a = 3, so answer = (1/3)arctan(x/3) + C A1

Example 7: Solve cosec x + 2cot x = 0 for 0 < x < 2π

1/sin x + 2cos x/sin x = 0 (convert to sin/cos) M1

(1 + 2cos x)/sin x = 0

For the fraction to be zero, numerator = 0: 1 + 2cos x = 0 ⟹ cos x = −1/2 M1

(sin x ≠ 0 for cosec to be defined, so denominator is not zero)

cos x = −1/2 ⟹ x = 2π/3 or x = 4π/3 A1 A1

Example 8: Prove cot A + tan A = cosec A · sec A

LHS = cos A/sin A + sin A/cos A M1

= (cos²A + sin²A)/(sin A cos A) (combine over common denominator)

= 1/(sin A cos A) (using sin²A + cos²A = 1) A1

= (1/sin A)(1/cos A) = cosec A · sec A = RHS ✓ A1

Common Mistakes to Avoid

Mistake 1: Confusing sec x with sin x / cos x

✗ sec x = sin x / cos x (WRONG — that is tan x)

✓ sec x = 1 / cos x (secant is the reciprocal of cosine)

Memory tip: sec goes with cos (both have the letter c). cosec goes with sin.

Mistake 2: Incorrect method when solving sec x = k

✗ Applying "sec" to both sides: sec⁻¹(sec x) = sec⁻¹(k), then writing x = sec⁻¹(k) without using a calculator correctly

✓ Convert first: sec x = k ⟹ cos x = 1/k, then solve cos x = 1/k using the standard cos curve

Mistake 3: Using the wrong Pythagorean identity

✗ sec²x = tan²x + cos²x (completely wrong)

✗ sec²x = sin²x + 1

✓ sec²x = 1 + tan²x (divide sin²+cos²=1 by cos²)

✓ cosec²x = 1 + cot²x (divide sin²+cos²=1 by sin²)

Mistake 4: Invalid inputs to arcsin and arccos

✗ arcsin(2) exists (WRONG — sin x only reaches max 1, so arcsin only accepts values in [−1, 1])

✓ arcsin is only defined for inputs in [−1, 1]. If asked to evaluate arcsin(1.5), the answer is "undefined" or "no solution".

Mistake 5: Confusing arctan(1) with π/2

✗ arctan(1) = π/2 (WRONG — that would mean tan(π/2) = 1, but tan(π/2) is undefined)

✓ arctan(1) = π/4 = 45°, because tan(45°) = 1

The range of arctan is (−π/2, π/2) but π/2 is NOT attained — it is an asymptote.

Mistake 6: Forgetting domain restrictions when solving

✗ From cosec x = k, writing sin x = 1/k then accepting x = 0° or x = 180° as solutions

✓ cosec x is undefined at x = 0°, 180°, 360°... Always check that your solution does not sit at an asymptote of the original reciprocal function.

Mistake 7: Confusing the derivatives of arcsin and arctan

✗ d/dx(arctan x) = 1/√(1 − x²) (that is the derivative of arcsin)

✗ d/dx(arcsin x) = 1/(1 + x²) (that is the derivative of arctan)

