← Trigonometric Functions (ME-T1, T2, T3)

NSWMaths Extension 1Syllabus dot point

How does the t-substitution $t = \tan \frac{\theta}{2}$ help simplify and solve trigonometric equations?

Use the t-formula (Weierstrass substitution) to express $\sin \theta$ , $\cos \theta$ and $\tan \theta$ as rational functions of $t = \tan \frac{\theta}{2}$

A focused answer to the HSC Maths Extension 1 dot point on the t-formula. Derivation of the t-substitution, the rational expressions for $\sin$ , $\cos$ and $\tan$ in terms of $t$ , and its use in solving and simplifying trig equations, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy7 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to know the t-formula (sometimes called the Weierstrass or half-angle substitution): set $t = \tan \frac{\theta}{2}$ and write $\sin \theta$ , $\cos \theta$ and $\tan \theta$ as rational functions of $t$ . This converts certain trig equations and integrals into algebraic ones.

The answer

The substitution

Let $t = \tan \frac{\theta}{2}$ . Then

\sin \theta = \frac{2 t}{1 + t^2}, \qquad \cos \theta = \frac{1 - t^2}{1 + t^2}, \qquad \tan \theta = \frac{2 t}{1 - t^2}.

These come from the double-angle identities and a right-triangle picture.

Derivation

Let $\phi = \frac{\theta}{2}$ , so $\theta = 2 \phi$ and $t = \tan \phi$ .

From the double-angle identities,

\sin 2 \phi = 2 \sin \phi \cos \phi.

Divide numerator and denominator by $\cos^2 \phi$ after multiplying by $\frac{\cos^2 \phi}{\cos^2 \phi}$ :

\sin 2 \phi = \frac{2 \sin \phi \cos \phi}{1} \cdot \frac{1/\cos^2 \phi}{\sec^2 \phi} = \frac{2 \tan \phi}{1 + \tan^2 \phi} = \frac{2 t}{1 + t^2}.

Similarly $\cos 2 \phi = \cos^2 \phi - \sin^2 \phi$ . Divide by $\cos^2 \phi$ on top and bottom:

\cos 2 \phi = \frac{1 - \tan^2 \phi}{1 + \tan^2 \phi} = \frac{1 - t^2}{1 + t^2}.

For tangent, $\tan 2 \phi = \frac{\sin 2 \phi}{\cos 2 \phi} = \frac{2 t}{1 - t^2}$ , valid when $1 - t^2 \neq 0$ .

When to use the t-formula

The t-formula is most useful for:

Equations of the form $a \sin \theta + b \cos \theta = c$ or $a \tan \theta + b = c \sin \theta + \dots$ where standard methods get stuck.
Integrals of rational functions of $\sin \theta$ and $\cos \theta$ .
Proving identities where every term can be rewritten in $t$ .

Limitations

The substitution $t = \tan \frac{\theta}{2}$ is undefined at $\theta = \pi + 2 n \pi$ (where $\cos \frac{\theta}{2} = 0$ ).

Always check whether $\theta = \pi + 2 n \pi$ is a solution to the original equation; the t-formula may miss it.

Algebraic conversion

To solve $a \sin \theta + b \cos \theta = c$ with t-formula:

a \cdot \frac{2 t}{1 + t^2} + b \cdot \frac{1 - t^2}{1 + t^2} = c.

Multiply through by $1 + t^2$ :

2 a t + b (1 - t^2) = c (1 + t^2).

Rearrange to a quadratic in $t$ :

(b + c) t^2 - 2 a t + (c - b) = 0.

Solve for $t$ , then recover $\theta$ via $\theta = 2 \arctan t$ .

Worked example

Solve a sinusoidal equation

Solve $\sin \theta + 2 \cos \theta = 1$ for $\theta \in [0, 2 \pi)$ .

t-formula gives

\frac{2 t}{1 + t^2} + 2 \cdot \frac{1 - t^2}{1 + t^2} = 1.

Multiply through by $1 + t^2$ : $2 t + 2(1 - t^2) = 1 + t^2$ .

$2 t + 2 - 2 t^2 = 1 + t^2$ , so $3 t^2 - 2 t - 1 = 0$ .

Factor: $(3 t + 1)(t - 1) = 0$ , so $t = -\frac{1}{3}$ or $t = 1$ .

$\theta = 2 \arctan(-\frac{1}{3}) = -2 \arctan \frac{1}{3} \approx -0.6435$ rad, plus the period.

Or $\theta = 2 \arctan 1 = \frac{\pi}{2}$ .

Adjust to $[0, 2 \pi)$ and check $\theta = \pi$ separately (where t-formula is undefined; substitute: $\sin \pi + 2 \cos \pi = 0 + 2(-1) = -2 \neq 1$ , so $\theta = \pi$ is not a solution).

In $[0, 2 \pi)$ : $\theta = \frac{\pi}{2}$ and $\theta \approx 2 \pi - 0.6435 \approx 5.640$ rad.

Verify the identity

Show $\frac{1 - \cos \theta}{\sin \theta} = \tan \frac{\theta}{2}$ .

LHS: $\frac{1 - \cos \theta}{\sin \theta} = \frac{1 - (1 - t^2)/(1 + t^2)}{2 t / (1 + t^2)} = \frac{(1 + t^2) - (1 - t^2)}{2 t} = \frac{2 t^2}{2 t} = t = \tan \frac{\theta}{2}$ . As required.

Express a sum

Express $\sin \theta + \cos \theta$ in terms of $t = \tan \frac{\theta}{2}$ .

$\sin \theta + \cos \theta = \frac{2 t}{1 + t^2} + \frac{1 - t^2}{1 + t^2} = \frac{2 t + 1 - t^2}{1 + t^2}.$

This is a useful form when the equation factorises nicely after the substitution.

Integral via t-formula

Evaluate $\int \frac{d \theta}{1 + \sin \theta}$ .

Set $t = \tan \frac{\theta}{2}$ , so $\sin \theta = \frac{2 t}{1 + t^2}$ and $d\theta = \frac{2}{1 + t^2} \, dt$ .

$\int \frac{1}{1 + \frac{2 t}{1 + t^2}} \cdot \frac{2}{1 + t^2} \, dt = \int \frac{2}{(1 + t^2) + 2 t} \, dt = \int \frac{2}{(1 + t)^2} \, dt = -\frac{2}{1 + t} + C$ .

Back-substitute: $-\frac{2}{1 + \tan(\theta/2)} + C$ .