9709 P3 §3.3 — Trigonometry II: W3–W4 Comprehensive Review

💡 Lead In: Identity Lightning Round

The Scenario

Before the module review, let's warm up by recalling the identities you now own. The ones that come to mind instantly are the ones you will use fluently in a test.

Review · Activating Prior Knowledge

Quick check — pick the correct statement in each question before the review begins.

Recall 1 — Which compound-angle expansion is correct?

$\sin(A+B)=\sin A+\sin B$

$\cos(A-B)=\cos A\cos B-\sin A\sin B$

$\sin(A+B)=\sin A\cos B+\cos A\sin B$

$\tan(A+B)=\dfrac{\tan A-\tan B}{1+\tan A\tan B}$

$\sin(A+B)=\sin A\cos B+\cos A\sin B$; similarly $\cos(A+B)=\cos A\cos B-\sin A\sin B$ and $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.

Recall 2 — Which double-angle identity is correct?

$\cos 2A=1+\sin^2 A$

$\cos 2A=\cos^2 A-\sin^2 A=2\cos^2 A-1=1-2\sin^2 A$

$\sin 2A=\sin A\cos A$

$\tan 2A=\dfrac{2\tan A}{1+\tan^2 A}$

$\cos 2A$ has three equal forms: $\cos^2 A-\sin^2 A=2\cos^2 A-1=1-2\sin^2 A$. Also $\sin 2A=2\sin A\cos A$ and $\tan 2A=\dfrac{2\tan A}{1-\tan^2 A}$.

Recall 3 — Which statement about the R-form / Pythagorean identities is correct?

$a\cos\theta+b\sin\theta=R\cos(\theta-\alpha)$ with $R=\sqrt{a^2+b^2}$; and $\sec^2\theta=1+\tan^2\theta$

$R=\dfrac{a}{b}$; and $\sec^2\theta=1-\tan^2\theta$

$R=a^2+b^2$; and $\cosec^2\theta=1-\cot^2\theta$

$R=a+b$; and $\sin^2\theta+\cos^2\theta=\sin\theta$

$a\cos\theta+b\sin\theta=R\cos(\theta-\alpha)$ with $R=\sqrt{a^2+b^2}$, and the Pythagorean identities $\sin^2\theta+\cos^2\theta=1$, $\sec^2\theta=1+\tan^2\theta$, $\cosec^2\theta=1+\cot^2\theta$ — the full toolkit used in today's review.

The Question

Which of these identities do you recall instantly, and which still feel slow? Aim to recall all of them within a minute — fluid recall is the clearest sign you are ready for the review.

💡 The Bridge (click to reveal)

This warm-up activates the full identity toolkit we deploy in the review problems: compound angles, double angles, the R-form, and the Pythagorean identities. The review will ask you to select the right one for each question.

📋 Review Points — W3–W4 at a Glance

1 Reciprocal functions: $\sec$, $\cosec$, $\cot$

Definitions: $\sec\theta = \dfrac{1}{\cos\theta}$, $\cosec\theta = \dfrac{1}{\sin\theta}$, $\cot\theta = \dfrac{\cos\theta}{\sin\theta}$. Each is undefined where its denominator is zero. Graphs are reciprocals of $\cos$, $\sin$, $\tan$ — vertical asymptotes replace zeros, and vice versa.

Domain traps: $\sec\theta$ undefined at $\theta = 90^{\circ}, 270^{\circ}, \ldots$ · $\cosec\theta$ undefined at $\theta = 0^{\circ}, 180^{\circ}, \ldots$ · $\cot\theta$ undefined at $\theta = 0^{\circ}, 180^{\circ}, \ldots$

2 Pythagorean identities

Three forms to commit to memory:

Identities: $$\sin^2\theta + \cos^2\theta \equiv 1 \qquad \sec^2\theta \equiv 1 + \tan^2\theta \qquad \cosec^2\theta \equiv 1 + \cot^2\theta$$

Used to convert between $\tan$ and $\sec$ (or $\cot$ and $\cosec$) in identity proofs and equation-solving.

3 Compound angle formulae

Formulae: $$\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B$$ $$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$$ $$\tan(A \pm B) = \dfrac{\tan A \pm \tan B}{1 \mp \tan A \tan B}$$

Watch the sign swaps in $\cos$ and $\tan$: when the original is $+$, the cross-term is $-$.

