Questions on compound angles and the harmonic form

Question 1

Find the exact value of cos 75°.

Accepted 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.

Question 2

Given that sin A = 3/5 and A is acute, find the exact value of sin 2A.

Accepted 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.

Question 3

Write down the three forms of cos 2A, and use one of them to show that cos² A = (1 + cos 2A)/2.

Accepted Answer

Setting B = A in the compound cosine formula gives cos 2A = cos² A − sin² A, and replacing sin² A by 1 − cos² A, or cos² A by 1 − sin² A, gives the other two: cos 2A = 2 cos² A − 1 and cos 2A = 1 − 2 sin² A. B1 for all three forms. Take the middle one, the only one carrying cosine alone, and rearrange it: 2 cos² A = 1 + cos 2A. M1 for the rearrangement. Dividing by 2 gives cos² A = (1 + cos 2A)/2, as required. A1 for the printed result. The companion, from the third form, is sin² A = (1 − cos 2A)/2. Both turn a squared trigonometric function into one of first power in a doubled angle, which is how the integration unit later handles cos² A.

Question 4

Given that sin A = 2/3 and A is obtuse, find the exact value of cos 2A, and explain why the fact that A is obtuse does not affect your answer.

Accepted Answer

Choose the form of cos 2A that carries the sine alone, since the sine is what the question gives: cos 2A = 1 − 2 sin² A = 1 − 2 × 4/9 = 1/9. B1 for 1/9. That form uses sin A only, and uses it squared, so neither the sign of cos A nor the quadrant A lies in enters the calculation at all. B1 for the reason. Had sin 2A been asked for instead, cos A would be needed, and the word obtuse would then fix its sign as negative.

Question 5

Given that cos A = 1/3 and A is acute, find the exact values of cos 2A and sin 2A.

Accepted 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°, so cos 2A and cos A need not have the same sign.

Question 6

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.

Accepted 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. The harmonic form here has tan α = 12/5; the reciprocal 5/12 gives a different angle and a wrong solution set.

Question 7

Solve sin 2x = sin x for 0 ≤ x < 360°.

Accepted 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.

Question 8

Prove that sin 3A ≡ 3 sin A − 4 sin³ A.

Accepted Answer

Split the angle: 3A = 2A + A, and expand with the compound sine formula. sin 3A = sin 2A cos A + cos 2A sin A. M1 for splitting the angle and expanding. Now put in the double angle formulae, taking for cos 2A the form that carries the sine alone so that everything will end up in sines: sin 3A = (2 sin A cos A) cos A + (1 − 2 sin² A) sin A = 2 sin A cos² A + sin A − 2 sin³ A. M1 for both double angle substitutions. Replace the remaining cos² A by 1 − sin² A: 2 sin A (1 − sin² A) + sin A − 2 sin³ A = 2 sin A − 2 sin³ A + sin A − 2 sin³ A. A1 for an expression in sines only. Collecting the like terms gives 3 sin A − 4 sin³ A, as required. A1 for the collected result.

Question 9

Prove that cot θ − tan θ ≡ 2 cot 2θ.

Accepted Answer

Work down the left-hand side, writing both terms in sine and cosine: cot θ − tan θ = cos θ/sin θ − sin θ/cos θ. Over the common denominator sin θ cos θ that is (cos² θ − sin² θ)/(sin θ cos θ). M1 for the single fraction. The numerator is cos 2θ, and the denominator is ½ sin 2θ, since sin 2θ = 2 sin θ cos θ. A1 for cos 2θ over ½ sin 2θ. So the expression is 2 cos 2θ/sin 2θ = 2 cot 2θ, as required. A1 for the conclusion. Both double angle formulae turn up in the same line, one above and one below, and the factor of 2 comes from the half in the denominator rather than from anything in the original expression.

Question 10

Solve cos 2x + 3 sin x = 2 for 0 ≤ x < 360°, stating which of the three forms of cos 2x you use and why.

