The substitution $t = \tan\frac{\theta}{2}$ is one of the most powerful tricks in trigonometry. It converts every trig function into a rational expression in $t$, turning transcendental equations into polynomial ones you can solve with algebra. Known in higher mathematics as the Weierstrass substitution, it appears in calculus, integration, and equation-solving at every level.
Let $t = \tan\frac{\theta}{2}$. What is $\tan\theta$ in terms of $t$? Recall the double angle formula: $\tan(2A) = \frac{2\tan A}{1-\tan^2 A}$. Set $A = \frac{\theta}{2}$ and substitute $\tan\frac{\theta}{2} = t$. Write your prediction.
One substitution, $t = \tan\frac{\theta}{2}$, converts all three trig functions into rational expressions in $t$. This transforms trig equations into polynomial equations, which you can solve with algebra. It is called the Weierstrass substitution.
The derivation for $\tan\theta$ uses the double angle formula directly. The derivations for $\sin\theta$ and $\cos\theta$ use a right triangle with opposite $t$, adjacent $1$, hypotenuse $\sqrt{1+t^2}$. Once you have $\sin\frac{\theta}{2} = \frac{t}{\sqrt{1+t^2}}$ and $\cos\frac{\theta}{2} = \frac{1}{\sqrt{1+t^2}}$, multiply using double angle for sine and cosine.
Let $t = \tan\frac{\theta}{2}$. Consider a right triangle with opposite $t$, adjacent $1$, so hypotenuse $\sqrt{1+t^2}$. Reading off:
Now apply double angle formulas:
Verification: $\sin^2\theta + \cos^2\theta = \left(\frac{2t}{1+t^2}\right)^2 + \left(\frac{1-t^2}{1+t^2}\right)^2 = \frac{4t^2 + 1 - 2t^2 + t^4}{(1+t^2)^2} = \frac{(1+t^2)^2}{(1+t^2)^2} = 1$ ✓
t = {2}: = 2t{1+t^2}, = 1-t^2{1+t^2}, = 2t{1-t^2}; Derivation: right triangle with opposite t, adjacent 1; then use double angle formulas
Pause, copy the three $t$-formulas into your book: $\sin\theta = \frac{2t}{1+t^2}$, $\cos\theta = \frac{1-t^2}{1+t^2}$, $\tan\theta = \frac{2t}{1-t^2}$, where $t = \tan\frac{\theta}{2}$.
Quick check: Using $t = \tan\frac{\theta}{2} = 1$, what is $\sin\theta$?
We just saw that $t = \tan\frac{\theta}{2}$ gives $\sin\theta = \frac{2t}{1+t^2}$, $\cos\theta = \frac{1-t^2}{1+t^2}$, $\tan\theta = \frac{2t}{1-t^2}$. That raises a question: at $\theta = \pi$, $\tan\frac{\theta}{2}$ is undefined, does this mean solutions at $\theta = \pi$ are simply missed? This card answers it → yes, so we must check $\theta = \pi$ directly in the original equation before applying the substitution, then solve the resulting polynomial for $t$.
When does the substitution fail? The substitution $t = \tan\frac{\theta}{2}$ is undefined when $\frac{\theta}{2} = \frac{\pi}{2} + n\pi$, i.e. when $\theta = \pi + 2n\pi$. At these values, $\tan\frac{\theta}{2}$ does not exist. Always check whether $\theta = \pi$ is in your domain before applying the t-formulas.
Strategy for simplifying trig expressions:
The t-substitution converts a transcendental equation into a polynomial, which algebra can solve.
t-substitution is undefined at = + 2n, always check the domain; Strategy: substitute, simplify algebraically, solve the polynomial, then = 2(t)
Pause, copy the domain-check protocol into your book: always test $\theta = \pi$ in the original equation first; if satisfied, record that solution; then apply the $t$-substitution and back-substitute via $\theta = 2\arctan t$.
Did you get this? True or false: the t-substitution is valid for all real values of $\theta$.
Worked examples · 3 in a row, reveal as you go
If $t = \tan\frac{\theta}{2} = \dfrac{1}{2}$, find $\sin\theta$ and $\cos\theta$.
Express $3\sin\theta + 4\cos\theta$ in terms of $t = \tan\dfrac{\theta}{2}$. Simplify your answer.
Show that $\dfrac{1-\cos\theta}{\sin\theta} = \tan\dfrac{\theta}{2}$.
Fill the gap: If $t = \tan\frac{\theta}{2} = 2$, then $\sin\theta = \frac{2 \times 2}{1 + 4} = $ .
Misconceptions to fix · the traps that cost marks
Did you get this? True or false: $\cos\theta = \dfrac{1+t^2}{1-t^2}$ is the correct t-formula for cosine.