✓ d/dx(arcsin x) = 1/√(1 − x²) — note the square root and the minus

✓ d/dx(arctan x) = 1/(1 + x²) — note the plus and no square root

Key Formulas — Further Trigonometry

Formula / Concept	Detail
sec x	= 1/cos x (undefined when cos x = 0)
cosec x	= 1/sin x (undefined when sin x = 0)
cot x	= cos x/sin x = 1/tan x (undefined when sin x = 0)
sec²x = 1 + tan²x	Pythagorean identity (divide sin²+cos²=1 by cos²)
cosec²x = 1 + cot²x	Pythagorean identity (divide sin²+cos²=1 by sin²)
arcsin x	Domain: [−1, 1] Range: [−π/2, π/2]
arccos x	Domain: [−1, 1] Range: [0, π]
arctan x	Domain: ℝ Range: (−π/2, π/2)
arcsin(1/2)	= π/6 (30°)
arccos(0)	= π/2 (90°)
arctan(1)	= π/4 (45°)
d/dx(arcsin x)	= 1/√(1 − x²) for \|x\| < 1
d/dx(arctan x)	= 1/(1 + x²) for all x
∫ 1/(1+x²) dx	= arctan x + C
∫ 1/√(1−x²) dx	= arcsin x + C
∫ 1/(a²+x²) dx	= (1/a) arctan(x/a) + C
Chain rule: arctan(f(x))	d/dx = f'(x)/(1 + [f(x)]²)
Chain rule: arcsin(f(x))	d/dx = f'(x)/√(1 − [f(x)]²)

Proof Bank

Proof 1: tan²x + 1 = sec²x

Start with the fundamental identity: sin²x + cos²x = 1

Divide every term by cos²x (valid when cos x ≠ 0):
  sin²x/cos²x + cos²x/cos²x = 1/cos²x

Apply definitions:
  (sin x/cos x)² + 1 = (1/cos x)²
  tan²x + 1 = sec²x   ✓

Rearrangements: sec²x − tan²x = 1, or (sec x − tan x)(sec x + tan x) = 1

Proof 2: d/dx(arctan x) = 1/(1 + x²) using implicit differentiation

Let y = arctan x, so tan y = x. Define

Differentiate both sides with respect to x (implicit differentiation):
  d/dx(tan y) = d/dx(x)
  sec²y · dy/dx = 1 Chain rule

So: dy/dx = 1/sec²y

Now express sec²y in terms of x. Since tan y = x:
  sec²y = 1 + tan²y = 1 + x² Identity

Therefore: dy/dx = 1/(1 + x²)   ✓ Substitute back

Bonus Proof: d/dx(arcsin x) = 1/√(1 − x²)

Let y = arcsin x, so sin y = x.

Differentiate implicitly: cos y · dy/dx = 1

So dy/dx = 1/cos y

Express cos y in terms of x. Since sin y = x:
cos y = √(1 − sin²y) = √(1 − x²) (positive root since y ∈ [−π/2, π/2])

Therefore: dy/dx = 1/√(1 − x²) ✓

Interactive Visualiser — Reciprocal Trig Functions

Toggle the functions below to compare the original and reciprocal trig curves. Vertical asymptotes are shown as dashed red lines.

Colour key: cos x (blue) · sec x (purple) · sin x (orange) · cosec x (green) · cot x (red)

Exercise 1 — Reciprocal Trig Values (10 Questions)

Exercise 2 — Reciprocal Identities (10 Questions)

Exercise 3 — Solving Equations with Reciprocal Trig (10 Questions)

Exercise 4 — Inverse Trigonometric Functions (10 Questions)

Exercise 5 — Differentiation and Integration (10 Questions)

Practice — 30 Questions

Challenge — 15 Hard Questions

Exam Style Questions

Question 1 [4 marks]

(a) Show that the equation 2cosec²x + cot x − 3 = 0 can be written as 2cot²x + cot x − 1 = 0. [2]

(b) Hence solve 2cosec²x + cot x − 3 = 0 for 0° < x < 360°. [2]

(a) Replace cosec²x = 1 + cot²x [M1]: 2(1+cot²x) + cotx − 3 = 2 + 2cot²x + cotx − 3 = 2cot²x + cotx − 1 [A1] ✓
(b) (2cotx − 1)(cotx + 1) = 0 [M1]
cotx = 1/2 ⟹ tanx = 2: x = 63.4°, 243.4° [A1]
cotx = −1 ⟹ tanx = −1: x = 135°, 315° [A1] (allow 1 mark for 2 out of 4)

Question 2 [5 marks]

