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 and quote sin θ, cos θ and tan θ in terms of t = tan(θ/2).
- Prove trig identities by converting everything to t.
- Solve a cos x + b sin x = c by reducing it to a quadratic in t.
WHAT YOU PROBABLY THINK
sin θ, cos θ and tan θ are three independent quantities; no single variable can carry all three.
One variable to rule the ratios
Set t = tan(θ/2) and the double angle formulae do the rest. Writing sin θ = 2 sin(θ/2)cos(θ/2) and dividing top and bottom by cos²(θ/2) gives the first of the three:
One number t now generates all three ratios, which retires the opening claim. The three expressions are the sides of a right triangle with legs 2t and 1 − t² and hypotenuse 1 + t², since (2t)² + (1 − t²)² = (1 + t²)².
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 follow by converting both sides to t and matching the algebra. The exam's favourite shape asks for a proof that some mix of trig functions equals an expression in tan(θ/2); substitute, simplify the fractions, and the two sides meet in the middle.
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, then recover x = 2 arctan t. One caution: t = tan(x/2) is undefined at x = π, so check x = π separately whenever the interval contains it.
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 (2t − 1)² = 0.
The repeated root t = 1/2 gives θ = 2 arctan(1/2) ≈ 0.927.
A repeated root was inevitable: 5 = √(3² + 4²) is the expression's maximum, so the line y = 5 touches the curve rather than crossing it. One θ, hit tangentially.
YOUR TURN
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: 1 + 0 and 0 + 1.
θ = π is invisible to the substitution since tan(π/2) is undefined; testing it directly gives −1 + 0 ≠ 1, so nothing was missed here, but the check itself is part of the method.
THE EXAM BIT
- Quote all three t-formulae before substituting; the derivation is only required when asked.
- Multiply through by 1 + t² early; it is never zero, so no solutions are created or lost.
- Always test x = π separately when the interval contains it; t cannot reach it.
- Recover angles with x = 2 arctan t, then bring them into the demanded interval.
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
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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, take 2 arctan t, and test x = π 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.
CHECK YOUR PROGRESS
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- Derive and quote sin θ, cos θ and tan θ in terms of t = tan(θ/2).
- Prove trig identities by converting everything to t.
- Solve a cos x + b sin x = c by reducing it to a quadratic in t.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.