The HSC trig question isn't testing whether you remember one formula, it's testing whether you can choose the right technique under exam pressure. This lesson puts you through a set of carefully chosen HSC-style problems covering auxiliary angle, t-formulae, inverse trig, and combined methods. You'll build the decision-making habit that turns a "I sort of know it" into a reliable mark.
You encounter $3\sin x + 4\cos x = 2$ on an HSC exam. Without reaching for a formula sheet which technique would you choose first, and what would your first line of working look like?
Every Module 7 exam question fits into one of four buckets. Identify the bucket first, the algebra follows naturally.
Bucket 1, Auxiliary angle: expression of the form $a\sin x + b\cos x$ (or $= c$). Convert to $R\sin(x+\alpha)$ or $R\cos(x-\alpha)$.
Bucket 2, t-formulae: rational expression in $\sin x$ and $\cos x$, or equation needing substitution $t = \tan\tfrac{x}{2}$.
Bucket 3, Inverse trig: evaluate $\sin^{-1}$, $\cos^{-1}$, $\tan^{-1}$, prove identities, or solve $\sin^{-1}(f(x)) = k$.
Bucket 4, Combined / harder: factorise, substitute, or chain two methods (e.g. auxiliary then quadratic).
The most common Module 7 exam question type is: solve $a\sin x + b\cos x = c$ for $x \in [0, 2\pi]$. Three-step method:
Example: Solve $3\sin x + 4\cos x = 2$ for $x \in [0, 2\pi]$.
$R = \sqrt{9+16} = 5$; $\tan\alpha = \tfrac{4}{3}$ so $\alpha = \tan^{-1}\!\tfrac{4}{3} \approx 0.9273$ rad.
$5\sin(x+\alpha) = 2 \Rightarrow \sin(x+\alpha) = 0.4$
$x + \alpha = \sin^{-1}(0.4) \approx 0.4115$ or $\pi - 0.4115 \approx 2.7301$
$x \approx 0.4115 - 0.9273 \approx -0.5158$ (add $2\pi$) $\approx 5.767$ rad, or $x \approx 2.7301 - 0.9273 \approx 1.803$ rad.
Both values lie in $[0, 2\pi]$: $x \approx 1.80$ and $x \approx 5.77$.
a x + b x = R(x+) with R = a^2+b^2, = b/a; Alternatively: R(x-) with = a/b (same R, useful when leads)
Pause, copy both auxiliary angle forms into your book: $R\sin(x+\phi)$ with $\tan\phi = b/a$ and $R\cos(x-\alpha)$ with $\tan\alpha = a/b$, use whichever matches the question's function.
Quick check: For $\sqrt{3}\sin x + \cos x$, what is the value of $R$?
We just saw both auxiliary angle forms: $R\sin(x+\phi)$ for equations where $\sin$ leads, and $R\cos(x-\alpha)$ for $\cos$-leading equations. That raises a question: when the equation contains $\sin x$ and $\cos x$ but can't easily be grouped into $a\sin x + b\cos x$ (e.g. after expanding a squared trig expression), which method avoids the grouping step entirely? This card answers it → the $t$-substitution replaces all trig functions at once, but remember to check $x = \pi$ separately beforehand.
Use $t = \tan\tfrac{x}{2}$ when the equation involves both $\sin x$ and $\cos x$ in a way that doesn't suit the auxiliary angle (e.g. rational expressions, or when the equation is quadratic after substitution).
Key steps:
Example: Solve $2\cos x - \sin x + 1 = 0$, $x \in [0, 2\pi]$.
Sub $t$: $2\cdot\dfrac{1-t^2}{1+t^2} - \dfrac{2t}{1+t^2} + 1 = 0$. Multiply by $(1+t^2)$:
$2(1-t^2) - 2t + (1+t^2) = 0 \Rightarrow -t^2 - 2t + 3 = 0 \Rightarrow t^2 + 2t - 3 = 0$
$(t+3)(t-1)=0 \Rightarrow t = -3$ or $t = 1$.
$t=1 \Rightarrow x = 2\tan^{-1}(1) = \tfrac\pi2$; $t=-3 \Rightarrow x = 2\tan^{-1}(-3) \approx -2.498$, add $2\pi \approx 3.785$ rad.
Check $x = \pi$: $2(-1) - 0 + 1 = -1 \neq 0$, so $x=\pi$ is not a solution. Answers: $x = \tfrac\pi2$ and $x \approx 3.79$ rad.
t-substitution: t = x2; x = 2t{1+t^2}; x = 1-t^2{1+t^2}; After solving for t, always check x= separately in the original equation
Pause, copy the $t$-substitution formulas into your book: $\sin x = \frac{2t}{1+t^2}$, $\cos x = \frac{1-t^2}{1+t^2}$; always check $x = \pi$ in the original equation before applying the substitution.
Did you get this? True or false: when using the t-substitution $t = \tan\tfrac{x}{2}$, the value $x = \pi$ must always be checked as a possible solution of the original equation.
Worked examples · 3 exam-style problems with full solutions
Solve $\sqrt{3}\sin x - \cos x = 1$ for $x \in [0, 2\pi]$. (4 marks)
Prove that $\tan^{-1}\!\dfrac{1}{2} + \tan^{-1}\!\dfrac{1}{3} = \dfrac{\pi}{4}$. (3 marks)
Solve $\sin x + 2\cos x = 2$ for $x \in [0, 2\pi]$ using the t-substitution. (4 marks)
Fill the gap: For $a\sin x + b\cos x$, the equation has solutions if and only if $|c| \leq $ where $R = \sqrt{a^2+b^2}$.
