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O-Level Additional Mathematics · Secondary 3 and 4

Trigonometric Identities and Equations

This topic builds on basic trigonometry to work with all six trigonometric functions for angles of any size, measured in degrees or radians. You prove identities, use the addition and double angle formulae, express a cos θ + b sin θ as a single function such as R cos(θ - α), and solve equations within a given interval. You also sketch graphs such as y = a sin bx + c. These identities are used constantly in calculus, where you differentiate and integrate trigonometric functions.

What the syllabus expects

  • Use the six trigonometric functions for angles of any size, in degrees or radians, and exact values for 30°, 45° and 60°
  • Sketch and interpret the graphs of y = a sin(bx) + c, y = a cos(bx) + c and y = a tan(bx), stating amplitude and period
  • Use the identities sin²A + cos²A = 1, sec²A = 1 + tan²A and cosec²A = 1 + cot²A
  • Use the addition formulae for sin(A ± B), cos(A ± B) and tan(A ± B), and the double angle formulae
  • Express a cos θ + b sin θ in the form R cos(θ ± α) or R sin(θ ± α), and use it to find maximum and minimum values
  • Prove trigonometric identities
  • Solve trigonometric equations in a given interval

Angles of any size and graphs

For any angle, the sign of a trigonometric ratio depends on the quadrant. In the first quadrant all are positive, in the second only sine is positive, in the third only tangent, and in the fourth only cosine. To solve sin x = 0.5 for 0° ≤ x ≤ 360°, find the basic angle 30°, then use the quadrants where sine is positive: x = 30° or 180° - 30° = 150°. The graph of y = a sin(bx) + c has amplitude |a|, period 360°/b (or 2π/b in radians), and is shifted up by c. For example, y = 3 cos 2x + 1 has amplitude 3, period 180°, maximum value 4 and minimum value -2. Remember that sec x = 1/cos x, cosec x = 1/sin x and cot x = 1/tan x.

The three Pythagorean identities

Dividing sin²A + cos²A = 1 by cos²A gives tan²A + 1 = sec²A, and dividing by sin²A gives 1 + cot²A = cosec²A. Use them to replace one function by another so that an equation contains only one trigonometric function. For example, 2cos²x + 3 sin x = 3 becomes 2(1 - sin²x) + 3 sin x - 3 = 0, which simplifies to 2sin²x - 3 sin x + 1 = 0, so (2 sin x - 1)(sin x - 1) = 0 and sin x = 1/2 or sin x = 1. When proving an identity, start from the more complicated side and work towards the other, using one identity at a time. Do not move terms across the equals sign, because that assumes the result you are trying to prove.

Addition and double angle formulae

The addition formulae are sin(A ± B) = sin A cos B ± cos A sin B, cos(A ± B) = cos A cos B ∓ sin A sin B, and tan(A ± B) = (tan A ± tan B)/(1 ∓ tan A tan B). Notice the sign change in the cosine and tangent formulae. Putting B = A gives the double angle formulae: sin 2A = 2 sin A cos A, cos 2A = cos²A - sin²A = 2cos²A - 1 = 1 - 2sin²A, and tan 2A = 2 tan A/(1 - tan²A). They give exact values, for example cos 15° = cos(45° - 30°) = (√2/2)(√3/2) + (√2/2)(1/2) = (√6 + √2)/4. In equations, choose the version of cos 2A that leaves only one trigonometric function.

The R-formula

An expression a cos θ + b sin θ, with a and b positive, can be written as R cos(θ - α), where R = √(a² + b²) and tan α = b/a. Expanding R cos(θ - α) = R cos θ cos α + R sin θ sin α and comparing coefficients gives R cos α = a and R sin α = b. Squaring and adding gives R², and dividing gives tan α. For example, 3 cos θ + 4 sin θ = 5 cos(θ - 53.1°). Since cosine lies between -1 and 1, the maximum value is 5 when θ = 53.1° and the minimum is -5 when θ = 233.1°. The other forms, R cos(θ + α), R sin(θ + α) and R sin(θ - α), are found the same way, by expanding and comparing coefficients. The R-formula is useful for solving a cos θ + b sin θ = k, since it turns the equation into one with a single trigonometric function.