4 Double angle formulae

Three forms of $\cos 2A$: $$\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$$ Plus: $$\sin 2A = 2\sin A \cos A \qquad \tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$$

Choosing the right $\cos 2A$ form is the key to most identity proofs: pick the form that matches what's already there.

5 R-form: $a\cos\theta + b\sin\theta = R\cos(\theta - \alpha)$

Expand $R\cos(\theta - \alpha)$, match coefficients, eliminate using $\sin^2\alpha + \cos^2\alpha = 1$:

Form: $$R = \sqrt{a^2 + b^2}, \qquad \tan\alpha = \dfrac{b}{a}$$ Signs: $R\cos\alpha = a$, $R\sin\alpha = b$ — use the signs of $a$ and $b$ to fix the quadrant of $\alpha$.

Used to: solve $a\cos\theta + b\sin\theta = c$, find maximum/minimum ($\pm R$), and sketch the combined wave.

6 Identity proof strategies

Strategy checklist:

Start from the more complex side. Simplify toward the simpler side.
Convert everything to $\sin$ and $\cos$ if $\sec$, $\cosec$, $\cot$ appear — it usually unlocks the path.
Choose the $\cos 2A$ form that matches the terms already present. If $\sin^2\theta$ appears, use $\cos 2\theta = 1 - 2\sin^2\theta$.
Common substitution: $1 + \cos 2\theta = 2\cos^2\theta$ and $1 - \cos 2\theta = 2\sin^2\theta$ — these convert "1 + cos2θ" into a single squared term.
Work one side at a time. Do not cross-multiply or jump sides mid-proof.

7 Equation-solving strategies

Choose the method by the structure of the equation:

Convert to a single function first (e.g. $\cosec^2 x + \cot^2 x = 2 \Rightarrow 1 + \cot^2 x + \cot^2 x = 2 \Rightarrow \cot^2 x = \tfrac{1}{2}$).
Use $\cos 2\theta$ to reduce a $\theta/2\theta$ mix to a single variable (e.g. $\cos 2\theta + \sin\theta = 0 \Rightarrow 1 - 2\sin^2\theta + \sin\theta = 0$).
Factorise wherever possible — do not divide by a factor that might be zero.
R-form when the equation is $a\cos\theta + b\sin\theta = c$.
Always state the range and list every solution in range (don't forget the second-quadrant solution to $\sin\theta = k$, etc.).

✏️ Worked Examples

Group · Guided Worked Example 1: Identity Proof + Equation Solving Exercise 3C Q12

(i) Prove the identity $\cosec 2\theta + \cot 2\theta \equiv \cot\theta$.

(ii) Hence solve the equation $\cosec 2\theta + \cot 2\theta = 2$, for $0^{\circ} \leq \theta \leq 360^{\circ}$.

Cambridge International AS & A Level Mathematics 9709 Paper 3 Q3 June 2009

👁 Reveal Solution

Part (i) — Proof of $\cosec 2\theta + \cot 2\theta \equiv \cot\theta$

Start from the LHS (more complex side):

$$\cosec 2\theta + \cot 2\theta = \dfrac{1}{\sin 2\theta} + \dfrac{\cos 2\theta}{\sin 2\theta} = \dfrac{1 + \cos 2\theta}{\sin 2\theta}$$

Use the double-angle identities: $1 + \cos 2\theta = 2\cos^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$:

$$= \dfrac{2\cos^2\theta}{2\sin\theta\cos\theta} = \dfrac{\cos\theta}{\sin\theta} = \cot\theta \quad \checkmark$$

Part (ii) — Solve $\cosec 2\theta + \cot 2\theta = 2$

By part (i), the equation becomes $\cot\theta = 2$, i.e. $\tan\theta = \dfrac{1}{2}$.

Principal value: $\theta = \arctan\!\left(\dfrac{1}{2}\right) \approx 26.565^{\circ} \approx 26.6^{\circ}$ (1 d.p.).

