Solving Equations with t-Formulae
The Weierstrass substitution $t = \tan\dfrac{\theta}{2}$ converts any trigonometric equation into a polynomial in $t$. Solve the polynomial, back-substitute, and handle the excluded angle $\theta = \pm\pi$ with care. One algebraic technique unlocks an entire family of trig equations.
From Lesson 6 you know the t-formulae. Before using them to solve an equation, predict: how many solutions does $3\sin\theta - \cos\theta = 1$ have in $[0°, 360°)$? Write your reasoning below.
Let $t = \tan\dfrac{\theta}{2}$. Then (from Lesson 6):
The three key identities are:
$\sin\theta = \dfrac{2t}{1+t^2}, \quad \cos\theta = \dfrac{1-t^2}{1+t^2}, \quad \tan\theta = \dfrac{2t}{1-t^2}$
These hold for all $\theta \neq \pm\pi$ (i.e. $t = \tan\frac{\theta}{2}$ is defined). Substitute into the equation, multiply through by $(1+t^2)$, and a rational trig equation becomes an ordinary polynomial.
Key facts
- $t = \tan\frac{\theta}{2}$ gives $\sin\theta = \frac{2t}{1+t^2}$, $\cos\theta = \frac{1-t^2}{1+t^2}$, $\tan\theta = \frac{2t}{1-t^2}$
- The substitution is undefined at $\theta = \pm\pi$ and must be checked separately
- Multiplying by $(1+t^2)$ converts a rational trig equation into a polynomial
Concepts
- Why back-substitution uses $\theta = 2\arctan(t) + 2k\pi$
- Why $\theta = \pi$ must be checked in the original equation, not via $t$
- When the t-method is more efficient than the auxiliary angle method
Skills
- Apply the t-substitution to solve linear and quadratic equations in $\sin\theta$ and $\cos\theta$
- Find all solutions within a specified domain
- Identify and check the excluded angle $\theta = \pi$
To solve any equation of the form $a\sin\theta + b\cos\theta = c$ using the t-substitution:
- Check $\theta = \pi$: Substitute $\theta = 180°$ into the original equation first. Record whether it is a solution.
- Substitute: Replace $\sin\theta$ and $\cos\theta$ with the t-formulae.
- Clear denominator: Multiply every term by $(1+t^2)$ to obtain a polynomial equation in $t$.
- Solve and back-substitute: Solve the polynomial. For each root $t$, find $\theta = 2\arctan(t) + 360°k$ within the domain.
This rearranges to a quadratic (or linear) equation in $t$. Factoring or the quadratic formula gives the values of $t$, and from each $t$ you recover $\theta$.
Let t = {2}: then = 2t{1+t^2}, = 1-t^2{1+t^2}, = 2t{1-t^2}; Step 0: always check = 180° (or ) in the original equation
Pause — copy the four-step $t$-method into your book: (0) check $\theta=\pi$ in original; (1) substitute $t = \tan\frac{\theta}{2}$; (2) clear denominator; (3) solve polynomial; (4) back-substitute $\theta = 2\arctan t$.
Quick check: After substituting $t = \tan\frac{\theta}{2}$ and multiplying by $(1+t^2)$, the equation $\sin\theta + \cos\theta = 1$ becomes:
We just saw the four-step method — always checking $\theta = \pi$ first since the substitution fails at $\theta = \pi$ — then substituting $t = \tan\frac{\theta}{2}$ to reduce to a polynomial equation. That raises a question: for a linear equation like $\sin\theta + \cos\theta = 1$, what polynomial does the $t$-substitution produce, and how many solutions should you expect? This card answers it → substituting gives $2t(1-t) = 0$, yielding $t = 0$ and $t = 1$, so $\theta = 0°$ and $\theta = 90°$.
Sometimes after clearing the denominator, the $t^2$ terms cancel and you get a linear equation. This happens when $a = c$ in $a\sin\theta + b\cos\theta = c$, or by coincidence. The process is faster — solve for $t$ directly.
Example: $\sin\theta + \cos\theta = 1$. Check $\theta = 180°$: $\sin 180° + \cos 180° = 0 - 1 = -1 \neq 1$. Not a solution.
Substitute: $\dfrac{2t}{1+t^2} + \dfrac{1-t^2}{1+t^2} = 1$. Multiply by $(1+t^2)$:
So $t = 0$ or $t = 1$. Back-substitute:
- $t = 0 \Rightarrow \tan\dfrac{\theta}{2} = 0 \Rightarrow \dfrac{\theta}{2} = 0°, 180° \Rightarrow \theta = 0°, 360°$
- $t = 1 \Rightarrow \tan\dfrac{\theta}{2} = 1 \Rightarrow \dfrac{\theta}{2} = 45° \Rightarrow \theta = 90°$
In $[0°, 360°]$: $\theta = 0°, 90°, 360°$ — i.e. three solutions (or two distinct points in $[0°, 360°)$: $0°$ and $90°$).
+ = 1: check = 180° first (fails), then sub → 2t(1-t)=0; t = 0 = 0°, 360°; t = 1 = 90°
Pause — copy the linear-case example into your book: $\sin\theta + \cos\theta = 1$ after $t$-substitution gives $2t(1-t) = 0$, yielding $t = 0$ ($\theta = 0°, 360°$) and $t = 1$ ($\theta = 90°$).
Did you get this? True or false: when using the t-substitution, $\theta = 180°$ must be tested in the original equation because the substitution is undefined there.
Worked examples · 3 in a row, reveal as you go
Solve $3\sin\theta - \cos\theta = 1$ for $0° \leq \theta \leq 360°$.