Activities · practice with the ideas
If $t = \tan\frac{\theta}{2} = 3$, find $\sin\theta$, $\cos\theta$ and $\tan\theta$. Verify your answers satisfy $\sin^2\theta + \cos^2\theta = 1$.
Express $5\cos\theta - 12\sin\theta$ entirely in terms of $t = \tan\frac{\theta}{2}$. Write your answer as a single fraction.
Show that $\dfrac{\sin\theta}{1+\cos\theta} = t$, i.e. $= \tan\frac{\theta}{2}$.
Why does the t-substitution fail at $\theta = \pi$? What is the value of $\sin\pi$ and $\cos\pi$ directly? Does the failure of the substitution cause any actual error in these values?
Use the t-substitution to solve $\sin\theta = \cos\theta$ for $\theta \in [0, 2\pi)$, $\theta \neq \pi$. First convert to a polynomial equation in $t$.
Earlier you were asked to find $\tan\theta$ in terms of $t = \tan\frac{\theta}{2}$ using the double angle formula. The answer is $\tan\theta = \frac{2t}{1-t^2}$.
Why is the t-substitution so powerful? Because it eliminates the transcendental nature of trig equations. Every combination $a\sin\theta + b\cos\theta + c = 0$ becomes a quadratic or higher-degree polynomial in $t$, and polynomials obey familiar algebraic rules. This is also why the substitution appears in integral calculus: $\int \frac{1}{a + b\cos\theta}\,d\theta$ becomes a standard rational integral in $t$. The $d\theta$ transforms too: since $t = \tan\frac{\theta}{2}$, we get $dt = \frac{1}{2}\sec^2\frac{\theta}{2}\,d\theta$, so $d\theta = \frac{2}{1+t^2}\,dt$.
Odd one out: Three of these are correct when $t = \tan\frac{\theta}{2}$. Which one is WRONG?
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. If $t = \tan\dfrac{\theta}{2} = 2$, find $\sin\theta$. (2 marks)
Q2. Express $3\sin\theta + 4\cos\theta$ in terms of $t = \tan\dfrac{\theta}{2}$. (2 marks)
Q3. Show that $\dfrac{1-\cos\theta}{\sin\theta} = \tan\dfrac{\theta}{2}$. (2 marks)
Activity answers:
1. $t = 3$: $\sin\theta = \frac{6}{10} = \frac{3}{5}$, $\cos\theta = \frac{1-9}{1+9} = \frac{-8}{10} = -\frac{4}{5}$, $\tan\theta = \frac{6}{1-9} = \frac{6}{-8} = -\frac{3}{4}$. Check: $\frac{9}{25} + \frac{16}{25} = 1$ ✓
2. $5\cos\theta - 12\sin\theta = \frac{5(1-t^2)}{1+t^2} - \frac{24t}{1+t^2} = \frac{5 - 5t^2 - 24t}{1+t^2} = \frac{-5t^2-24t+5}{1+t^2} = \frac{-(5t-1)(t+5)}{1+t^2}$
3. $\frac{\sin\theta}{1+\cos\theta} = \frac{2t/(1+t^2)}{1 + (1-t^2)/(1+t^2)} = \frac{2t/(1+t^2)}{(2)/(1+t^2)} = \frac{2t}{2} = t$ ✓
4. At $\theta = \pi$: $\tan(\pi/2)$ is undefined (vertical asymptote), so $t = \infty$. Directly: $\sin\pi = 0$, $\cos\pi = -1$. The t-formulas would give $\frac{2\cdot\infty}{1+\infty^2} \to 0$ (limit exists) and $\frac{1-\infty^2}{1+\infty^2} \to -1$ (limit exists), so the values are recoverable as limits, but the formulas are not directly applicable.
5. $\sin\theta = \cos\theta$: $\frac{2t}{1+t^2} = \frac{1-t^2}{1+t^2}$, so $2t = 1-t^2$, giving $t^2+2t-1=0$, $t = \frac{-2\pm\sqrt{8}}{2} = -1 \pm \sqrt{2}$. So $\tan\frac{\theta}{2} = \sqrt{2}-1$ or $-1-\sqrt{2}$, giving $\theta = \frac{\pi}{4}$ or $\frac{5\pi}{4}$.
Q1 (2 marks): $\sin\theta = \frac{2(2)}{1+4} = \frac{4}{5}$ [2].
Q2 (2 marks): $= \frac{6t+4-4t^2}{1+t^2} = \frac{-4t^2+6t+4}{1+t^2}$ [1]. Factor: $= \frac{-2(2t^2-3t-2)}{1+t^2} = \frac{-2(2t+1)(t-2)}{1+t^2}$ [1].
Q3 (2 marks): $\frac{1 - \frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}} = \frac{\frac{2t^2}{1+t^2}}{\frac{2t}{1+t^2}}$ [1] $= \frac{2t^2}{2t} = t = \tan\frac{\theta}{2}$ [1]. ■
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