Tangent is positive in Q1 and Q3, so the second solution in range is $\theta = 26.565^{\circ} + 180^{\circ} = 206.565^{\circ} \approx 206.6^{\circ}$ (1 d.p.).

$$\boxed{\theta \approx 26.6^{\circ}, \quad 206.6^{\circ} \text{ (1 d.p.)}}$$

Verification

At $\theta = 26.565^{\circ}$: $\cot\theta = \dfrac{1}{\tan 26.565^{\circ}} = \dfrac{1}{0.5} = 2$ &check;

At $\theta = 206.565^{\circ}$: $\tan(206.565^{\circ}) = \tan(26.565^{\circ}) = 0.5$ (period $180^{\circ}$), so $\cot\theta = 2$ &check;

Individual · Guided Worked Example 2: R-form & Equation Solving Exercise 3D Q13

(i) Express $7\cos\theta + 24\sin\theta$ in the form $R\cos(\theta - \alpha)$, where $R > 0$ and $0^{\circ} < \alpha < 90^{\circ}$, giving the exact value of $R$ and the value of $\alpha$ correct to 2 decimal places.

(ii) Hence solve the equation $7\cos\theta + 24\sin\theta = 15$, giving all solutions in the interval $0^{\circ} \leq \theta \leq 360^{\circ}$.

Cambridge International AS & A Level Mathematics 9709 Paper 3 Q4 June 2006

👁 Reveal Solution

Part (i) — Find $R$ and $\alpha$

Expand: $R\cos(\theta - \alpha) = R\cos\theta\cos\alpha + R\sin\theta\sin\alpha$.

Matching with $7\cos\theta + 24\sin\theta$: $R\cos\alpha = 7$ and $R\sin\alpha = 24$.

$R = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ (exact).

$\tan\alpha = \dfrac{R\sin\alpha}{R\cos\alpha} = \dfrac{24}{7}$, so $\alpha = \arctan\!\left(\dfrac{24}{7}\right) \approx 73.74^{\circ}$ (2 d.p.).

$$\boxed{7\cos\theta + 24\sin\theta = 25\cos(\theta - 73.74^{\circ})}$$

Part (ii) — Solve $7\cos\theta + 24\sin\theta = 15$

Substitute: $25\cos(\theta - 73.74^{\circ}) = 15$, so $\cos(\theta - 73.74^{\circ}) = 0.6$.

Let $u = \theta - 73.74^{\circ}$. Then $\cos u = 0.6$.

Principal value: $u = \arccos(0.6) \approx 53.13^{\circ}$.

Cosine is positive in Q1 and Q4: $u = 53.13^{\circ}$ or $u = 360^{\circ} - 53.13^{\circ} = 306.87^{\circ}$.

So $\theta = 73.74^{\circ} + 53.13^{\circ} \approx 126.87^{\circ}$ or $\theta = 73.74^{\circ} + 306.87^{\circ} = 380.61^{\circ} \equiv 20.61^{\circ}$ (subtract $360^{\circ}$).

$$\boxed{\theta \approx 20.61^{\circ}, \quad 126.87^{\circ} \text{ (2 d.p.)}}$$

Verification

At $\theta = 20.61^{\circ}$: $7\cos 20.61^{\circ} + 24\sin 20.61^{\circ} \approx 7(0.9361) + 24(0.3517) \approx 6.55 + 8.44 = 14.99 \approx 15$ &check;

At $\theta = 126.87^{\circ}$: $7\cos 126.87^{\circ} + 24\sin 126.87^{\circ} \approx 7(-0.6) + 24(0.8) = -4.2 + 19.2 = 15$ &check;

✏️ Concept Quiz — W3–W4 Objectives

Six multiple-choice questions covering every W3–W4 sub-topic. Options are shuffled each time the page loads.

Score: 0 / 6

1. Which expression is equal to $\sec 150^{\circ}$?

$-\dfrac{2}{\sqrt{3}}$ · wait, this is correct. Tap to confirm.

$-\dfrac{2}{\sqrt{3}}$

$-\dfrac{\sqrt{3}}{2}$

$2$

$\sec 150^{\circ} = \dfrac{1}{\cos 150^{\circ}}$. Since $\cos 150^{\circ} = -\dfrac{\sqrt{3}}{2}$, $\sec 150^{\circ} = -\dfrac{2}{\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}$. The trap $-\dfrac{\sqrt{3}}{2}$ is $\cos 150^{\circ}$ itself — $\sec$ is the reciprocal, not the value.

