Pearson Edexcel International Advanced Level Mathematics: Further trigonometry and R-formulae (YMA01) – Revision Notes

These notes condense Unit P3 section 2 (topics 2.1 to 2.3) of the Pearson Edexcel International Advanced Subsidiary/Advanced Level Mathematics specification, Issue 3, April 2019. The specification lists P3 as compulsory for IAL Mathematics and Pure Mathematics only; IAS weighting is N/A. Longer explanations and fuller worked examples are in the study guide. Then test yourself with the practice questions.

Course hub: Edexcel A Level Mathematics. Printable checklist. Unsure what to revise first? A free 10-minute diagnostic will show you. Related earlier work: P2 revision notes.

2.1 Reciprocal functions

Function	Definition	Period	Range	Undefined at
sec θ	1/cos θ	360°	|sec θ| ≥ 1	θ = 90° + 180°n
cosec θ	1/sin θ	360°	|cosec θ| ≥ 1	θ = 180°n
cot θ	1/tan θ = cos θ/sin θ	180°	all real	θ = 180°n

• Reciprocal graphs have vertical asymptotes where the original function is zero.
• Where the original function is ±1, the reciprocal touches it with the same value.
• sec and cosec equations: invert first, then solve cos or sin. If the value is between −1 and 1, there is no solution.

2.1 Inverse functions

Function	Output (radians)	Input
arcsin x	−π/2 to π/2, ends included	−1 to 1, ends included
arccos x	0 to π, ends included	−1 to 1, ends included
arctan x	−π/2 to π/2, ends excluded	any real number

• Degrees: replace π/2 by 90° and π by 180°.
• Graph of an inverse = reflection of the restricted graph in y = x.
• f(f⁻¹(x)) = x on the inverse’s domain. f⁻¹(f(x)) = x only on the restricted domain of f.

2.2 and 2.3 Identity table

Identity	Notes
sin² θ + cos² θ ≡ 1	Base identity
sec² θ ≡ 1 + tan² θ	Use when sec and tan appear together
cosec² θ ≡ 1 + cot² θ	Use when cosec and cot appear together
sin(A ± B) ≡ sin A cos B ± cos A sin B	Signs match
cos(A ± B) ≡ cos A cos B ∓ sin A sin B	Signs are opposite
tan(A ± B) ≡ (tan A ± tan B)/(1 ∓ tan A tan B)	Denominator sign opposite
sin 2A ≡ 2 sin A cos A
cos 2A ≡ cos² A − sin² A ≡ 2 cos² A − 1 ≡ 1 − 2 sin² A	Choose the form that leaves one function
tan 2A ≡ 2 tan A/(1 − tan² A)
cos² A ≡ (1 + cos 2A)/2, sin² A ≡ (1 − cos 2A)/2	Half-angle forms

Half angles: put A = θ/2. For example cos θ ≡ 1 − 2 sin²(θ/2), so sin²(θ/2) = (1 − cos θ)/2. The sign of the square root depends on the quadrant of θ/2.

Method boxes

Equation with a double angle

1. Replace the double angle: sin 2θ by 2 sin θ cos θ, cos 2θ by the form matching the other terms.
2. Collect everything on one side. Factorise (never divide by a trig function that could be zero).
3. Solve each factor. If the angle is kθ, the interval for kθ is k times as long as the interval for θ, so list every value in it before dividing by k.
4. Rule out impossible values (sine or cosine beyond ±1).

Equation with sec² or cosec²

1. Swap sec² for 1 + tan² (or cosec² for 1 + cot²) so a single function remains.
2. Solve the quadratic. Reject values that cannot occur.
3. Find all angles in the interval.

Proving an identity

1. Start with the more complicated side. Do not move terms across the ≡ sign.
2. Apply double angle or compound formulae. Convert to sin and cos if stuck.
3. Finish on the other side and write “= RHS”.

a cos θ + b sin θ in R-form

1. Choose the target form and expand it.
2. Write the two matching equations, for example r cos α = a and r sin α = b.
3. r = √(a² + b²). Find α from the ratio of the two equations, with α acute when both matches are positive.
4. To solve = c: divide by r, take the inverse, then list every value in the interval for the whole bracket before subtracting or adding α.
5. The expression has greatest value r and least value −r. For r cos(θ − α) the greatest value occurs when θ = α and the least when θ = α + 180°.

Form wanted	Coefficient of cos θ and of sin θ
r cos(θ − α)	a = r cos α, b = r sin α
r cos(θ + α)	a = r cos α, b = −r sin α
r sin(θ + α)	a = r sin α, b = r cos α
r sin(θ − α)	a = −r sin α, b = r cos α

Small worked reminders

Reminder 1. Solve 2 cot² x = 3 cosec x for 0 < x < 360°.

