Maths › Further Pure 1 › The t-formulae
The t-formulae
Write every trig function in terms of one variable, the tangent of the half angle, and identities become algebra while equations become quadratics.
Builds on Compound angles and the harmonic form and Reciprocal and inverse trigonometric functions.
IN THIS TOPIC
- Derive sin θ, cos θ and tan θ in terms of t = tan(θ/2), and quote them without hesitating.
- Prove trig identities by converting every term to t.
- Turn a cos x + b sin x = c into a quadratic in t and solve it.
- Test x = π by hand, since the substitution cannot reach it.
COMMON MISCONCEPTION
sin θ, cos θ and tan θ are three independent quantities; no single variable can carry all three.
One variable, three ratios
Put t = tan(θ/2) and the double angle formulae do everything else. Start from sin θ = 2 sin(θ/2)cos(θ/2) and cos θ = 2cos²(θ/2) − 1, then use cos²(θ/2) = 1/(1 + t²). Both ratios fall out in one step.
None of the three is in the booklet, so they have to come from you. One number now generates all three ratios. You can also read them off a right-angled triangle with legs 2t and 1 − t² and hypotenuse 1 + t², since (2t)² + (1 − t²)² = (1 + t²)². Sketch that triangle in the margin and you will stop misremembering which expression carries the minus sign. Treat it as a signed mnemonic rather than a genuine figure: outside 0 < θ < π/2 the legs 2t and 1 − t² can go negative, and the formulae keep the signs right even where no such triangle can be drawn.
WORKED EXAMPLE
A spot check at a known angle
Verify the t-formulae at θ = π/3.
t = tan(π/6) = 1/√3, so t² = 1/3 and 1 + t² = 4/3.
sin θ = (2/√3)/(4/3) = (2/√3)(3/4) = √3/2, correct.
cos θ = (2/3)/(4/3) = 1/2, correct. One value of t has produced both ratios exactly.
Identities, once everything is in t
A proof question converts both sides to t and lets the algebra meet somewhere in the middle. Work on the messier side first. Every denominator that appears is 1 + t² or 1 − t², and those cancel readily, so clear the fractions early instead of dragging them through six lines of working.
Try it on (1 − cos θ)/sin θ. The top becomes 1 − (1 − t²)/(1 + t²), which is 2t²/(1 + t²). The bottom is 2t/(1 + t²). Divide, the 1 + t² disappears, and what remains is t. So the whole expression collapses to tan(θ/2), a result worth carrying around.
One caution belongs here and not in the small print. Since tan(π/2) is undefined, t does not exist at θ = π itself, or at any odd multiple of π, though the formulae remain valid at every angle arbitrarily close. An identity proved this way still holds wherever both sides are defined, but any equation solved this way needs θ = π, and any other odd multiple in the interval, checked separately.
Equations that become quadratics
In a cos x + b sin x = c, substitute the t-formulae and multiply through by 1 + t². The result is a quadratic in t. Solve it, recover x = 2 arctan t, and bring the angles into the demanded interval.
WORKED EXAMPLE
A double root that means tangency
Solve 3 cos θ + 4 sin θ = 5 for 0 ≤ θ < 2π.
Substitute: 3(1 − t²) + 8t = 5(1 + t²), so 8t² − 8t + 2 = 0, which is 2(2t − 1)² = 0.
The repeated root t = 1/2 gives θ = 2 arctan(1/2) ≈ 0.927.
A repeated root was inevitable. Since √(3² + 4²) = 5, the level y = 5 is the expression's maximum, so the line touches the curve instead of cutting it. One θ, hit tangentially.
GUIDED PRACTICE
A two-solution equation
Solve cos θ + sin θ = 1 for 0 ≤ θ < 2π, using the t-formulae, and state any value of θ the substitution cannot see.
Show the working
Substitute: (1 − t²) + 2t = 1 + t², so 2t² − 2t = 0 and t = 0 or 1.
t = 0 gives θ = 0; t = 1 gives θ = π/2. Both check, since 1 + 0 and 0 + 1 each equal 1.
θ = π is invisible to the substitution. Testing it directly gives −1 + 0 ≠ 1, so nothing was lost here. The check itself is still part of the method.
ASSESSMENT FOCUS
- Quote all three t-formulae before substituting. The derivation is only wanted when the question asks for it.
- Multiply through by 1 + t² early. That factor is never zero, so no solutions are created or lost.
- Test x = π separately whenever the interval contains it; the substitution cannot produce it.
- Recover angles with x = 2 arctan t, then adjust into the demanded interval before writing the final line.
- In a proof, stop when both sides are the same expression in t and say so; do not keep manipulating.
CHECK YOURSELF
Given t = tan(θ/2) = 1/3, find sin θ and cos θ exactly.
Show a hint
1 + t² = 10/9; the t-triangle does the rest.
Show the answer
sin θ = (2/3)/(10/9) = 3/5 and cos θ = (8/9)/(10/9) = 4/5. The 3-4-5 triangle turns up whenever t = 1/3.
With t = tan(θ/2): sin θ = 2t/(1 + t²), cos θ = (1 − t²)/(1 + t²), tan θ = 2t/(1 − t²).
a cos x + b sin x = c becomes a quadratic in t. Solve it, then take x = 2 arctan t.
The substitution never reaches x = π, so test that value by hand.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the the t-formulae questions page.
CHECK YOUR PROGRESS
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- Derive sin θ, cos θ and tan θ in terms of t = tan(θ/2), and quote them without hesitating.
- Prove trig identities by converting every term to t.
- Turn a cos x + b sin x = c into a quadratic in t and solve it.
- Test x = π by hand, since the substitution cannot reach it.
Open the full revision checklist to see every objective in the course in one place.