2. If $\tan\theta = 3$ (with $\theta$ in Q3), what is $\sec\theta$?

$+\sqrt{10}$

$-\sqrt{10}$

$-3$

$+10$

$\sec^2\theta = 1 + \tan^2\theta = 1 + 9 = 10$, so $\sec\theta = \pm\sqrt{10}$. In Q3, both $\cos\theta$ and $\sec\theta$ are negative, so $\sec\theta = -\sqrt{10}$. The trap $+\sqrt{10}$ forgets the quadrant check.

3. Which is the correct expansion of $\cos(A - B)$?

$-2\sin A \sin B$

$-2\sin A \sin B$ · wait, none of these

$2\cos^2 A - 1 + 2\sin A \sin B$ (correct after simplification)

$\cos(A - B) = \cos A \cos B + \sin A \sin B$. The sign of the cross term is opposite to the original: $A - B$ gives $+ \sin A \sin B$. For $\cos(A + B)$, the cross term would be $- \sin A \sin B$.

4. To simplify $\dfrac{1 - \cos 2\theta}{\sin 2\theta}$, which form of $\cos 2\theta$ is most useful?

$2\cos^2\theta - 1$

$1 - 2\sin^2\theta$

$1 + \tan^2\theta$

$-1 + 2\sin^2\theta$

Use $1 - \cos 2\theta = 1 - (1 - 2\sin^2\theta) = 2\sin^2\theta$. Then $\dfrac{1 - \cos 2\theta}{\sin 2\theta} = \dfrac{2\sin^2\theta}{2\sin\theta\cos\theta} = \tan\theta$. Choosing the form that introduces $\sin^2\theta$ lets the $\sin\theta$ in the denominator cancel cleanly.

5. For $3\cos x + 4\sin x = R\cos(x - \alpha)$ (with $0^{\circ} < \alpha < 90^{\circ}$), find $R$ and $\alpha$.

$R = 7, \alpha = 53.1^{\circ}$

$R = 5, \alpha = 53.1^{\circ}$

$R = 5, \alpha = 36.9^{\circ}$

$R = \sqrt{7}, \alpha = 53.1^{\circ}$

$R = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5$. $\tan\alpha = \dfrac{b}{a} = \dfrac{4}{3}$, so $\alpha = \arctan(4/3) \approx 53.1^{\circ}$. The trap $36.9^{\circ}$ comes from computing $\arctan(3/4)$ — that swaps $a$ and $b$. Remember $\tan\alpha = b/a$ where $a$ is the $\cos\theta$ coefficient.

6. To solve $\cos 2\theta + \sin\theta = 0$ for $0^{\circ} \leq \theta \leq 360^{\circ}$, the best first step is…

Write $\cos 2\theta = \cos^2\theta - \sin^2\theta$ and factorise.

Use $\cos 2\theta = 1 - 2\sin^2\theta$ to get a quadratic in $\sin\theta$.

Take $\cos 2\theta$ to the other side and divide by $\sin\theta$.

Write everything in terms of $\cos\theta$ using $\sin\theta = \sqrt{1 - \cos^2\theta}$.

Substituting $\cos 2\theta = 1 - 2\sin^2\theta$ gives $1 - 2\sin^2\theta + \sin\theta = 0$, i.e. $2\sin^2\theta - \sin\theta - 1 = 0$, a quadratic in $\sin\theta$. Factorising: $(2\sin\theta + 1)(\sin\theta - 1) = 0$, giving $\sin\theta = -\tfrac{1}{2}$ or $\sin\theta = 1$. Using the $\sin^2\theta$ form (rather than $\cos^2\theta - \sin^2\theta$) reduces the equation to a single function — the standard strategy for any "$\cos 2\theta + \sin\theta$" mix.

🎉

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📝 More Practice — Review Selection

Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest. Each question is a review item drawn from across W3–W4.

Q1 · Exercise 3A Q2 (review, was W3-L1) Standard

Find the following, giving your answers as fractions or in surd form. You should not need your calculator.

