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

What this dot point is asking

NESA wants you to know the t-formula (also called the Weierstrass or half-angle substitution): set $t = \tan \frac{\theta}{2}$ and rewrite $\sin \theta$, $\cos \theta$ and $\tan \theta$ as rational functions of $t$. The point is a change of subject: trig equations and integrals that mix sines and cosines are hard, but once everything is written in $t$ they become ordinary algebra, usually a quadratic. You trade the trigonometry for a polynomial, solve that, then convert back with $\theta = 2\arctan t$. The one catch, which the exam loves, is that the substitution quietly excludes $\theta = \pi$ (plus full turns), so you must check that value by hand.

The answer

The substitution

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

These are not arbitrary; they fall straight out of a right triangle for the half-angle together with the double-angle identities.

Where the formulas come from: a half-angle triangle

Write $\phi = \frac{\theta}{2}$, so $t = \tan\phi$. Build a right triangle whose angle at the origin is $\phi$, with the side opposite equal to $t$ and the side adjacent equal to $1$. By Pythagoras the hypotenuse is $\sqrt{1 + t^2}$.

Reading the triangle gives sine and cosine of the half-angle directly.

Now feed these into the double-angle identities for $\theta = 2\phi$. For sine, $\sin\theta = 2\sin\phi\cos\phi = 2\cdot\frac{t}{\sqrt{1+t^2}}\cdot\frac{1}{\sqrt{1+t^2}} = \frac{2t}{1+t^2}$. For cosine, $\cos\theta = \cos^2\phi - \sin^2\phi = \frac{1 - t^2}{1+t^2}$. Dividing gives $\tan\theta = \frac{2t}{1-t^2}$.

The algebraic derivation (no triangle needed)

If you prefer pure algebra, start from the double-angle forms and divide top and bottom by $\cos^2\phi$. From $\sin 2\phi = 2\sin\phi\cos\phi$,

From $\cos 2\phi = \cos^2\phi - \sin^2\phi$, dividing by $\cos^2\phi + \sin^2\phi$ gives $\cos\theta = \frac{1 - \tan^2\phi}{1 + \tan^2\phi} = \frac{1 - t^2}{1 + t^2}$. Then $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{2t}{1 - t^2}$, valid when $1 - t^2 \neq 0$.

Turning an equation into a quadratic

To solve $a\sin\theta + b\cos\theta = c$ with the t-formula, substitute and clear the denominator:

Collect terms into a quadratic in $t$:

Solve for $t$ (factor or quadratic formula), then recover $\theta = 2\arctan t$ for each root. The general solution for $\theta$ may add further branches, so read the interval the question asks for.

When to reach for the t-formula

The t-formula is the right tool when:

1. The equation is $a\sin\theta + b\cos\theta = c$ and you want exact or numerical roots (it competes with the auxiliary-angle method; t-formula shines when the quadratic factorises).
2. You are integrating a rational function of $\sin\theta$ and $\cos\theta$, such as $\int\frac{d\theta}{1+\sin\theta}$, since $d\theta = \frac{2}{1+t^2}\,dt$ converts the whole integral to a rational function of $t$.
3. You are proving an identity in which every term can be rewritten in $t$ and then simplified.

The limitation you must always check

Because $t = \tan\frac{\theta}{2}$ is undefined when $\frac{\theta}{2} = \frac{\pi}{2} + n\pi$, that is $\theta = \pi + 2n\pi$, the substitution silently drops those angles. After solving in $t$, substitute $\theta = \pi$ (and $\theta = -\pi$ if relevant to the interval) into the original equation to see whether it is a solution the t-formula missed. This single check is worth a mark and is the most common t-formula trap.

How exam questions ask about the t-formula

• "Using $t = \tan\frac{\theta}{2}$, show that $\sin\theta = \ldots$" (or derive $\cos\theta$, $\tan\theta$): reproduce the half-angle triangle or the algebraic division.
• "Use the t-formula to solve $a\sin\theta + b\cos\theta = c$ in [interval]": substitute, clear to a quadratic in $t$, solve, convert with $\theta = 2\arctan t$, and check $\theta = \pi$.
• "Prove the identity ... using $t = \tan\frac{\theta}{2}$": substitute every term, put over the common denominator $1 + t^2$, and simplify.
• "Using $t = \tan\frac{\theta}{2}$, evaluate $\int (\text{rational in }\sin,\cos)\,d\theta$": substitute including $d\theta = \frac{2}{1+t^2}\,dt$, integrate the rational function, back-substitute.
• "Express $\sin\theta + \cos\theta$ in terms of $t$": add the rational forms over $1 + t^2$.

Edge cases worth knowing

• A repeated root gives one angle from the quadratic. If the quadratic in $t$ is a perfect square, like $(3t-1)^2 = 0$, there is a single $t$, hence one base angle from $2\arctan t$ (plus any from the $\theta = \pi$ check).
• $\cos\theta$ changes sign with $|t|$. $\cos\theta = \frac{1-t^2}{1+t^2}$ is positive for $|t| < 1$ and negative for $|t| > 1$; a quick number line settles the sign if you are unsure.
• The integral version needs $d\theta$ converted too. Forgetting $d\theta = \frac{2}{1+t^2}\,dt$ is the usual integral slip; it is what makes the integrand rational in $t$.