2(cosec² x − 1) = 3 cosec x
2 cosec² x − 3 cosec x − 2 = 0
(2 cosec x + 1)(cosec x − 2) = 0
cosec x = −1/2 is impossible. cosec x = 2, so sin x = 1/2
x = 30°, 150°

Reminder 2. Solve 5 cos θ + 2 sin θ = 3 for 0 ≤ θ < 360°.

r = √29, tan α = 2/5, α = 21.8°
cos(θ − 21.8°) = 3/√29,  θ − 21.8° = ±56.1°
θ = 77.9° or θ = −34.3° + 360° = 325.7°

Reminder 3. Solve √3 cos θ − sin θ = 1 for 0 ≤ θ < 2π.

r cos α = √3, r sin α = 1 for r cos(θ + α)
r = 2, tan α = 1/√3, α = π/6
2 cos(θ + π/6) = 1,  cos(θ + π/6) = 1/2
range: π/6 ≤ θ + π/6 < 13π/6
θ + π/6 = π/3 or 5π/3, so θ = π/6 or 3π/2

Graph sketches to know

• y = sec x: branches with minimum points (0, 1) and maximum points (π, −1); asymptotes at x = ±π/2, ±3π/2.
• y = cosec x: branches with minimum point (π/2, 1) and maximum point (3π/2, −1); asymptotes at x = 0, π, 2π.
• y = cot x: decreasing curve through (π/2, 0) with asymptotes at x = 0, π, 2π.
• y = arcsin x: increasing, from (−1, −π/2) through the origin to (1, π/2).
• y = arccos x: decreasing, from (−1, π) through (0, π/2) to (1, 0).
• y = arctan x: increasing through the origin, flattening towards y = ±π/2.

Must-know distinctions

• cos(A + B) versus sin(A + B): cosine changes the sign between the terms; sine keeps it.
• sec θ versus arccos x: sec is 1/cos. arccos is the inverse function. cos⁻¹ means arccos, not 1/cos.
• Reciprocal versus inverse: the reciprocal of sin is cosec. The inverse of sin is arcsin.
• Range of arcsin versus arccos: arcsin is centred on 0 (−π/2 to π/2), arccos runs from 0 to π.
• Period of cot versus cosec: cot repeats every 180°, cosec every 360°.
• Identity versus equation: an identity (≡) holds for all θ. An equation has solutions in an interval.

Quick self-test

1. Find the exact value of cosec(7π/6).
2. State the range of y = arccos x in radians.
3. Find the exact value of arctan(−1) in radians.
4. Write 5 − 3 sec² θ in terms of tan θ.
5. Find the exact value of sin 15° using sin(45° − 30°).
6. sin A = 3/5 and A is acute. Find sin 2A and cos 2A.
7. sin x = 1/3. Find cos 2x.
8. Write cos θ + 2 sin θ as r cos(θ − α) with α in degrees to 1 d.p. (state r and α).
9. State the greatest value of 6 + cos θ + 2 sin θ, as an exact value.
10. Solve sin 2θ = sin θ for 0 ≤ θ ≤ 360°.
11. Prove that (1 − cos 2x)/sin 2x ≡ tan x.
12. Explain why sec θ = 0.4 has no solutions.

Answers

1. cosec(7π/6) = 1/sin(7π/6) = 1/(−1/2) = −2.
2. 0 ≤ y ≤ π.
3. −π/4.
4. sec² θ = 1 + tan² θ, so 5 − 3(1 + tan² θ) = 2 − 3 tan² θ.
5. sin 45° cos 30° − cos 45° sin 30° = (√2/2)(√3/2) − (√2/2)(1/2) = (√6 − √2)/4.
6. cos A = 4/5. sin 2A = 2(3/5)(4/5) = 24/25. cos 2A = 16/25 − 9/25 = 7/25.
7. cos 2x = 1 − 2 sin² x = 1 − 2/9 = 7/9.
8. r = √(1 + 4) = √5, tan α = 2, α = 63.4°.
9. Greatest value of cos θ + 2 sin θ is √5, so greatest is 6 + √5.
10. 2 sin θ cos θ − sin θ = 0, so sin θ(2 cos θ − 1) = 0. sin θ = 0 gives 0°, 180°, 360°. cos θ = 1/2 gives 60°, 300°. Solutions: 0°, 60°, 180°, 300°, 360°.
11. LHS = 2 sin² x/(2 sin x cos x) = sin x/cos x = tan x = RHS, using 1 − cos 2x = 2 sin² x and sin 2x = 2 sin x cos x.
12. sec θ = 1/cos θ, so cos θ = 2.5. Cosine never exceeds 1, so there are no solutions.

Where marks are usually lost

• Dividing both sides of an equation by sin θ or cos θ and losing the solutions where it is zero.
• Finding one angle for kθ and not listing all values in the longer interval before dividing by k.
• Taking arcsin(sin x) as x without checking the restricted range.
• Wrong sign in cos(A ± B) or in the denominator of tan(A ± B).
• Choosing the wrong form of cos 2A, leaving two different functions in a quadratic.
• Matching a and b to the wrong terms in an R-form, so α comes out as 90° − α.
• Giving α in degrees in a question set in radians, or the reverse.
• Leaving a “prove” answer without a final line showing LHS = RHS, or changing both sides together.
• Rounding too early in R-form solutions, which shifts the third significant figure.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics, Specification, Issue 3, April 2019, Pearson Education Limited: Unit P3, section 2 (topics 2.1 to 2.3).