Revision for exact values, graphs, amplitude and period, identities, equations in degrees and radians, sec, cosec and cot (0606), and the addition formulae and 3D problems (4PM1). Both boards. Cambridge 0606 uses all six functions and the identities sec² = 1 + tan² and cosec² = 1 + cot². Edexcel 4PM1 adds the addition formulae and angles in three dimensions. Both use the sine and cosine rules from IGCSE Maths: 0606 prints them in its list of formulas; 4PM1 gives only the cosine rule.
Learn the exact values of sin, cos and tan at 30°, 45° and 60° (π/6, π/4, π/3).
sin²x + cos²x = 1 and tan x = sin x/cos x. Divide the first by cos²x or sin²x to get sec²x = 1 + tan²x and cosec²x = 1 + cot²x.
To solve an equation, find the principal value, then use the symmetry of the graph (or the CAST diagram) for the others in the range.
With a multiple angle such as sin 2x, change the range first (0° to 720° for 2x), solve, then divide.
y = a sin bx + c has amplitude a, period 360°/b and midline y = c.
sin(A + B) = sin A cos B + cos A sin B; cos(A + B) = cos A cos B − sin A sin B.
Watch out: Never divide both sides by sin x or cos x: you lose the solutions where it is zero. Factorise instead.
Worked example· 2 marks
Solve \(2\sin x = 1\) for \(0^\circ \leq x \leq 360^\circ\).
B1 x = 30°
B1 x = 150°
sin x = ½, so the principal value is x = 30°.
Sine is also positive in the second quadrant: x = 180° − 30° = 150°.
Answer: \(x = 30^\circ\) or \(x = 150^\circ\)
Trigonometry questions
Original exam-style questions, written for this site and each re-solved independently. Tags show the courses whose content a question tests and whether it can be done without a calculator (Cambridge 0606 Paper 1 is non-calculator).
Question 1 · 3 marks06064PM1no calculator
Solve \(\tan 2x = \sqrt{3}\) for \(0^\circ \leq x \leq 180^\circ\).
Show the mark scheme and worked solution
M1 2x = 60° (and 2x in 0° to 360°)
A1 x = 30°
A1 x = 120°
2x runs from 0° to 360°.
tan 2x = √3 gives 2x = 60° or 2x = 60° + 180° = 240°.
x = 30° or x = 120°.
Answer: \(x = 30^\circ\) or \(x = 120^\circ\)
Question 2 · 4 marks0606
Solve \(\sec^2\theta - 3\tan\theta + 1 = 0\) for \(0^\circ \leq \theta \leq 180^\circ\), giving answers to 1 decimal place where necessary.
Show the mark scheme and worked solution
M1 sec²θ = 1 + tan²θ
A1 tan²θ − 3tanθ + 2 = 0
A1 θ = 45°
A1 θ = 63.4°
1 + tan²θ − 3 tan θ + 1 = 0, so tan²θ − 3 tan θ + 2 = 0.
(tan θ − 1)(tan θ − 2) = 0.
tan θ = 1: θ = 45°. tan θ = 2: θ = 63.43…° = 63.4°. (The next solutions, 225° and 243.4°, are outside the range.)
Answer: \(\theta = 45^\circ\) or \(\theta = 63.4^\circ\)
Question 3 · 3 marks0606no calculator
For the curve \(y = 4\sin 3x + 1\), where \(x\) is in degrees, state
(a) the amplitude (b) the period (c) the maximum and minimum values of \(y\).
Show the mark scheme and worked solution
B1 (a) 4
B1 (b) 120°
B1 (c) max 5, min −3
The amplitude is the multiplier of sin: 4.
The period of sin 3x is 360° ÷ 3 = 120°.
y runs from 1 − 4 = −3 to 1 + 4 = 5.
Answer: (a) \(4\) (b) \(120^\circ\) (c) maximum \(5\), minimum \(-3\)
Question 4 · 4 marks4PM1no calculator
\(A\) and \(B\) are acute angles with \(\tan A = \frac{3}{4}\) and \(\sin B = \frac{8}{17}\). Work out the exact value of \(\sin(A + B)\).
Show the mark scheme and worked solution
B1 sin A = 3/5, cos A = 4/5
B1 cos B = 15/17
M1 sin A cos B + cos A sin B
A1 77/85
tan A = 3/4 with A acute: a 3, 4, 5 triangle, so sin A = 3/5 and cos A = 4/5.
IGCSE Math Revision