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Compound angles and the harmonic form questions
Sine and cosine do not distribute over a sum, so compound angle formulae are needed for sin(A ± B) and cos(A ± B). From them come the double angle identities, a supply of new exact values, the half-angle forms that integration uses, and the harmonic form, which writes any combination of sine and cosine as a single wave with a readable amplitude and phase shift.
7 original questions · 30 marks · the compound angles and the harmonic form notes · Trigonometry
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
Find the exact value of cos 75°.
Worked answer
Write 75° as 45° + 30° and expand. cos 75° = cos 45° cos 30° − sin 45° sin 30° = (√2/2)(√3/2) − (√2/2)(½) = (√6 − √2)/4. M1 for the expansion with the correct middle sign, A1 for the simplified surd. Cosine flips the sign in the middle, so a plus inside the bracket becomes a minus outside.Given that sin A = 3/5 and A is acute, find the exact value of sin 2A.
Worked answer
From the 3-4-5 triangle, cos A = 4/5, positive because A is acute. Then sin 2A = 2 sin A cos A = 2 × (3/5)(4/5) = 24/25. B1 for cos A, M1 for quoting the double angle formula, A1 for 24/25. Doubling the angle is not doubling the sine. An answer of 6/5 gives that mistake away instantly, since no sine can exceed 1.Given that cos A = 1/3 and A is acute, find the exact values of cos 2A and sin 2A.
Worked answer
The cosine-only form settles the first part in one step. cos 2A = 2 cos2 A − 1 = 2/9 − 1 = −7/9. For sin 2A you do need sin A, and sin A = √(1 − 1/9) = √(8/9) = 2√2/3, taking the positive root because A is acute. Then sin 2A = 2 sin A cos A = 2 × (2√2/3)(1/3) = 4√2/9. M1 A1 for cos 2A, B1 for sin A, M1 A1 for sin 2A. Notice that cos 2A comes out negative while A itself is acute. Doubling has carried the angle past 90°, and students who expect the sign to follow A talk themselves out of a correct answer.Express 5 sin θ + 12 cos θ in the form R sin (θ + α), where R > 0 and 0 < α < π/2, giving α to 3 decimal places. Hence write down the maximum value of 5 sin θ + 12 cos θ and find, to 3 decimal places, the smallest positive value of θ at which that maximum occurs.
Worked answer
Expanding R sin (θ + α) = R cos α sin θ + R sin α cos θ and matching coefficients gives R cos α = 5 and R sin α = 12. Square and add for R = √(25 + 144) = 13; divide for tan α = 12/5, so α = 1.176. Hence 5 sin θ + 12 cos θ = 13 sin (θ + 1.176). The maximum is 13, reached when the sine equals 1, so θ + 1.176 = π/2 and θ = 0.395. The marks run M1 for the method for R, A1 for 13, M1 for tan α, A1 for 1.176, B1 for the maximum, A1 for θ. Writing tan α = 5/12 is the standard slip, and it costs the last three marks in one go.Solve sin 2x = sin x for 0 ≤ x < 360°.
Worked answer
Expand the double angle to get 2 sin x cos x = sin x, gather everything on one side and factorise, giving sin x (2 cos x − 1) = 0. From sin x = 0 come x = 0° and 180°; from cos x = ½ come x = 60° and 300°. All four are needed: 0°, 60°, 180°, 300°. M1 for the double angle expansion, M1 for the factorisation, A1 A1 for the four solutions. Dividing both sides by sin x at the start looks tidy and silently destroys two of them, and that is exactly why the factorisation is the step worth showing.Solve 3 cos θ − 4 sin θ = 2 for 0 ≤ θ < 2π, giving your answers to 3 decimal places.
Worked answer
Fold the left-hand side first. Expanding R cos (θ + α) = R cos α cos θ − R sin α sin θ and matching gives R cos α = 3 and R sin α = 4, so R = 5 and tan α = 4/3, that is α = 0.9273. The equation becomes 5 cos (θ + 0.9273) = 2, so cos (θ + 0.9273) = 0.4. Now θ + 0.9273 runs over [0.9273, 0.9273 + 2π), and cosine takes the value 0.4 there at 1.1593 and at 2π − 1.1593 = 5.1239. Subtracting α gives θ = 0.232 and θ = 4.197. M1 for the expansion and matching, A1 for R, M1 A1 for α, A1 A1 for the two solutions. The trap is stopping at the principal value the calculator returns. Work out the full interval for θ + α, list every solution inside it, and subtract α only at the very end.Show that sin (A + B) + sin (A − B) = 2 sin A cos B. Hence find the exact value of sin 75° + sin 15°.
Worked answer
Expand both compound angles: (sin A cos B + cos A sin B) + (sin A cos B − cos A sin B). The cos A sin B terms cancel and the two sin A cos B terms add, leaving 2 sin A cos B as required. For the second part, 75° and 15° average 45° and differ by 30°, so take A = 45° and B = 30°. Then sin 75° + sin 15° = 2 sin 45° cos 30° = 2 × (√2/2)(√3/2) = √6/2. M1 for expanding both compound angles, A1 for the printed identity, M1 for choosing the two angles, A1 for the exact value. Half the sum with half the difference finds them every time.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
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