Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 2: Pure Mathematics 2 · official syllabus
TRIGONOMETRY LAB
Learn it by playing it
Four short missions on the reciprocal functions, the Pythagorean identities, the compound-angle expansions, and merging two waves into one with the R-form. When a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
Meet the reciprocal trio
sec, cosec and cot are the familiar ratios flipped: 1/cos, 1/sin and 1/tan. The dashed bar sits at sec theta = 2 and your solid bar is 1/cos theta. Slide theta until the bars coincide - and watch the value surge as cos theta shrinks.
Before you slide: as theta approaches 90 degrees, what does sec theta = 1/cos theta do?
your bar: sec 30 = 1.1547 - target is 2. cos theta is still above 1/2, so sec theta is still below 2 - increase theta.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
The reciprocal functions: sec, cosec and cot 2.3
sec x = 1/cos x, cosec x = 1/sin x and cot x = 1/tan x = cos x/sin x. Each one is defined everywhere its denominator is nonzero, and each takes its sign from its parent function, so the CAST diagram still decides signs in every quadrant.
Standard values come straight from the parent ratios: cos 60 = 1/2 gives sec 60 = 2, sin 30 = 1/2 gives cosec 30 = 2, and tan 45 = 1 gives cot 45 = 1.
The reciprocal functions are undefined where their denominator is zero: sec x has no value at x = 90 + 180n degrees, cosec x and cot x have no value at x = 180n degrees, and tan x itself is undefined at 90 + 180n degrees.
Because sec and cosec are reciprocals of quantities between -1 and 1, their own values satisfy |sec x| >= 1 and |cosec x| >= 1. cot x can take any real value, just like tan x.
Graphs of the six functions for angles of any magnitude 2.3
sin x and cos x repeat every 360 degrees and stay between -1 and 1; sec x and cosec x inherit the 360-degree period but leave a gap: their graphs live outside the band -1 to 1, with vertical asymptotes wherever the parent curve crosses zero.
tan x and cot x repeat every 180 degrees and take every real value. tan x climbs through each period with asymptotes at 90 + 180n degrees; cot x falls through each period with asymptotes at 180n degrees.
For an angle of any magnitude, reduce it to an acute reference angle and fix the sign from the quadrant. cos 210 = -cos 30 = -root 3 / 2, so sec 210 = -2 / root 3.
Sketching y = sec x from y = cos x is a reliable exam move: where cos x = 1 or -1 the curves touch, and where cos x approaches 0 the sec curve flies off to an asymptote.
The Pythagorean identities 2.3
Everything starts from sin^2 x + cos^2 x = 1. Dividing every term by cos^2 x gives 1 + tan^2 x = sec^2 x; dividing by sin^2 x gives 1 + cot^2 x = cosec^2 x. These are identities: they hold for every angle where both sides are defined.
The identities move between functions in one step. If tan x = 3 with x acute, then sec^2 x = 10, so cos x = 1 / root 10 and sin x = 3 / root 10 - no triangle drawing required.
Watch the sign when taking roots: sec^2 x = 10 gives sec x = plus or minus root 10, and only the quadrant of x decides which. Squaring information always loses the sign.
The identities rewrite equations into a single function. 2 sec^2 x - tan x = 5 becomes 2(1 + tan^2 x) - tan x = 5, the quadratic 2 tan^2 x - tan x - 3 = 0, which factors as (2 tan x - 3)(tan x + 1) = 0, so tan x = 3/2 or tan x = -1.
sin(A + B) = sin A cos B + cos A sin B and sin(A - B) = sin A cos B - cos A sin B. The sine expansion keeps the sign and mixes the functions.
cos(A + B) = cos A cos B - sin A sin B and cos(A - B) = cos A cos B + sin A sin B. The cosine expansion flips the sign and pairs like with like. tan(A + B) = (tan A + tan B)/(1 - tan A tan B), with the denominator sign flipped for tan(A - B).
A sum of angles never distributes: sin(A + B) is not sin A + sin B. At A = 45, B = 30 the true value is sin 75 = (root 6 + root 2)/4, about 0.966, while sin 45 + sin 30 is about 1.207 - impossible for a sine.
Exact values for 75 and 15 degrees come from 45 +/- 30: sin 75 = (root 6 + root 2)/4, cos 75 = (root 6 - root 2)/4 and tan 75 = 2 + root 3. The same expansions simplify expressions like cos(x - 30) - 3 sin(x - 60), which reduces to 2 root 3 cos x - sin x.
Double-angle formulae 2.3
Setting B = A in the compound expansions gives sin 2A = 2 sin A cos A, cos 2A = cos^2 A - sin^2 A and tan 2A = 2 tan A / (1 - tan^2 A).
cos 2A has three forms: cos^2 A - sin^2 A = 2 cos^2 A - 1 = 1 - 2 sin^2 A. Choose the one that matches the rest of the problem: an equation in sin x wants 1 - 2 sin^2 x, an equation in cos x wants 2 cos^2 x - 1.
Exact evaluation: if sin x = 1/3, then cos 2x = 1 - 2(1/3)^2 = 7/9, with no need to find x itself.
Equations mixing angles, like cos 2x + sin x = 0, become quadratics after the right form: 1 - 2 sin^2 x + sin x = 0 rearranges to 2 sin^2 x - sin x - 1 = 0, so sin x = 1 or sin x = -1/2.
The R-form: combining a sin x + b cos x 2.3
Any a sin x + b cos x (with a, b positive) is a single wave: R sin(x + alpha) where R = root(a^2 + b^2), cos alpha = a/R and sin alpha = b/R. It can equally be written R cos(x - alpha) with cos alpha = a/R and sin alpha = b/R - the form the question asks for decides.
3 sin x + 4 cos x = 5 sin(x + 53.13) because R = root(9 + 16) = 5 and tan alpha = 4/3. The largest value is 5 and the smallest is -5: the two parts peak at different angles, so their coefficients never simply add.
The R-form settles whether an equation has solutions at all: a sin x + b cos x = c is solvable exactly when |c| <= R, because a single wave never exceeds its amplitude.
To solve, rewrite and invert. 3 cos x + 2 sin x = 1 becomes root 13 cos(x - 33.69) = 1, so cos(x - 33.69) = 1/root 13, giving x - 33.69 = plus or minus 73.90 plus 360n, and the solutions in 0 to 360 degrees are x = 107.6 and x = 319.8.
Solving trigonometric equations in a given interval 2.3
The routine is always the same: rewrite in one function of one angle, solve for that function, find the base angle, then fan out every solution the interval allows - adding the period and using symmetry, never just the calculator's first answer.
tan x + cot x = 4 is one function in disguise: multiplying through by tan x gives tan^2 x - 4 tan x + 1 = 0, so tan x = 2 +/- root 3. Recognising tan 75 = 2 + root 3 and tan 15 = 2 - root 3 gives x = 15, 75, 195 and 255 degrees in 0 to 360.
After 2 sec^2 x - tan x = 5 becomes (2 tan x - 3)(tan x + 1) = 0, the solutions in 0 to 360 degrees come from tan x = 3/2 (56.3 and 236.3 degrees) and tan x = -1 (135 and 315 degrees).
Always finish with a check: substitute each answer into the original equation, confirm it lies inside the required interval, and count the solutions against what the interval length predicts.
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