Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 1: Pure Mathematics 1 · official syllabus
Study notes
Use these worked notes to check exact formulas, follow the reasoning and learn the exam wording.
Angles, units and the unit circle 1.5
Radians and degrees describe the same turn: π radians = 180°. Multiply degrees by π/180 to get radians; multiply radians by 180/π to get degrees. Keep the calculator mode consistent with the question.
On the unit circle, the point at angle θ has coordinates (cos θ, sin θ). This explains the signs in each quadrant: both positive in the first, only sine positive in the second, both negative in the third, only cosine positive in the fourth.
For the same position after a complete turn, add or subtract 360° or 2π. For example, sin 390° = sin 30° = 1/2, and cos(-π/3) = 1/2.
Sine is odd: sin(-θ) = -sin θ. Cosine is even: cos(-θ) = cos θ. Tangent is odd and repeats after 180° or π, wherever it is defined.
Exact values and related angles 1.5
For θ = 0°, 30°, 45°, 60°, 90°, the sine values are 0, 1/2, √2/2, √3/2, 1; cosine has the same values in reverse order. Tangent is sine divided by cosine: 0, 1/√3, 1, √3, then undefined at 90°.
Use a reference angle and the quadrant sign. For 150°, the reference angle is 30° and cosine is negative, so cos 150° = -√3/2. At 225°, sine and cosine are both -√2/2, so tan 225° = 1.
In radians, π/6, π/4 and π/3 are 30°, 45° and 60°. Thus sin(3π/4) = √2/2 and tan(5π/6) = -1/√3. Keep exact radicals rather than rounding them.
Graphs: amplitude, period and translations 1.5
The graphs y = sin x and y = cos x have range [-1,1] and period 360° or 2π. Tangent has period 180° or π, no maximum or minimum, and vertical asymptotes where cosine is zero.
For y = A sin(Bx + C) + D with B nonzero, the amplitude is |A|, the midline is y = D and the range is [D - |A|, D + |A|]. The period is 360°/|B| or 2π/|B|, and the horizontal shift is -C/B. The same rules apply to cosine.
For y = 1 - cos 2x, the midline is 1, amplitude 1, range [0,2] and period 180° or π. Key points in degrees are (0,0), (45,1), (90,2), (135,1), (180,0). A negative coefficient reflects the cosine wave in its midline.
For y = tan(x + π/4), the graph shifts left by π/4. In one period, a zero is at x = -π/4 and the adjacent asymptotes are x = -3π/4 and x = π/4. Do not join separate tangent branches across an asymptote.
Inverse trigonometric values 1.5
The inverse notations sin⁻¹, cos⁻¹ and tan⁻¹ mean inverse trigonometric relations here, not reciprocals. Their principal ranges are [-90°,90°], [0°,180°] and (-90°,90°), respectively, or the equivalent radian intervals.
sin⁻¹(1/2) = 30° is a principal value, but sin x = 1/2 has another solution 150° in [0°,360°]. A calculator gives a starting angle, not every solution in the interval.
cos⁻¹(-1/2) = 120° = 2π/3. tan⁻¹(-1) = -45° = -π/4; in [0°,360°), the corresponding tangent solutions are 135° and 315°.
Identities and safe rearrangement 1.5
The identities sin²x + cos²x = 1 and tan x = sin x/cos x (where cos x is nonzero) are tools for rewriting expressions in one function. An identity holds wherever both sides are defined; an equation holds only at its solutions.
Example: (1 - cos²x)/sin x = sin²x/sin x = sin x, provided sin x is nonzero. The simplified expression does not make the original defined at sin x = 0.
To prove an identity, start on one side and rewrite it into the other. For example, sin x tan x + cos x = sin²x/cos x + cos²x/cos x = 1/cos x, with cos x nonzero.
Do not divide an equation by sin x or cos x without considering when that factor is zero. Factoring preserves those cases: 2sin²x - sin x = sin x(2sin x - 1) = 0.
Solving equations on a stated interval 1.5
First isolate the trigonometric function, then find the reference angle, use signs or symmetry, list every angle in the stated interval, and check its endpoints. If the isolated sine or cosine is outside [-1,1], there are no real solutions.
For 2sin x = 1 on 0° ≤ x ≤ 360°, sin x = 1/2 gives x = 30° and 150°. For cos x = -√3/2 on -π ≤ x ≤ π, the solutions are x = -5π/6 and 5π/6.
For sin 2x = √3/2 on 0° ≤ x ≤ 180°, solve for 2x in [0°,360°]: 2x = 60° or 120°, so x = 30° or 60°. Expand the interval before solving a multiple-angle equation.
For 2sin²x - sin x = 0 on 0° ≤ x ≤ 360°, sin x = 0 or 1/2. Hence x = 0°, 30°, 150°, 180°, 360°. Both 0° and 360° belong because the interval includes both endpoints.
For 3sin²x + 5cos x - 1 = 0 on 0° ≤ x ≤ 360°, replace sin²x by 1 - cos²x. This gives 3cos²x - 5cos x - 2 = 0 = (3cos x + 1)(cos x - 2). Reject cos x = 2; cos x = -1/3 gives x approximately 109.5° and 250.5°. Check the requested rounding.
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