Questions on the t-formulae

Question 1

Write down sin θ, cos θ and tan θ in terms of t = tan(θ/2).

Accepted Answer

sin θ = 2t/(1 + t²), cos θ = (1 − t²)/(1 + t²) and tan θ = 2t/(1 − t²). B1 B1 B1 for the three formulae. All three share the same right triangle, with legs 2t and 1 − t² and hypotenuse 1 + t².

Question 2

Given that t = tan(θ/2) = 2, find sin θ and cos θ exactly.

Accepted Answer

1 + t² = 5, so sin θ = 4/5 and cos θ = (1 − 4)/5 = −3/5. M1 for 1 + t² = 5, A1 for sin θ, A1 for cos θ. A negative cosine with a positive sine puts θ in the second quadrant, which a sketch confirms since t > 1 means θ/2 exceeds 45°.

Question 3

Given that θ is acute with sin θ = 3/5 and cos θ = 4/5, find t = tan(θ/2) exactly.

Accepted Answer

From cos θ = (1 − t²)/(1 + t²): 4(1 + t²) = 5(1 − t²), so 9t² = 1 and t = ±1/3. Since θ is acute, θ/2 is acute and t is positive: t = 1/3. M1 for setting up either formula, A1 for 1/3 with the sign justified. Quicker still, t = sin θ/(1 + cos θ) = (3/5)/(9/5) = 1/3, and the same half-angle form is worth knowing for the reverse direction.

Question 4

With t = tan(θ/2), express 1 + cos θ and 1 − cos θ in terms of t, and hence write down (1 − cos θ)/(1 + cos θ) in terms of t.

Accepted Answer

1 + cos θ = 1 + (1 − t²)/(1 + t²) = 2/(1 + t²), and 1 − cos θ = 2t²/(1 + t²). B1 for both. Dividing, the (1 + t²) cancels and (1 − cos θ)/(1 + cos θ) = t², that is tan²(θ/2). B1. The two expressions are the half-angle formulae 2cos²(θ/2) and 2sin²(θ/2) written in t, and their ratio being tan²(θ/2) is the same fact from the other side.

Question 5

Use the t-formulae to solve 3 cos θ − 4 sin θ = 5 for 0 ≤ θ < 2π, giving the answer to 3 significant figures.

Accepted Answer

Substituting and multiplying by 1 + t²: 3(1 − t²) − 8t = 5(1 + t²), so 8t² + 8t + 2 = 0, that is (2t + 1)² = 0. The repeated root t = −1/2 gives θ = 2 arctan(−1/2) = −0.927, which is 5.36 radians in the required interval. M1 for substituting the t-formulae, M1 for clearing the denominator, A1 for 8t² + 8t + 2 = 0, M1 for solving, A1 for t = −1/2, A1 for 5.36. A repeated root was inevitable here. Since √(3² + 4²) = 5 is the greatest value the left side can take, the line is reached tangentially rather than crossed, so the two solutions coincide.

Question 6

Explain why the substitution t = tan(θ/2) cannot find a solution at θ = π, and state what should be done about it.

Accepted Answer

tan(π/2) is undefined, so θ = π corresponds to no finite value of t, so the substitution cannot reach it at all. Whenever the interval contains π, substitute θ = π into the original equation by hand and check separately whether it is a solution. B1 for tan(π/2) being undefined, B1 for no finite t reaching θ = π, B1 for testing θ = π separately.

Question 7

Show that tan 22.5° = √2 − 1, using the t-formula for tan θ with θ = 45°.

Accepted Answer

With t = tan 22.5°, the formula gives tan 45° = 2t/(1 − t²) = 1, so 2t = 1 − t², that is t² + 2t − 1 = 0. Solving gives t = (−2 ± √8)/2 = −1 ± √2. Since 22.5° is acute, t is positive, so t = √2 − 1 ≈ 0.414. M1 for the t-formula with θ = 45°, A1 for t² + 2t − 1 = 0, M1 for solving the quadratic, A1 for √2 − 1 with the sign justified.

Question 8

Starting from the double angle formulae, derive the expressions for sin θ and cos θ in terms of t = tan(θ/2).

