A-Level Maths trigonometric identities: sec, compound angles and R sin(θ + α)

Where trig identities sit in each specification

The key ideas

Divide sin^2 θ + cos^2 θ = 1 by cos^2 θ and you get tan^2 θ + 1 = sec^2 θ; divide by sin^2 θ and you get 1 + cot^2 θ = cosec^2 θ. These let you turn an equation with two different functions into a quadratic in one function, which you can factorise.

The compound angle formulae are sin(A + B) = sin A cos B + cos A sin B and cos(A + B) = cos A cos B - sin A sin B (signs flip for A - B, and the cos formula has the opposite sign inside). Put A = B to get the double angle formulae: sin 2A = 2 sin A cos A and cos 2A = cos^2 A - sin^2 A = 2cos^2 A - 1 = 1 - 2sin^2 A. Choose the version of cos 2A that leaves only one function in the equation.

Any a sin θ + b cos θ can be written as R sin(θ + α) with R = sqrt(a^2 + b^2) and tan α = b/a. This turns a two-function equation into one sine, and gives the maximum value R and minimum -R immediately. When solving, adjust the interval for θ + α before finding solutions, then subtract α.

Worked example 1: solve 2 sec^2 θ - tan θ = 5 for 0° ≤ θ ≤ 180°

1. Replace sec^2 θ with 1 + tan^2 θ: 2 + 2tan^2 θ - tan θ = 5.
2. Rearrange: 2tan^2 θ - tan θ - 3 = 0, which factorises as (2tan θ - 3)(tan θ + 1) = 0.
3. So tan θ = 1.5 or tan θ = -1.
4. tan θ = 1.5 gives θ = 56.3° (the next solution, 236.3°, is outside the interval).
5. tan θ = -1 gives θ = 180° - 45° = 135°.
6. Answer: θ = 56.3° or 135°. Check θ = 135°: sec^2 = 2, tan = -1, so 4 + 1 = 5.

Worked example 2: write 3 sin θ + 4 cos θ as R sin(θ + α), then solve 3 sin θ + 4 cos θ = 2.5 for 0° ≤ θ ≤ 360°

1. Expand R sin(θ + α) = R sin θ cos α + R cos θ sin α and compare: R cos α = 3 and R sin α = 4.
2. R = sqrt(3^2 + 4^2) = 5 and tan α = 4/3, so α = 53.13°.
3. The equation becomes 5 sin(θ + 53.13°) = 2.5, so sin(θ + 53.13°) = 0.5.
4. θ + 53.13° lies between 53.13° and 413.13°, so θ + 53.13° = 150° or 390° (30° is too small).
5. Answer: θ = 96.9° or 336.9°. Both satisfy the original equation to 1 decimal place.

Worked example 3: prove that (1 - cos 2x)/sin 2x ≡ tan x

1. Start with the more complicated side, the left.
2. Use cos 2x = 1 - 2sin^2 x, so 1 - cos 2x = 2sin^2 x.
3. Use sin 2x = 2 sin x cos x.
4. The left side becomes 2sin^2 x/(2 sin x cos x) = sin x/cos x = tan x, which is the right side.

Common mistakes that cost marks

• Dividing both sides by cos θ or sin θ and losing the solutions where it is zero. Factorise instead.
• Finding one solution from the calculator and missing the others in the interval.
• Not adjusting the interval for θ + α or 2θ before solving.
• Writing sin(A + B) = sin A + sin B.
• Using degrees when the interval is in radians, or the reverse.
• In a proof, working on both sides at once or starting from the statement you are asked to prove.

Exam technique and how a tutor helps

Trig equations are often split into a 'show that' part, which turns the equation into a quadratic, and a 'hence solve' part. Even if you cannot do the first part, use the given result to attempt the second, because the marks are separate. Give angles to the accuracy asked, usually 1 decimal place in degrees or 3 significant figures in radians.

In proofs, write one line per identity used and finish with the exact target expression. Sketching the graph over the interval takes seconds and shows how many solutions to expect.

In one-to-one lessons a tutor spends time on the choice of identity, because students usually know the formulae but pick a version that leaves two functions. The tutor sets equations with deliberately awkward intervals and checks every solution set against a sketch on the shared whiteboard until finding all solutions becomes routine.

Self-check: can you do these?

• Find the exact value of cos 75°. (Answer: (sqrt6 - sqrt2)/4)
• Solve sin 2x = cos x for 0° ≤ x ≤ 360°. (Answer: 30°, 90°, 150°, 270°)
• State the maximum value of 3 sin θ + 4 cos θ. (Answer: 5)
• Write 2cos^2 x - 1 as a single trig function. (Answer: cos 2x)
• Simplify (sec^2 x - 1)/tan x. (Answer: tan x)