A Level Maths 9709 · Paper 1 Pure Mathematics 1 · 1.5 · ~10 min read
Trigonometry: graphs, identities and equations
Graphs of sine, cosine and tangent in degrees and radians, exact values, principal values of sin⁻¹, cos⁻¹ and tan⁻¹, the two key identities, and how to find every solution of a trig equation in an interval.
Trig graphs & transformationsExact valuesCAST & related anglesPrincipal valuestan θ = sin θ/cos θsin²θ + cos²θ = 1Proving identitiesSolving equations
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Paper tag = where it is examinedFORMULA LIST (MF19) given in the examM1A1B1 how marks are awardedhighlight = comes up again and againTest yourself at the end
In 30 seconds
- sin and cos repeat every 360° (2π); tan repeats every 180° (π). Transformations change the amplitude, period and position exactly as in Functions and transformations.
- Exact values at 30°, 45° and 60° come from two triangles. Learn them; they are not in the formula list.
- sin⁻¹, cos⁻¹ and tan⁻¹ give one value only, the principal value. Every other solution comes from the graph or the CAST diagram.
- Two identities do almost all of Paper 1: tan θ ≡ sin θ/cos θ and sin²θ + cos²θ ≡ 1.
- To solve an equation, reduce it to one trig function, find the principal value, then list every solution in the interval. Never divide by sin x or cos x: factorise.
By the end you can
- Sketch and use the graphs of sin, cos and tan, including transformed graphs.
- Recall exact values and use related angles in all four quadrants.
- Use principal values of the inverse trig functions.
- Prove identities and solve equations using tan θ ≡ sin θ/cos θ and sin²θ + cos²θ ≡ 1.
- Find all solutions in a given interval, in degrees or radians, including multiple angles like 2x.
1Graphs
sin, cos and tan
| Function | Period | Range | Key features |
|---|---|---|---|
| y = sin x | 360° or 2π | −1 ≤ y ≤ 1 | zeros at 0, π, 2π; max at π/2 |
| y = cos x | 360° or 2π | −1 ≤ y ≤ 1 | max at 0 and 2π; zeros at π/2, 3π/2 |
| y = tan x | 180° or π | all real values | asymptotes at x = 90° (π/2), 270° (3π/2); zeros at 0, π |
- y = a sin(bx) + c: amplitude ǀaǀ, period 360°/b (or 2π/b), centre line y = c. Range from c − ǀaǀ to c + ǀaǀ.
- Example: y = 1 − cos 2x has period π and range 0 ≤ y ≤ 2. It is y = cos x stretched by ½ parallel to the x-axis, reflected in the x-axis, then translated by 01.
Worked example
The function f(x) = a + b cos x has maximum value 7 and minimum value −1. Find a and b, given b > 0.
- maximum = a + b = 7, minimum = a − b = −1 M1
- a = 3, b = 4 A1 A1
2Exact values and related angles
The two triangles
- Half an equilateral triangle (sides 1, 2, √3) gives 30° and 60°. A right-angled isosceles triangle (sides 1, 1, √2) gives 45°.
| θ | 0 | 30° = π/6 | 45° = π/4 | 60° = π/3 | 90° = π/2 |
|---|---|---|---|---|---|
| sin θ | 0 | ½ | 1/√2 = √2/2 | √3/2 | 1 |
| cos θ | 1 | √3/2 | 1/√2 = √2/2 | ½ | 0 |
| tan θ | 0 | 1/√3 = √3/3 | 1 | √3 | undefined |
- Related angles: sin(180° − θ) = sin θ; cos(180° − θ) = −cos θ; tan(180° + θ) = tan θ; cos(360° − θ) = cos θ; sin(−θ) = −sin θ; cos(−θ) = cos θ.
- Example: sin 240° = −sin 60° = −√3/2 (3rd quadrant: only tan is positive).
Principal values
| Inverse function | Principal value range |
|---|---|
| sin⁻¹x | −90° ≤ sin⁻¹x ≤ 90° (−π/2 to π/2) |
| cos⁻¹x | 0° ≤ cos⁻¹x ≤ 180° (0 to π) |
| tan⁻¹x | −90° < tan⁻¹x < 90° (−π/2 to π/2) |
- These ranges make each inverse a function: sin x is one-one on −π/2 ≤ x ≤ π/2, so it has an inverse there. They are in the formula list FORMULA LIST (MF19).
- Example: cos⁻¹(−½) = 2π/3 (not −π/3, which is outside 0 to π). sin⁻¹(−½) = −π/6.
3Identities
The two you need
tan θ ≡ sin θ / cos θ FORMULA LIST (MF19)
valid whenever cos θ ≠ 0sin²θ + cos²θ ≡ 1 FORMULA LIST (MF19)
Pythagoras on the unit circle; rearranges to sin²θ ≡ 1 − cos²θ, cos²θ ≡ 1 − sin²θ- Proving an identity: start from one side (usually the more complicated one) and work to the other. Don't move terms across the ≡ sign.
- Useful moves: write tan as sin/cos; combine fractions over a common denominator; replace 1 with sin²θ + cos²θ; factorise.
Worked example
Prove that tan θ + 1/tan θ ≡ 1/(sin θ cos θ).