Solve $5\sin\theta + 12\cos\theta = 6$ for $0 \leq \theta \leq 2\pi$.
Solve $\tan\theta + 2\sin\theta = 0$ for $-180° \leq \theta \leq 180°$, using the t-substitution where appropriate.
Fill the gap: After substituting $t = \tan\frac{\theta}{2}$ into $2\sin\theta + \cos\theta = 1$ and multiplying by $(1+t^2)$, the equation becomes $4t + 1 - t^2 = 1 + t^2$, which simplifies to $4t - $ $= 0$.
Misconceptions · the 3 traps that cost marks
Did you get this? True or false: if $t = \tan\frac{\theta}{2}$ gives $t = -1$, then $\theta = -90°$ is always the only solution (regardless of domain).
Activities · practice with the ideas
Check whether $\theta = 180°$ is a solution of $5\sin\theta - 3\cos\theta = 3$.
Use the t-substitution to convert $2\sin\theta - \cos\theta = 0$ into a polynomial in $t$. Do not solve yet.
Solve $2\sin\theta - \cos\theta = 0$ for $0° \leq \theta \leq 360°$. (Continue from Activity 2.)
Show that if $t = \tan\frac{\theta}{2}$ and $\cos\theta = \frac{1}{2}$, then $t = \pm\frac{1}{\sqrt{3}}$ and hence $\theta = \pm 60°$.
Solve $\sin\theta + \sqrt{3}\cos\theta = 1$ for $0 \leq \theta \leq 2\pi$ using the t-substitution. State all solutions.
Odd one out: Three of these are correct steps when applying the t-substitution. Which one is NOT a correct step?
You estimated the number of solutions to $3\sin\theta - \cos\theta = 1$ in $[0°, 360°]$.
The exact solutions are $\theta = 180°$ and $\theta = 2\arctan\!\left(\frac{1}{3}\right) \approx 36.87°$ — two solutions. The key insight: one solution came from the $\theta = 180°$ check (not from the algebra) and one from the linear equation $t = \frac{1}{3}$. Missing the first one is the most common exam error.
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. Use the t-substitution to show that $\sin\theta = 1$ has only one solution in $[0°, 360°)$. State that solution. (2 marks)
Q2. Solve $5\sin\theta - 3\cos\theta = 3$ for $0° \leq \theta \leq 360°$. (3 marks)
Q3. Solve $\sin\theta + \sqrt{3}\cos\theta = 1$ for $0 \leq \theta \leq 2\pi$, giving exact answers where possible. (3 marks)
Comprehensive answers (click to reveal)
Activity answers:
1. $5(0) - 3(-1) = 3$ ✓ → $\theta = 180°$ is a solution.
2. $\frac{4t}{1+t^2} - \frac{1-t^2}{1+t^2} = 0$ → multiply: $4t - 1 + t^2 = 0$ → $t^2 + 4t - 1 = 0$.
3. $t = \frac{-4 \pm \sqrt{20}}{2} = -2 \pm \sqrt{5}$. $t_1 \approx 0.236 \Rightarrow \theta \approx 26.57°$; $t_2 \approx -4.236 \Rightarrow \theta \approx 2(-76.7°) = -153.4° + 360° = 206.6°$. Solutions: $\approx 26.6°$ and $\approx 206.6°$.
4. $\frac{1-t^2}{1+t^2} = \frac{1}{2}$ → $2(1-t^2) = 1+t^2$ → $3t^2 = 1$ → $t = \pm\frac{1}{\sqrt{3}}$ → $\theta = \pm 60°$.
5. Check $\theta = \pi$: $0 + \sqrt{3}(-1) = -\sqrt{3} \neq 1$. Sub: $2t + \sqrt{3}(1-t^2) = 1+t^2$ → $(\sqrt{3}+1)t^2 - 2t + (1-\sqrt{3}) = 0$. Solutions: $\theta = \frac{\pi}{6}$ and $\theta = \frac{3\pi}{2}$ (approximately).
Q1 (2 marks): Check $\theta = 180°$: $\sin 180° = 0 \neq 1$ [1]. Sub: $\frac{2t}{1+t^2} = 1 \Rightarrow 2t = 1+t^2 \Rightarrow t^2 - 2t + 1 = 0 \Rightarrow (t-1)^2 = 0 \Rightarrow t = 1 \Rightarrow \theta = 90°$ [1].
Q2 (3 marks): $\theta = 180°$ ✓ [1]. Sub: $10t - 3(1-t^2) = 3(1+t^2)$ → $10t - 3 + 3t^2 = 3 + 3t^2$ → $10t = 6$ → $t = 0.6$ [1]. $\theta = 2\arctan(0.6) \approx 61.93°$ [1]. Solutions: $\theta \approx 61.93°$ and $180°$.
Q3 (3 marks): $\theta = \pi$ check fails [1]. Sub + clear: $2t + \sqrt{3}(1-t^2) = 1+t^2$ → $(1+\sqrt{3})t^2 - 2t + (1-\sqrt{3}) = 0$. By inspection $t = 1$ is a root (check: $(1+\sqrt{3}) - 2 + (1-\sqrt{3}) = 0$ ✓) and $t = \frac{1-\sqrt{3}}{1+\sqrt{3}} = -\frac{\sqrt{3}-1}{\sqrt{3}+1}$. So $\theta = \frac{\pi}{6}$ and $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$ — or using $t$ values: $\theta = 2\arctan(1) = \frac{\pi}{2}$ … [2]. Award marks for correct method and at least two valid solutions.
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering t-substitution questions. Lighter alternative to the boss.
Mark lesson as complete
Tick when you've finished the practice and review.