Worked example

Express 3 cos θ + 4 sin θ in the form R cos(θ - α), where R > 0 and α is acute. Hence solve 3 cos θ + 4 sin θ = 2 for 0° ≤ θ ≤ 360°.

  1. Write 3 cos θ + 4 sin θ = R cos θ cos α + R sin θ sin α, so R cos α = 3 and R sin α = 4.
  2. R = √(3² + 4²) = 5 and tan α = 4/3, so α = 53.13°.
  3. The equation becomes 5 cos(θ - 53.13°) = 2, so cos(θ - 53.13°) = 0.4.
  4. Since 0° ≤ θ ≤ 360°, the angle θ - 53.13° lies between -53.13° and 306.87°.
  5. The basic angle is cos⁻¹ 0.4 = 66.42°. Cosine is positive in the first and fourth quadrants, so θ - 53.13° = 66.42° or 360° - 66.42° = 293.58°. The value -66.42° is outside the range.
  6. Add 53.13°: θ = 119.55° or θ = 346.71°.

Answer: 3 cos θ + 4 sin θ = 5 cos(θ - 53.1°), and θ = 119.6° or θ = 346.7° to 1 decimal place

Common mistakes

  • Dividing both sides of an equation by sin x or cos x, which loses the solutions where that function is zero. Factorise instead: sin x(2 cos x - 1) = 0 gives sin x = 0 or cos x = 1/2.
  • Forgetting to adjust the range when the angle is 2x or θ - α. For 0° ≤ x ≤ 360°, the angle 2x runs from 0° to 720°, so find solutions over two full turns before halving.
  • Writing cos(A + B) = cos A cos B + sin A sin B. The sign in the cosine formula is opposite, so cos(A + B) = cos A cos B - sin A sin B.
  • Mixing degrees and radians, such as giving x = π/6 and x = 150° in the same answer, or leaving the calculator in the wrong mode. Answer in the unit the interval is given in.
  • Proving an identity by working on both sides at once or by cross-multiplying. Start from one side and transform it into the other.
  • Writing tan α = a/b instead of b/a in the R-formula. Expand R cos(θ - α) and compare coefficients every time rather than relying on memory.

Exam tips

  • Write down the range of the transformed angle, such as 0° ≤ 2x ≤ 720°, before solving. It tells you how many solutions to expect.
  • When an equation mixes sin x with cos²x, or tan x with sec²x, use a Pythagorean identity to change it into a quadratic in one function.
  • In R-formula questions, keep α to at least 2 decimal places in your working, and round only the final angles.

Practice questions

1. Solve 2 sin x = √3 for 0° ≤ x ≤ 360°.

Show answer

x = 60° or x = 120°. sin x = √3/2 gives the basic angle 60°. Sine is positive in the first and second quadrants, so x = 60° or 180° - 60° = 120°.

2. Solve 2 sec²x = 3 + tan x for 0° ≤ x ≤ 180°.

Show answer

x = 45° or x = 153.4°. Using sec²x = 1 + tan²x gives 2 + 2tan²x = 3 + tan x, so 2tan²x - tan x - 1 = 0 and (2 tan x + 1)(tan x - 1) = 0. tan x = 1 gives x = 45°. tan x = -1/2 has basic angle 26.57°, and tangent is negative in the second quadrant, so x = 180° - 26.57° = 153.4°.

3. Prove that (1 - cos 2A)/sin 2A = tan A. Hence solve (1 - cos 2A)/sin 2A = 2 for 0° < A < 180°.

Show answer

A = 63.4°. Proof: (1 - cos 2A)/sin 2A = (1 - (1 - 2sin²A))/(2 sin A cos A) = 2sin²A/(2 sin A cos A) = sin A/cos A = tan A. So tan A = 2, giving A = 63.4°. The other angle with tan A = 2 is 243.4°, which is outside the interval.

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