- LHS = sin θ/cos θ + cos θ/sin θ M1
- = (sin²θ + cos²θ)/(sin θ cos θ) M1
- = 1/(sin θ cos θ) = RHS A1
Worked example
Prove that (1 − cos²θ)/(1 − sin²θ) ≡ tan²θ.
- LHS = sin²θ/cos²θ M1
- = (sin θ/cos θ)² = tan²θ = RHS A1
Common mistakeWorking on both sides at once and meeting in the middle with no clear chain. Write one continuous chain: LHS = … = … = RHS.
4Solving equations
The method
- 1 · Rearrange to one trig function equal to a number: sin x = k, cos 2x = k or tan(x − 30°) = k.
- 2 · Adjust the interval for the angle inside: if 0° ≤ x ≤ 180°, then 0° ≤ 2x ≤ 360°.
- 3 · Principal value from the calculator, then all solutions in the adjusted interval using the graph or CAST.
- 4 · Undo the inside (divide by 2, add 30°) and check each answer lies in the original interval.
Worked example
Solve sin 2x = 0.6 for 0° ≤ x ≤ 180°.
- 0° ≤ 2x ≤ 360° B1
- 2x = 36.87° or 180° − 36.87° = 143.13° M1
- x = 18.4° or 71.6° A1 A1
Worked example
Solve tan(x − 30°) = −1 for 0° ≤ x ≤ 360°.
- −30° ≤ x − 30° ≤ 330° B1
- tan⁻¹(−1) = −45° (outside); add 180°: 135°, 315° M1
- x = 165° or 345° A1 A1
Common mistakeRounding the inner angle first. If 2x = 36.869…°, keep all the digits and halve them; rounding 2x to 36.9° can change the final 1 d.p. This was flagged in the June 2024 report. Also note that the second solution of sin 2x = k is 180° − 2x, not 90° − x.
Equations that need an identity
- Mixed sin and cos, one of them squared: replace with sin²x + cos²x = 1 to get a quadratic in one function.
- sin x and cos x to the same power: divide by cos x to get tan x, but only after checking cos x = 0 isn't a solution.
- A common factor: factorise. Dividing by sin x throws away the solutions of sin x = 0.
Worked example
Solve 2 sin²x − cos x − 1 = 0 for 0 ≤ x ≤ 2π.
- 2(1 − cos²x) − cos x − 1 = 0 ⇒ 2cos²x + cos x − 1 = 0 M1
- (2cos x − 1)(cos x + 1) = 0 M1
- cos x = ½ ⇒ x = π/3, 5π/3; cos x = −1 ⇒ x = π A1 A1
- x = π/3, π, 5π/3
Worked example
Solve 3 sin²x = sin x cos x for 0° ≤ x ≤ 180°.
- sin x(3 sin x − cos x) = 0 M1
- sin x = 0 ⇒ x = 0°, 180° A1
- 3 sin x = cos x ⇒ tan x = ⅓ ⇒ x = 18.4° A1
- x = 0°, 18.4°, 180°
Worked example
Solve 2 sin x = 5 cos x for 0° ≤ x ≤ 360°.
- cos x = 0 would force sin x = 0 too, which is impossible, so dividing by cos x is safe: tan x = 2.5 M1
- x = 68.2° or 68.2° + 180° = 68.2°, 248.2° A1 A1
Common mistakeGiving cos x = −¼ as an "answer". That is a step, not a solution: find the angles. And never keep a value like cos x = 3, which has no solutions.
Linking questionQuadratics in sin x or cos x use the methods of Quadratics; radian measure is in Circular measure; and Paper 3 extends these ideas to sec, cosec, cot and R sin(θ + α) in Further trigonometry.
Test yourself — cover the answers first
Quick check
- Write down the exact value of cos 150°.
- What is the principal value of tan⁻¹(−√3)?
- State the period of y = tan 3x in degrees.
- If 0 ≤ x ≤ π, what interval does 2x lie in?
- Why must you not divide 2 sin x cos x = sin x by sin x?
Exam-style
- Solve 3 cos²θ + 8 sin θ = 0 for 0° ≤ θ ≤ 360°.
- (a) Show that 3 sin θ tan θ = 8 can be written as 3cos²θ + 8cos θ − 3 = 0. (b) Hence solve 3 sin θ tan θ = 8 for 0° < θ < 360°.
- The function f is defined by f(x) = 4 − 3 sin 2x for 0 ≤ x ≤ π. State the range of f and solve f(x) = 5.5.
Show answers
- −√3/2
- −π/3 (−60°)
- 60°
- 0 ≤ 2x ≤ 2π
- you lose the solutions sin x = 0
- 3(1 − sin²θ) + 8 sin θ = 0 ⇒ 3s² − 8s − 3 = 0 ⇒ (3s + 1)(s − 3) = 0 ⇒ sin θ = −⅓ ⇒ θ = 199.5°, 340.5°
- (a) 3 sin²θ/cos θ = 8 ⇒ 3(1 − cos²θ) = 8 cos θ ⇒ 3cos²θ + 8cos θ − 3 = 0
(b) (3cos θ − 1)(cos θ + 3) = 0 ⇒ cos θ = ⅓ (cos θ = −3 impossible) ⇒ θ = 70.5°, 289.5° - 1 ≤ f(x) ≤ 7; sin 2x = −½ ⇒ 2x = 7π/6, 11π/6 ⇒ x = 7π/12, 11π/12
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