Accepted Answer

Write sin θ = 2 sin(θ/2)cos(θ/2) = 2 tan(θ/2)cos²(θ/2). Since sec²(θ/2) = 1 + tan²(θ/2), cos²(θ/2) = 1/(1 + t²), and so sin θ = 2t/(1 + t²). For the cosine take cos θ = 2cos²(θ/2) − 1 = 2/(1 + t²) − 1 = (2 − (1 + t²))/(1 + t²) = (1 − t²)/(1 + t²). M1 for turning sin θ into 2 tan(θ/2)cos²(θ/2), M1 for cos²(θ/2) = 1/(1 + t²) via sec², A1 for sin θ = 2t/(1 + t²), A1 for cos θ = (1 − t²)/(1 + t²). Dividing the two results then gives tan θ = 2t/(1 − t²) at no extra cost.

Question 9

Use the t-formulae to solve 3 sin θ + 4 cos θ = 2 for 0 ≤ θ < 2π, giving your answers to 3 significant figures.

Accepted Answer

Substituting and multiplying by 1 + t²: 6t + 4(1 − t²) = 2(1 + t²), so 6t² − 6t − 2 = 0, that is 3t² − 3t − 1 = 0. M1 for substituting and clearing the denominator, A1 for the quadratic. Then t = (3 ± √21)/6 = 1.2638 or −0.2638. A1 for both roots. θ = 2 arctan t gives 1.8028 and −0.5158, and adding 2π to the second puts it in range: θ = 1.80 or 5.77. A1. The interval contains π, which the substitution cannot reach, so check it: 3 sin π + 4 cos π = −4, not 2, so nothing has been missed. Two distinct roots were expected here, since 2 is below √(3² + 4²) = 5 and the line is crossed rather than touched.

Question 10

Use the t-formulae to prove that sin θ/(1 + cos θ) ≡ tan(θ/2) and that (1 − cos θ)/sin θ ≡ tan(θ/2), stating where each identity is defined.

Accepted Answer

sin θ/(1 + cos θ) = [2t/(1 + t²)]/[2/(1 + t²)] = t, since 1 + cos θ = 2/(1 + t²). M1 for substituting, A1 for the first identity, valid where 1 + cos θ ≠ 0, that is θ not an odd multiple of π.
(1 − cos θ)/sin θ = [2t²/(1 + t²)]/[2t/(1 + t²)] = t, cancelling one factor of t, which needs t ≠ 0. M1 for substituting, A1 for the second identity, valid where sin θ ≠ 0, that is θ not a multiple of π. Both say tan(θ/2) = sin θ/(1 + cos θ) = (1 − cos θ)/sin θ, which is how a half-angle tangent is recovered from sin θ and cos θ without any square roots, and the two forms multiply to (1 − cos θ)/(1 + cos θ) = tan²(θ/2).

Question 11

When the t-formulae are substituted into a cos θ + b sin θ = c and the denominator is cleared, show that the result is (c + a)t² − 2bt + (c − a) = 0. Explain when this equation is linear rather than quadratic, and what that signals about the solutions of the original equation.

Accepted Answer

a(1 − t²) + 2bt = c(1 + t²) gives a − at² + 2bt = c + ct², and collecting: (c + a)t² − 2bt + (c − a) = 0. B1 for the rearrangement. The t² coefficient vanishes exactly when c = −a, and only then is the equation linear, with the single root t = (c − a)/(2b) = −a/b. B1. In that case θ = π satisfies the original equation, since a cos π + b sin π = −a = c, and π is the one value the substitution cannot reach: the missing second root has gone to θ = π, where t is undefined. B1 for the explanation.

Question 12

Prove the identity (1 + cosec θ)/cot θ ≡ (1 + tan(θ/2))/(1 − tan(θ/2)).

