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The t-formulae questions
Writing t = tan(x/2) expresses sin x, cos x and tan x as rational functions of the single variable t. Identities then become algebra and trigonometric equations become quadratics, which is often the quickest route to a complete solution.
6 original questions · 25 marks · the t-formulae notes · Further Pure 1
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Write down sin θ, cos θ and tan θ in terms of t = tan(θ/2).
Worked 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.Given that t = tan(θ/2) = 2, find sin θ and cos θ exactly.
Worked 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°.Use the t-formulae to solve 3 cos θ − 4 sin θ = 5 for 0 ≤ θ < 2π, giving the answer to 3 significant figures.
Worked 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.Explain why the substitution t = tan(θ/2) cannot find a solution at θ = π, and state what should be done about it.
Worked 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.Show that tan 22.5° = √2 − 1, using the t-formula for tan θ with θ = 45°.
Worked 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.Prove the identity (1 + cosec θ)/cot θ ≡ (1 + tan(θ/2))/(1 − tan(θ/2)).
Worked 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.
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