PYQ Vault

IMAT Mathematics · Trigonometry

Graphs, Equations and the Sine and Cosine Rules

Sine and cosine are waves between minus 1 and 1 that repeat every 360°; equations in them have several solutions in a range, and the sine and cosine rules solve any triangle.

Why this matters

The 2023 paper asked how many solutions a trigonometric equation has in a given range, and 2025 asked which inequality in sin x or cos x holds for every real x. The sine and cosine rules have not been asked yet but are on the syllabus.

Concept 1 of 3: Graphs of sine, cosine and tangent: period and amplitude

As the point goes round the unit circle again and again, its height traces the sine wave and its x-coordinate traces the cosine wave. After one full turn everything repeats, so the period is 360°. Multiplying the angle by bb makes the point go round bb times faster, so the wave repeats bb times as often.

Definition

  • y=sin⁡xy = \sin x and y=cos⁡xy = \cos x: waves between −1-1 and 11, period 360∘360^\circ (2π2\pi). sin⁡x\sin x starts at 0; cos⁡x\cos x starts at its maximum 1.
  • y=tan⁡xy = \tan x: period 180∘180^\circ, takes every real value, and has vertical asymptotes at x=90∘,270∘,…x = 90^\circ, 270^\circ, \ldots
  • For y=asin⁡(bx)+cy = a\sin(bx) + c (with b>0b > 0), and the same for cosine: the amplitude is ∣a∣|a|, the period is 360∘/b360^\circ / b, the midline is y=cy = c, and the values run from c−∣a∣c - |a| to c+∣a∣c + |a|.

Period and amplitude

y=asin⁡(bx)+c:amplitude ∣a∣,period 360∘by = a\sin(bx) + c: \quad \text{amplitude } |a|, \quad \text{period } \frac{360^\circ}{b}

Worked example

For y=3sin⁡(2x)+1y = 3\sin(2x) + 1, find the amplitude, the period and the greatest and least values.
Practice this conceptself-check · 4 quick reps

The period of sin(bx) is 360° divided by b

A larger bb means faster repetition and a shorter period: sin⁡(3x)\sin(3x) has period 120∘120^\circ, not 1080∘1080^\circ. The amplitude is ∣a∣|a|, a positive number, even when aa is negative.

Concept 2 of 3: Solving trigonometric equations in a given range

A wave crosses any horizontal line between its lowest and highest values twice in every period, so an equation like sin⁡x=k\sin x = k usually has two solutions per turn. Find one angle from the exact values, use the symmetry of the wave for the second, then add whole periods until you leave the range. If the line is outside −1-1 to 11 there are no solutions at all.

Definition

  • Sine: sin⁡x=k\sin x = k gives x=αx = \alpha and x=180∘−αx = 180^\circ - \alpha, plus multiples of 360∘360^\circ.
  • Cosine: cos⁡x=k\cos x = k gives x=αx = \alpha and x=360∘−αx = 360^\circ - \alpha, plus multiples of 360∘360^\circ.
  • Tangent: tan⁡x=k\tan x = k gives x=αx = \alpha plus multiples of 180∘180^\circ.
  • Multiple angles: for sin⁡(bx)=k\sin(bx) = k, solve for u=bxu = bx over the range multiplied by bb, then divide by bb.
  • Quadratic in sin x: put s=sin⁡xs = \sin x, solve the quadratic, and reject any root outside [−1,1][-1, 1].
  • Inequality for every x: with s=sin⁡xs = \sin x, the expression only needs checking for −1≤s≤1-1 \le s \le 1.

General solutions

sin⁡x=k: x=α, 180∘−αcos⁡x=k: x=±αtan⁡x=k: x=α(+whole periods)\sin x = k: \ x = \alpha,\ 180^\circ - \alpha \qquad \cos x = k: \ x = \pm\alpha \qquad \tan x = k: \ x = \alpha \quad (+\text{whole periods})

Worked example

Solve 2sin⁡2x−3sin⁡x+1=02\sin^2 x - 3\sin x + 1 = 0 for 0∘≤x≤360∘0^\circ \le x \le 360^\circ.
Practice this conceptself-check · 4 quick reps

Solve for the whole angle first

In cos⁡(2x)=k\cos(2x) = k the angle 2x2x runs over twice the range of xx. Solving for xx directly over one turn misses half the solutions; solve for 2x2x over the doubled range, then halve.

Throw away values of sin x outside minus 1 to 1

A quadratic in sin⁡x\sin x can have a root such as 2 or −3-3. It gives no angles at all, so do not count it when you count solutions.

Concept 3 of 3: The sine rule and the cosine rule for any triangle

SOH CAH TOA needs a right angle. In any other triangle, the sine rule links each side to the angle facing it, and the cosine rule is Pythagoras with a correction term for an angle that is not 90°. When that angle is 90°, cos⁡90∘=0\cos 90^\circ = 0 and the correction vanishes.

Definition

In a triangle with sides a,b,ca, b, c facing angles A,B,CA, B, C:

  • Sine rule: asin⁡A=bsin⁡B=csin⁡C\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}. Use it when you know a side and the angle facing it.
  • Cosine rule: c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C. Use it with two sides and the angle between them, or with all three sides.
  • Area: 12absin⁡C\tfrac{1}{2}ab\sin C, using two sides and the angle between them.
  • An obtuse angle has a negative cosine, so the side facing it is longer than Pythagoras alone would give.

Sine rule, cosine rule and area

asin⁡A=bsin⁡B=csin⁡Cc2=a2+b2−2abcos⁡CArea=12absin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \qquad c^2 = a^2 + b^2 - 2ab\cos C \qquad \text{Area} = \tfrac{1}{2}ab\sin C

Worked example

A triangle has a=7a = 7 cm, b=8b = 8 cm and the angle between them C=60∘C = 60^\circ. Find cc and the area.
Practice this conceptself-check · 3 quick reps

The cosine rule uses the angle between the two sides

In c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C, the angle CC is the one between sides aa and bb, facing side cc. Using any other angle gives a wrong side, and forgetting the minus sign turns an obtuse angle into an acute one.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • Graphs of sine, cosine and tangent: period and amplitude

    Period and amplitude

    y=asin⁡(bx)+c:amplitude ∣a∣,period 360∘by = a\sin(bx) + c: \quad \text{amplitude } |a|, \quad \text{period } \frac{360^\circ}{b}
  • Solving trigonometric equations in a given range

    General solutions

    sin⁡x=k: x=α, 180∘−αcos⁡x=k: x=±αtan⁡x=k: x=α(+whole periods)\sin x = k: \ x = \alpha,\ 180^\circ - \alpha \qquad \cos x = k: \ x = \pm\alpha \qquad \tan x = k: \ x = \alpha \quad (+\text{whole periods})
  • The sine rule and the cosine rule for any triangle

    Sine rule, cosine rule and area

    asin⁡A=bsin⁡B=csin⁡Cc2=a2+b2−2abcos⁡CArea=12absin⁡C\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \qquad c^2 = a^2 + b^2 - 2ab\cos C \qquad \text{Area} = \tfrac{1}{2}ab\sin C

Watch out for (4)

Drill the past-year questions on this topic

Open the bank with this topic pre-filtered.