Prove the identity: (1 + cosecθ)(1 − sinθ) = cosθ · cotθ [5]

LHS = (1 + 1/sinθ)(1 − sinθ) [M1 converting cosec]
= (sinθ + 1)/sinθ · (1 − sinθ) [A1]
= (1 − sin²θ)/sinθ [M1 multiplying out]
= cos²θ/sinθ [A1 using 1−sin²θ = cos²θ]
= cosθ · (cosθ/sinθ) = cosθ · cotθ = RHS ✓ [A1]

Question 3 [5 marks]

(a) Differentiate y = x·arctan x with respect to x. [2]

(b) Hence find ∫ arctan x dx, giving your answer in terms of arctan x and ln. [3]

(a) Using product rule: dy/dx = arctan x + x · 1/(1+x²) [M1 A1]
(b) Rearranging: ∫arctan x dx = x·arctan x − ∫ x/(1+x²) dx [M1]
∫ x/(1+x²) dx = (1/2)ln(1+x²) + C [M1 recognising derivative of ln(1+x²)/2]
∴ ∫arctan x dx = x·arctan x − (1/2)ln(1+x²) + C [A1]

Question 4 [4 marks]

Solve the equation sec²x = 3 + tan x for 0° ≤ x ≤ 360°, giving exact solutions where appropriate. [4]

Replace sec²x = 1 + tan²x [M1]: 1 + tan²x = 3 + tanx
tan²x − tanx − 2 = 0 [A1]
(tanx − 2)(tanx + 1) = 0 [M1]
tanx = 2: x = 63.4°, 243.4° [A1 for both]
tanx = −1: x = 135°, 315° [A1 for both]

Question 5 [4 marks]

Find ∫₀^(√3) 1/(1+x²) dx, giving your answer exactly. [4]

∫₀^(√3) 1/(1+x²) dx = [arctan x]₀^(√3) [M1 A1]
= arctan(√3) − arctan(0) [M1]
= π/3 − 0 = π/3 [A1]

Question 6 [5 marks]

The function f is defined by f(x) = arctan(2x − 1) for x ∈ ℝ.

(a) State the range of f. [1]

(b) Find f'(x). [2]

(c) Find the gradient of the curve y = f(x) at the point where x = 1. [2]

(a) Range: (−π/2, π/2) [B1]
(b) Let u = 2x−1, du/dx = 2. f'(x) = 2/(1+(2x−1)²) [M1 A1]
(c) At x=1: f'(1) = 2/(1+(2−1)²) = 2/(1+1) = 2/2 = 1 [M1 A1]

Question 7 [6 marks]

Prove the identity: sec⁴θ − tan⁴θ ≡ sec²θ + tan²θ [3]

Hence show that sec⁴θ − tan⁴θ ≡ 2sec²θ − 1. [3]

LHS = (sec²θ − tan²θ)(sec²θ + tan²θ) [M1 difference of two squares]
= 1 · (sec²θ + tan²θ) [M1 using sec²θ − tan²θ = 1]
= sec²θ + tan²θ = RHS ✓ [A1]
Now: sec²θ + tan²θ = sec²θ + (sec²θ − 1) [M1 using tan²θ = sec²θ−1]
= 2sec²θ − 1 [A1] ✓ [A1 for complete argument]

Question 8 [5 marks]

(a) By writing cosec x = 1/sin x and cot x = cos x/sin x, solve cosec x = 2 + cot x for 0 < x < 2π. [3]

(b) State how many solutions the equation has in the interval 0 < x < 4π. [2]

(a) 1/sinx = 2 + cosx/sinx [M1]
1 = 2sinx + cosx [M1 multiply by sinx, sinx ≠ 0]
This is of the form Rsin(x+α): R=√5, α=arctan(1/2)≈0.4636 rad
√5 sin(x + 0.4636) = 1 → sin(x+0.4636) = 1/√5 [M1]
x + 0.4636 = 0.4636 → x = 0 (excluded) or x + 0.4636 = π − 0.4636 = 2.678 → x = 2.21 rad [A1]
Check x=0 excluded; one solution x ≈ 2.21 rad [A1]
(b) The equation has period 2π, so in (0, 4π) there are 2 solutions [M1 A1]

