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O-Level Trigonometric functions, identities and equations
What the O-Level syllabus expects for Trigonometric functions, identities and equations, and how to practise it.
What the syllabus expects
- Handling all six trigonometric functions across angles of any size, whether measured in degrees or radians
- Working with the principal values of sin^-1 x, cos^-1 x and tan^-1 x
- Knowing the exact trigonometric values at the special angles 30 deg, 45 deg, 60 deg (or pi/6, pi/4, pi/3)
- Recognising the amplitude, period and symmetry features tied to the sine and cosine functions
- Sketching graphs of y = a sin(bx) + c, y = a sin(x/b) + c, y = a cos(bx) + c, y = a cos(x/b) + c and y = a tan(bx), with a real, b a positive integer and c an integer
- Applying the relations tan A = sin A / cos A, cot A = cos A / sin A, sin^2 A + cos^2 A = 1, sec^2 A = 1 + tan^2 A and cosec^2 A = 1 + cot^2 A
- Applying the expansions of sin(A +/- B), cos(A +/- B) and tan(A +/- B)
- Applying the double-angle formulae for sin 2A, cos 2A and tan 2A
- Rewriting a cos theta + b sin theta as R cos(theta +/- alpha) or R sin(theta +/- alpha)
- Reducing trigonometric expressions to simpler form
- Solving straightforward trigonometric equations within a stated interval
Scope: The general solution is not required - Proving simple trigonometric identities
- Modelling with trigonometric functions
How it's examined
Questions on this topic most often ask you to find, solve, prove, show. About 18% of the past-paper style questions in Rae's bank for this subject sit in this topic.
Worked examples
Example 1 (4 marks)
Without a calculator, prove that tan(π/12) = 2 − √3.
Show the worked answer
π/12 = 15° = 45° − 30°. tan(45° − 30°) = (tan45° − tan30°)/(1 + tan45°·tan30°) = (1 − 1/√3)/(1 + 1/√3). Multiply numerator and denominator by √3: (√3 − 1)/(√3 + 1). Rationalise by (√3 − 1): [(√3 − 1)²]/[(√3)² − 1] = (3 − 2√3 + 1)/(3 − 1) = (4 − 2√3)/2 = 2 − √3. Hence tan(π/12) = 2 − √3.
Example 2 (4 marks)
Prove that cot x − cot x tan² x + tan x = (1 + cos 2x)/sin 2x.
Show the worked answer
LHS = cot x - cot x tan² x + tan x. Note cot x tan² x = (1/tan x)(tan² x) = tan x. So LHS = cot x - tan x + tan x = cot x = cos x / sin x. RHS = (1 + cos 2x)/sin 2x. Use 1 + cos 2x = 2cos² x and sin 2x = 2 sin x cos x: RHS = 2cos² x /(2 sin x cos x) = cos x / sin x = cot x. LHS = RHS. (proved)
Example 3 (3 marks)
Show that sin(theta) / (1 + cos(theta)) is equal to tan(theta/2).
Show the worked answer
Use double-angle forms: sin(theta) = 2 sin(theta/2) cos(theta/2) 1 + cos(theta) = 2 cos²(theta/2) Therefore sin(theta) / (1 + cos(theta)) = [2 sin(theta/2) cos(theta/2)] / [2 cos²(theta/2)] = sin(theta/2) / cos(theta/2) = tan(theta/2) (shown)
More worked questions on this topic
- Show, without the use of a calculator, that tan 285° = (√3 + 1)/(1 − √3). (3 marks)
- Show that (cos A − sin A)(1 + sin A cos A) divided by sin³A equals cot³A − 1. (4 marks)
- Given cos A = 1/2 with 0° < A < 90°, and sin B = −1/√2 with 180° < B < 270°, evaluate cos(A − B (4 marks)
- At a harbour, the water depth d metres at t hours after 08 00 is described by the sinusoidal mo (3 marks)
- Hence find the least value of 1/((3cosθ + 5sinθ)² + 1) for 0 ≤ θ ≤ π/2. (2 marks)
More O-Level Additional Maths (A-Maths) topics
Quadratic functions · Equations and inequalities · Surds · Polynomials and partial fractions · Binomial expansions · Exponential and logarithmic functions · all of O-Level Additional Maths (A-Maths)