(i) $\cot 135^{\circ}$ (ii) $\sec 150^{\circ}$ (iii) $\cosec 240^{\circ}$ (iv) $\sec 210^{\circ}$ (v) $\cot 270^{\circ}$ (vi) $\cosec 225^{\circ}$

💡 Hint

Use the ASTC diagram to fix the sign, and the special-angle values ($30^{\circ}$, $45^{\circ}$, $60^{\circ}$) for the magnitude. Then $\cot = \dfrac{\cos}{\sin}$, $\sec = \dfrac{1}{\cos}$, $\cosec = \dfrac{1}{\sin}$.

👁 Reveal Solution

(i) $\cot 135^{\circ}$

$\cos 135^{\circ} = -\dfrac{\sqrt{2}}{2}$, $\sin 135^{\circ} = \dfrac{\sqrt{2}}{2}$. So $\cot 135^{\circ} = \dfrac{-\sqrt{2}/2}{\sqrt{2}/2} = \boxed{-1}$.

(ii) $\sec 150^{\circ}$

$\cos 150^{\circ} = -\dfrac{\sqrt{3}}{2}$. So $\sec 150^{\circ} = \dfrac{1}{\cos 150^{\circ}} = \boxed{-\dfrac{2}{\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}}$.

(iii) $\cosec 240^{\circ}$

$\sin 240^{\circ} = -\dfrac{\sqrt{3}}{2}$. So $\cosec 240^{\circ} = \dfrac{1}{\sin 240^{\circ}} = \boxed{-\dfrac{2}{\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}}$.

(iv) $\sec 210^{\circ}$

$\cos 210^{\circ} = -\dfrac{\sqrt{3}}{2}$. So $\sec 210^{\circ} = \boxed{-\dfrac{2}{\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}}$.

(v) $\cot 270^{\circ}$

$\cos 270^{\circ} = 0$, $\sin 270^{\circ} = -1$. So $\cot 270^{\circ} = \dfrac{0}{-1} = \boxed{0}$.

(vi) $\cosec 225^{\circ}$

$\sin 225^{\circ} = -\dfrac{\sqrt{2}}{2}$. So $\cosec 225^{\circ} = \dfrac{1}{-\sqrt{2}/2} = \boxed{-\sqrt{2}}$.

Q2 · Exercise 3B Q7 (review, was W3-L4) Medium

The angle $\alpha$ lies between $0^{\circ}$ and $90^{\circ}$ and is such that $2\tan^2\alpha + \sec^2\alpha = 5 - 4\tan\alpha$.

(i) Show that $3\tan^2\alpha + 4\tan\alpha - 4 = 0$ and hence find the exact value of $\tan\alpha$.

(ii) It is given that the angle $\beta$ is such that $\cot(\alpha + \beta) = 6$. Without using a calculator, find the exact value of $\cot\beta$.

💡 Hint

Part (i): replace $\sec^2\alpha$ with $1 + \tan^2\alpha$ (Pythagorean identity) and rearrange.

Part (ii): from $\cot(\alpha+\beta) = 6$ we get $\tan(\alpha+\beta) = \tfrac{1}{6}$. Use the compound-angle formula for $\tan$ to solve for $\tan\beta$, then take the reciprocal.

👁 Reveal Solution

Part (i) — Show that $3\tan^2\alpha + 4\tan\alpha - 4 = 0$

Use $\sec^2\alpha = 1 + \tan^2\alpha$:

$$2\tan^2\alpha + (1 + \tan^2\alpha) = 5 - 4\tan\alpha$$

$$3\tan^2\alpha + 1 = 5 - 4\tan\alpha$$

$$3\tan^2\alpha + 4\tan\alpha - 4 = 0 \quad \checkmark$$

Factorise: $(3\tan\alpha - 2)(\tan\alpha + 2) = 0$, so $\tan\alpha = \dfrac{2}{3}$ or $\tan\alpha = -2$.

Since $0^{\circ} < \alpha < 90^{\circ}$, $\tan\alpha > 0$, so $\boxed{\tan\alpha = \dfrac{2}{3}}$.

Part (ii) — Find $\cot\beta$ given $\cot(\alpha + \beta) = 6$

From $\cot(\alpha + \beta) = 6$, $\tan(\alpha + \beta) = \dfrac{1}{6}$.