Past Paper Questions — Cambridge 9709 Pure 3

Past Paper 1 — Cambridge 9709/31 (adapted)

Solve the equation cosec θ = 3 + cot θ for 0 < θ < 2π, giving your answers in radians to 3 significant figures. [5]

1/sinθ = 3 + cosθ/sinθ [M1 converting to sin/cos]
1 = 3sinθ + cosθ [M1 multiplying through by sinθ, sinθ≠0]
Write as Rsin(θ+α): R = √10, α = arctan(1/3) = 0.3217 rad [M1]
√10 sin(θ+0.322) = 1 → sin(θ+0.322) = 1/√10 [A1]
θ+0.322 = 0.3217 → θ ≈ 0 (excluded, sinθ=0 at boundary) [check]
θ+0.322 = π−0.3217 = 2.820 → θ = 2.498 ≈ 2.50 rad [A1]
Also check: θ+0.322 = 2π+0.3217 → θ ≈ 6.28 (outside range)
Solution: θ ≈ 2.50 rad

Past Paper 2 — Cambridge 9709/32 (adapted)

(i) Prove that sec²A − cosec²A = (sec²A)(cosec²A)(sin²A − cos²A). [3]

(ii) Hence, given that sin²A − cos²A = 1/2, find the value of sec²A − cosec²A. [2]

(i) RHS = (1/cos²A)(1/sin²A)(sin²A − cos²A) [M1]
= (sin²A − cos²A)/(sin²A cos²A)
= sin²A/(sin²A cos²A) − cos²A/(sin²A cos²A) [M1]
= 1/cos²A − 1/sin²A = sec²A − cosec²A = LHS ✓ [A1]
(ii) sec²A − cosec²A = sec²A · cosec²A · (1/2) [M1]
But sec²A = 1+tan²A and using sin²A−cos²A=1/2 with sin²A+cos²A=1
gives sin²A = 3/4, cos²A = 1/4, so sec²A=4, cosec²A=4/3
sec²A − cosec²A = 4 − 4/3 = 8/3 [A1]

Past Paper 3 — Cambridge 9709/33 (adapted)

Find ∫₀^(1/2) 3/√(1−x²) dx, giving your answer exactly. [3]

∫₀^(1/2) 3/√(1−x²) dx = 3[arcsin x]₀^(1/2) [M1 A1]
= 3(arcsin(1/2) − arcsin(0))
= 3(π/6 − 0) = π/2 [A1]

Past Paper 4 — Cambridge 9709/31 (adapted)

Given that y = arctan(x²), find dy/dx. Hence find the exact value of ∫₀¹ 2x/(1+x⁴) dx. [5]

y = arctan(x²), dy/dx = 2x/(1+(x²)²) = 2x/(1+x⁴) [M1 A1]
Therefore ∫₀¹ 2x/(1+x⁴) dx = [arctan(x²)]₀¹ [M1]
= arctan(1) − arctan(0) [M1]
= π/4 − 0 = π/4 [A1]

Past Paper 5 — Cambridge 9709/32 (adapted)

Solve the equation 3sec²x − 5tanx − 1 = 0 for 0° ≤ x ≤ 360°, giving answers to 1 decimal place. [5]

Replace sec²x = 1+tan²x [M1]: 3(1+tan²x) − 5tanx − 1 = 0
3tan²x − 5tanx + 2 = 0 [A1]
(3tanx − 2)(tanx − 1) = 0 [M1]
tanx = 2/3: x = 33.7°, 213.7° [A1]
tanx = 1: x = 45°, 225° [A1]

Further Trigonometry A-Level Pure 3