Accepted Answer

Write the left side over sin θ and cos θ: (1 + 1/sin θ)/(cos θ/sin θ) = (sin θ + 1)/cos θ. Now substitute the t-formulae. The numerator becomes (2t + 1 + t²)/(1 + t²) = (1 + t)²/(1 + t²) and the denominator (1 − t²)/(1 + t²) = (1 − t)(1 + t)/(1 + t²). Dividing, the (1 + t²) and one factor of (1 + t) cancel, leaving (1 + t)/(1 − t) as required. Cancelling 1 + t needs t ≠ −1, and that value is already excluded: t = −1 makes cot θ zero, so the left side is undefined there and the identity is being proved only where both sides exist. M1 for putting the left side over a common denominator, A1 for (sin θ + 1)/cos θ, M1 for substituting the t-formulae, A1 for the numerator, A1 for the denominator, A1 for the cancelled result with t ≠ −1 recorded.

Question 13

Use the t-formulae to solve sin θ − cos θ = 1 for 0 ≤ θ < 2π, remembering to test any value of θ the substitution cannot reach.

Accepted Answer

Substituting and multiplying by 1 + t²: 2t − (1 − t²) = 1 + t². The t² terms cancel, leaving 2t = 2, so t = 1 and θ = 2 arctan 1 = π/2. That is not the whole story: the substitution never reaches θ = π, so test it by hand. sin π − cos π = 0 − (−1) = 1, so θ = π is a solution as well. The solutions are θ = π/2 and θ = π. M1 for substituting the t-formulae, M1 for multiplying by 1 + t², A1 for t = 1, A1 for θ = π/2, M1 for testing θ = π in the original equation, A1 for both solutions in the final answer. The collapse from quadratic to linear is the tell-tale: one of the two expected roots has escaped to θ = π, where t is undefined, and it takes the t² term with it.

Question 14

Prove the identity (1 − cos θ + sin θ)/(1 + cos θ + sin θ) ≡ tan(θ/2).

Accepted Answer

Convert the left side to t = tan(θ/2). The numerator is 1 − (1 − t²)/(1 + t²) + 2t/(1 + t²) = (1 + t² − 1 + t² + 2t)/(1 + t²) = 2t(1 + t)/(1 + t²). The denominator is 1 + (1 − t²)/(1 + t²) + 2t/(1 + t²) = (2 + 2t)/(1 + t²) = 2(1 + t)/(1 + t²). Dividing, the factor 2(1 + t)/(1 + t²) cancels throughout, leaving t = tan(θ/2), as required. The cancellation needs t ≠ −1, and that value is already excluded: at t = −1 the denominator 1 + cos θ + sin θ is 1 + 0 − 1 = 0, so the left side is undefined there and the identity is claimed only where both sides exist. M1 for substituting the t-formulae, A1 for the numerator 2t(1 + t)/(1 + t²), A1 for the denominator 2(1 + t)/(1 + t²), M1 for dividing and cancelling, A1 for tan(θ/2) with t ≠ −1 noted.

Question 15

Starting from tan 2A = 2 tan A/(1 − tan² A), derive the t-formula tan θ = 2t/(1 − t²) where t = tan(θ/2). Show that, for a given value of tan θ, the values of t satisfy a quadratic whose roots have product −1, and explain what the second root represents. Check your explanation with θ = 60°.

Accepted Answer

Put A = θ/2 in the double angle formula: tan θ = 2 tan(θ/2)/(1 − tan²(θ/2)) = 2t/(1 − t²). B1 for the derivation.
Writing T = tan θ and clearing the denominator: T(1 − t²) = 2t, so Tt² + 2t − T = 0. M1 for the quadratic in t. Its roots multiply to −T/T = −1. A1.
If one root is t₁ = tan(θ/2), the other is −1/t₁ = −cot(θ/2) = tan(θ/2 + 90°). That is the half of θ + 180°, and tan θ = tan(θ + 180°), so the second root is the half-angle tangent of the other angle in the range with the same tangent. M1 for −1/t₁ as −cot(θ/2), A1 for the identification with θ + 180°.
With θ = 60°, T = √3 and the quadratic is √3 t² + 2t − √3 = 0, with roots t = (−2 ± 4)/(2√3), that is 1/√3 and −√3. B1 for the check: 1/√3 = tan 30° and −√3 = tan 120° = tan(30° + 90°), and 120° is half of 240° = 60° + 180°, whose tangent is also √3. On its own, a value of tan θ fixes t only up to this pair of roots, because tangent repeats every half turn; sin θ and cos θ together fix t uniquely.