Exam-Style Trigonometry

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:

1. Find $R$ and $\alpha$: $R = \sqrt{a^2+b^2}$; use $\tan\alpha = b/a$ with $\alpha \in (0, \tfrac\pi2)$ when $a,b>0$.
2. Rewrite the equation as $\sin(x+\alpha) = c/R$. Check $|c/R| \leq 1$.
3. Solve $x + \alpha = \theta$ for all $\theta$ in the extended domain, then subtract $\alpha$ and select values in the required domain.

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.

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:

1. Substitute — clear denominators, obtain a polynomial in $t$.
2. Solve the polynomial for $t$.
3. Convert back: $x = 2\tan^{-1}(t)$, and check whether $x = \pi$ (i.e. $\cos x = -1$) is a solution of the original that was lost.
4. List all solutions in the given domain.

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.