Most tested Papers 1 & 2

Trigonometry: Sine/Cosine Rules, Radians, Exact Values and Equations

The sine and cosine rules, unit-circle graphs, and a small set of exact values extend right-angled ratios to any angle. The TMUA rewards spotting the identity, symmetry, or rule that collapses a problem to a single line; with no calculator, exact surds and radian multiples of pi dominate.

Part of the TMUA syllabus — revision for the Test of Mathematics for University Admission (TMUA), the UAT-UK maths test used by Cambridge, Oxford, Imperial, UCL, LSE, Warwick and Durham. No calculator; multiple choice.

Key points

  • Use the sine rule when you have a side paired with its opposite angle, plus one more piece (a/sin(A) = b/sin(B) = c/sin(C)).
  • Use the cosine rule for two sides and the included angle, or for all three sides to recover an angle (a2 = b2 + c2 - 2 × b × c × cos(A)).
  • Area from two sides and the angle between them needs no height (area = (1/2) × a × b × sin(C)).
  • Convert angles through pi radians = 180 degrees; the staples are 30 = pi/6, 45 = pi/4, 60 = pi/3, 90 = pi/2.
  • Arc and sector formulas require radians; a segment is the sector minus the triangle (s = r × theta, sector = (1/2) × r2 × theta).
  • The two core identities let you rewrite one ratio via the others (sin2(x) + cos2(x) = 1 and tan(x) = sin(x)/cos(x)).
  • Learn exact values off the special triangles: sin(30) = 1/2, cos(30) = √(3)/2, tan(45) = 1, sin(60) = √(3)/2.
  • Graph symmetry fixes how many solutions an interval holds: sin and cos have period 2pi (360 degrees), tan has period pi (180 degrees); sin is odd, cos is even.

Diagram

bacarea = (1/2) a b sin C
With two sides and the included angle, area = (1/2) a b sin(C); the cosine rule then gives the third side c.

Formulae

a/sin(A) = b/sin(B) = c/sin(C)

Sine rule: a known side opposite a known angle, plus one further side or angle. Beware the ambiguous case when finding an angle.

a2 = b2 + c2 - 2 × b × c × cos(A)

Cosine rule: two sides and the included angle to find the third side, or three sides to find any angle.

area = (1/2) × a × b × sin(C)

Area of a triangle from two sides and the angle between them (no perpendicular height needed).

s = r × theta , sector = (1/2) × r2 × theta

Arc length and sector area with theta in radians; subtract (1/2) × r2 × sin(theta) to get the segment.

sin2(x) + cos2(x) = 1

Convert between sin and cos, or eliminate one ratio to reduce an equation to a single trig function.

tan(x) = sin(x)/cos(x)

Rewrite equations that mix tan with sin or cos, or turn a tan equation into a sin/cos ratio.

Definitions

Radian
The angle subtended at a circle's centre by an arc equal in length to the radius; pi radians = 180 degrees, so one full turn is 2pi. Radians are required for arc-length and sector formulas.
Ambiguous case (SSA)
When the sine rule gives sin(A) = k with 0 < k < 1, both A and 180 - A can be geometrically valid, sometimes producing two different triangles; always check the angles still total under 180 degrees.
Period
The smallest positive p for which f(x + p) = f(x). sin and cos repeat every 2pi (360 degrees); tan repeats every pi (180 degrees). The period tells you how many solutions to expect in a given interval.
Segment
The region of a circle between a chord and its arc. Its area equals the sector area minus the area of the triangle formed by the two radii and the chord.

Worked examples

1

Find all solutions of 2 × sin2(x) - 3 × cos(x) = 0 for 0 ≤ x ≤ 360 degrees.

  1. 1

    Replace sin2 using sin2(x) = 1 - cos2(x) and write c = cos(x):

    2(1 - c2) - 3c = 0.

  2. 2

    Expand and tidy into a quadratic:

    2c2 + 3c - 2 = 0.

  3. 3

    Factorise the quadratic:

    (2c - 1)(c + 2) = 0.