Common misconceptions · the 3 traps that cost marks
Did you get this? True or false: $\sin^{-1}\!\left(\sin\dfrac{2\pi}{3}\right) = \dfrac{2\pi}{3}$.
Activities · practice with the ideas
Write $5\sin x + 12\cos x$ in the form $R\sin(x + \alpha)$. State $R$ exactly and $\alpha$ to 2 d.p.
Solve $5\sin x + 12\cos x = 6.5$ for $x \in [0, 2\pi]$, using the result from Q1.
Evaluate $\sin^{-1}\!\left(\sin\dfrac{5\pi}{6}\right)$ exactly. Justify your answer.
Solve $3 - 5\cos x = \sin x \cdot \tan\tfrac{x}{2}$ using a t-substitution for $x \in (0, 2\pi)$.
A particle's displacement is $d = 3\sin(2t) + 4\cos(2t)$ cm. Find the maximum displacement and the first time it occurs.
Odd one out: Three of these statements are correct. Which one is NOT?
Earlier you chose a technique for $3\sin x + 4\cos x = 2$. The ideal approach is the auxiliary angle method: $R = 5$, $\alpha = \tan^{-1}\!\tfrac43 \approx 0.93$ rad, leading to $\sin(x+\alpha) = 0.4$. Did your instinct match? If you reached for t-formulae that would also work, but the auxiliary angle is faster here because the equation is linear in $\sin x$ and $\cos x$ with no rational structure.
Pick your answer, then rate your confidence. That tells the system what to drill next.
Q1. Express $\sin x - \sqrt{3}\cos x$ in the form $R\sin(x - \alpha)$ where $R > 0$ and $0 < \alpha < \dfrac{\pi}{2}$. State $R$ and $\alpha$ exactly. (2 marks)
Q2. Solve $\sin x - \sqrt{3}\cos x = 1$ for $x \in [0, 2\pi]$, using your result from Q1. (3 marks)
Q3. Prove that $\tan^{-1}\!\dfrac{1}{3} + \tan^{-1}\!\dfrac{1}{2} = \dfrac{\pi}{4}$. (3 marks)
Activity answers:
1. $R = \sqrt{25+144} = 13$; $\alpha = \tan^{-1}\!\tfrac{12}{5} \approx 1.18$ rad; $5\sin x + 12\cos x = 13\sin(x + \alpha)$.
2. $\sin(x + 1.18) = 0.5 \Rightarrow x+1.18 = 0.524$ (outside domain) or $\pi - 0.524 = 2.618$; $x \approx 1.44$ rad or $x+1.18 = 0.524 + 2\pi \approx 6.807 \Rightarrow x \approx 5.63$ rad.
3. $\tfrac{5\pi}{6} \in [\tfrac\pi2, \pi]$ so $\sin^{-1}(\sin\tfrac{5\pi}{6}) = \pi - \tfrac{5\pi}{6} = \dfrac{\pi}{6}$.
4. Let $t = \tan\tfrac x2$: $3 - 5\cdot\tfrac{1-t^2}{1+t^2} = \tfrac{2t}{1+t^2}\cdot t$; multiply by $(1+t^2)$: $3+3t^2 - 5+5t^2 = 2t^2 \Rightarrow 6t^2 - 2 = 0 \Rightarrow t = \pm\tfrac{1}{\sqrt3}$; $x = 2\tan^{-1}(\pm\tfrac{1}{\sqrt3}) = \pm\tfrac\pi3$. In $(0,2\pi)$: $x = \tfrac\pi3$ and $x = \tfrac{5\pi}{3}$.
5. $R = \sqrt{9+16} = 5$ cm; max displacement $= 5$ cm. Occurs when $2t + \tan^{-1}\!\tfrac43 = \tfrac\pi2 \Rightarrow t = \tfrac{1}{2}(\tfrac\pi2 - \tan^{-1}\!\tfrac43) \approx \tfrac{1}{2}(1.571 - 0.927) \approx 0.32$ s.
Q1 (2 marks): $R = \sqrt{1+3} = 2$ [1]; $\tan\alpha = \sqrt{3}$ so $\alpha = \dfrac{\pi}{3}$ [1]. Answer: $2\sin(x - \tfrac\pi3)$.
Q2 (3 marks): $2\sin(x-\tfrac\pi3) = 1 \Rightarrow \sin(x-\tfrac\pi3) = \tfrac12$ [1]. $x - \tfrac\pi3 = \tfrac\pi6$ or $\tfrac{5\pi}{6}$ [1]. $x = \tfrac\pi2$ or $x = \tfrac{7\pi}{6}$ [1].
Q3 (3 marks): $\tan(\alpha+\beta) = \dfrac{\tfrac13+\tfrac12}{1-\tfrac16} = \dfrac{\tfrac56}{\tfrac56} = 1$ [2]. Since $\alpha, \beta > 0$ and $\alpha+\beta \in (0,\pi)$, conclude $\alpha + \beta = \dfrac\pi4$ [1]. $\blacksquare$
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