Accepted Answer

The equation already carries sin x, so take the form of cos 2x written in sines alone, cos 2x = 1 − 2 sin² x; either other form would bring cos² x into an equation that has no cosine in it and would need converting back. B1 for the choice with that reason. Substituting gives 1 − 2 sin² x + 3 sin x = 2, which tidies to 2 sin² x − 3 sin x + 1 = 0. M1 for the quadratic in sin x. It factorises as (2 sin x − 1)(sin x − 1) = 0, so sin x = ½ or sin x = 1. A1 for both values. From sin x = ½ come x = 30° and x = 150°. A1 for both. From sin x = 1 comes x = 90° and nothing else, since the sine reaches its maximum once per turn. A1 for 90°.

Question 11

Given that cos (θ + 30°) = 2 cos (θ − 30°), find the exact value of tan θ.

Accepted Answer

Expand both sides with the compound cosine formula, remembering that the sign in the middle is the opposite of the sign inside the bracket: cos θ cos 30° − sin θ sin 30° = 2(cos θ cos 30° + sin θ sin 30°). M1 for both expansions with the correct middle signs. Putting in cos 30° = √3/2 and sin 30° = ½ gives (√3/2) cos θ − ½ sin θ = √3 cos θ + sin θ. A1 for the substituted equation. Collect the cosines on one side and the sines on the other: (√3/2) cos θ − √3 cos θ = sin θ + ½ sin θ, that is −(√3/2) cos θ = (3/2) sin θ. M1 for collecting the two kinds of term. Dividing by cos θ is safe, because cos θ = 0 would force sin θ = 0 too and no angle has both, and it gives tan θ = −(√3/2) ÷ (3/2) = −√3/3. A1 for the exact value. Writing it as −1/√3 is the same number, and either form is exact. The negative sign puts θ in the second or the fourth quadrant, at 150° and 330° in a single turn.

Question 12

Solve 3 cos θ − 4 sin θ = 2 for 0 ≤ θ < 2π, giving your answers to 3 decimal places.

Accepted 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.

Question 13

Show that sin (A + B) + sin (A − B) = 2 sin A cos B. Hence find the exact value of sin 75° + sin 15°.

Accepted 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.

Question 14

Express 3 sin θ + √3 cos θ in the form R sin (θ + α), where R > 0 and 0 < α < π/2, giving R and α exactly. Hence find the greatest value of (3 sin θ + √3 cos θ)², and all the values of θ in 0 ≤ θ < 2π at which that greatest value occurs.

Accepted Answer

Expanding R sin (θ + α) = R cos α sin θ + R sin α cos θ and matching the coefficients gives R cos α = 3 and R sin α = √3. M1 for the expansion with both coefficients matched. Squaring and adding, R² = 9 + 3 = 12, so R = 2√3. A1 for R. Dividing, tan α = √3/3, and α lies in the first quadrant, so α = π/6. A1 for α. Hence 3 sin θ + √3 cos θ = 2√3 sin (θ + π/6), and its square is 12 sin² (θ + π/6). M1 for squaring the harmonic form. A square of a sine is greatest when the sine is 1 or −1, so the greatest value is 12. A1 for 12. That happens when θ + π/6 = π/2 or 3π/2, both of which lie in the widened interval, so θ = π/3 or 4π/3. A1 for both values. Squaring doubles the count of maxima: the expression itself reaches 2√3 once and −2√3 once in a turn, and the square treats those two alike.

Question 15

Prove that (sin 2θ)/(1 + cos 2θ) ≡ tan θ. Hence show that tan 22.5° = √2 − 1.

Accepted Answer

Replace the numerator by 2 sin θ cos θ, and choose for the denominator the form of cos 2θ that makes 1 + cos 2θ collapse: with cos 2θ = 2 cos² θ − 1 the denominator becomes 2 cos² θ. M1 for both double angle substitutions. The expression is now (2 sin θ cos θ)/(2 cos² θ). A1 for the single fraction. Cancelling 2 cos θ, which is not zero wherever tan θ is defined, leaves sin θ/cos θ = tan θ, as required. A1 for the conclusion. Now take θ = 22.5°, so 2θ = 45°, and the identity reads tan 22.5° = sin 45°/(1 + cos 45°) = (√2/2)/(1 + √2/2) = √2/(2 + √2). M1 for substituting θ = 22.5° and reaching that surd. Rationalise by multiplying top and bottom by 2 − √2: √2(2 − √2)/((2 + √2)(2 − √2)) = (2√2 − 2)/(4 − 2) = (2√2 − 2)/2 = √2 − 1, as required. A1 for the rationalised value.