  4. 4
    Reject c = -2 since |cos| ≤ 1, leaving:
    cos(x) = 1/2
  5. 5

    Read both solutions in the interval (cos is positive in quadrants 1 and 4):

    x = 60 or 300

Answer: x = 60 degrees and x = 300 degrees.

2

In triangle ABC, AB = 6, AC = 5 and the angle at A is 60 degrees. Find the exact length of BC and the exact area of the triangle.

  1. 1

    Apply the cosine rule at A:

    BC2 = 62 + 52 - 2 × 6 × 5 × cos(60)
  2. 2

    Substitute the exact value cos(60) = 1/2:

    BC2 = 61 - 30 = 31
  3. 3

    Take the positive square root:

    BC = √(31)
  4. 4

    Use area from two sides and the included angle:

    area = (1/2) × 6 × 5 × sin(60)
  5. 5

    Substitute sin(60) = √(3)/2:

    area = 15 × √(3) / 2

Answer: BC = √(31) and area = 15 × √(3) / 2.

Common mistakes

  • ×Forgetting the ambiguous case: after the sine rule gives sin(A) = k, the obtuse angle 180 - A may also fit; test whether the angle sum stays below 180 degrees.
  • ×Applying s = r × theta or (1/2) × r2 × theta with theta in degrees: these arc and sector formulas only work in radians.
  • ×Dividing an equation by cos(x) or sin(x): this silently discards the roots where that factor is zero, so factorise instead of dividing.
  • ×Wrong quadrant sign: sin is positive in quadrants 1 and 2, cos in 1 and 4, tan in 1 and 3; one sign slip flips the whole answer.
  • ×Mismatching side and angle in the cosine rule: the side on the left must be opposite the angle used, so a2 pairs with angle A.

No-calculator tips

  • ✓Memorise the two special triangles, sides (1, 1, √(2)) and (1, √(3), 2); every exact sin, cos and tan of 30, 45 and 60 reads straight off them.
  • ✓Keep answers as surds and multiples of pi; the options are exact, so a stray decimal usually signals a slip.
  • ✓Sketch the graph or unit circle first to count expected solutions: period 360 for sin/cos, 180 for tan.
  • ✓Convert degrees to radians by multiplying by pi/180 and cancelling, e.g. 135 × pi/180 = 3pi/4.
  • ✓Eliminate options by sign and size: sin and cos must lie in [-1, 1], and an obtuse angle always has a negative cosine.

Test yourself

Original practice questions, no calculator. Work each out before revealing the answer.

Q1.In triangle ABC, angle A = 120 degrees, side b = 3 and side c = 5. Find the length of side a (the side opposite angle A).

  • A. √(19)
  • B. √(34)
  • C. 7
  • D. 8
Show answer

Answer: C — 7

By the cosine rule a2 = b2 + c2 - 2bc*cos A = 9 + 25 - 2*3*5*cos(120) = 34 - 30*(-1/2) = 34 + 15 = 49, so a = 7. The trap √(19) comes from taking cos(120) = +1/2 (a sign slip); √(34) drops the -2bc*cos A term entirely.

Q2.A sector of a circle has radius 8 and angle pi/4 radians. What is the perimeter of the sector?

  • A. 2*pi
  • B. 2*pi + 8
  • C. 8*pi
  • D. 2*pi + 16
Show answer

Answer: D — 2*pi + 16

Arc length = r*theta = 8*(pi/4) = 2*pi, and the perimeter also includes the two straight radii, so it is 2*pi + 2*8 = 2*pi + 16. The 2*pi option forgets both radii, 2*pi + 8 adds only one, and 8*pi is the sector area (1/2*r2*theta), not its perimeter.

Q3.How many solutions does 2*cos2(x) - cos(x) - 1 = 0 have in the interval 0 ≤ x < 2*pi?

  • A. 4
  • B. 3
  • C. 2
  • D. 1
Show answer

Answer: B — 3

Factorising gives (2*cos x + 1)(cos x - 1) = 0, so cos x = -1/2 (x = 2*pi/3, 4*pi/3) or cos x = 1 (x = 0 only, since x = 2*pi is excluded), giving 3 solutions. Answer 4 wrongly counts x = 2*pi as well; answer 2 drops the cos x = 1 branch.

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