Compound angle: $\tan(\alpha + \beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$. Let $t = \tan\beta$:

$$\dfrac{\tfrac{2}{3} + t}{1 - \tfrac{2}{3}t} = \dfrac{1}{6}$$

Cross-multiply: $6\!\left(\dfrac{2}{3} + t\right) = 1 - \dfrac{2}{3}t$, i.e. $4 + 6t = 1 - \dfrac{2}{3}t$.

Multiply by 3: $12 + 18t = 3 - 2t$, so $20t = -9$, giving $\tan\beta = -\dfrac{9}{20}$.

Therefore $\boxed{\cot\beta = -\dfrac{20}{9}}$.

Verification

$\tan\alpha = 2/3 \approx 0.667$ → $\alpha \approx 33.69^{\circ}$. With $\cot\beta = -20/9$, $\tan\beta = -9/20 = -0.45$ → $\beta \approx 180^{\circ} - 24.23^{\circ} = 155.77^{\circ}$ (Q2 since $\tan < 0$).

$\alpha + \beta \approx 33.69^{\circ} + 155.77^{\circ} = 189.46^{\circ}$, and $\cot(189.46^{\circ}) = \cot(9.46^{\circ}) \approx 6.006 \approx 6$ &check;

Q3 · Exercise 3C Q3 (review, was W3-L5) Medium

By first writing $\sin 3\theta$ as $\sin(2\theta + \theta)$, express $\sin 3\theta$ in terms of $\sin\theta$. Hence solve the equation $\sin 3\theta = \sin\theta$ for $0 \leq \theta \leq 2\pi$.

💡 Hint

Step 1: expand $\sin(2\theta + \theta)$ using the compound-angle formula. Use $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 1 - 2\sin^2\theta$ (the form with $\sin^2\theta$) so everything ends up in terms of $\sin\theta$.

Step 2: rearrange $\sin 3\theta = \sin\theta$ to one side and factorise.

👁 Reveal Solution

Derive $\sin 3\theta$ in terms of $\sin\theta$

$\sin 3\theta = \sin(2\theta + \theta) = \sin 2\theta \cos\theta + \cos 2\theta \sin\theta$.

Substitute $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 1 - 2\sin^2\theta$:

$$= 2\sin\theta\cos^2\theta + (1 - 2\sin^2\theta)\sin\theta$$

$$= 2\sin\theta\cos^2\theta + \sin\theta - 2\sin^3\theta$$

Replace $\cos^2\theta = 1 - \sin^2\theta$:

$$= 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta$$

$$= 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta$$

$$\boxed{\sin 3\theta = 3\sin\theta - 4\sin^3\theta}$$

Solve $\sin 3\theta = \sin\theta$ for $0 \leq \theta \leq 2\pi$

Using the identity: $3\sin\theta - 4\sin^3\theta = \sin\theta$, so $2\sin\theta - 4\sin^3\theta = 0$.

Factorise: $2\sin\theta(1 - 2\sin^2\theta) = 0$.

So either $\sin\theta = 0$ or $\sin^2\theta = \dfrac{1}{2}$, i.e. $\sin\theta = \pm\dfrac{1}{\sqrt{2}}$.

Case 1: $\sin\theta = 0 \Rightarrow \theta = 0, \pi, 2\pi$.

Case 2: $\sin\theta = \dfrac{1}{\sqrt{2}} \Rightarrow \theta = \dfrac{\pi}{4}, \dfrac{3\pi}{4}$.

Case 3: $\sin\theta = -\dfrac{1}{\sqrt{2}} \Rightarrow \theta = \dfrac{5\pi}{4}, \dfrac{7\pi}{4}$.

$$\boxed{\theta = 0,\ \dfrac{\pi}{4},\ \dfrac{3\pi}{4},\ \pi,\ \dfrac{5\pi}{4},\ \dfrac{7\pi}{4},\ 2\pi}$$

Verification

At $\theta = \pi/4$: $\sin 3\theta = \sin(3\pi/4) = \dfrac{\sqrt{2}}{2} = \sin(\pi/4)$ &check;

At $\theta = 5\pi/4$: $\sin 3\theta = \sin(15\pi/4) = \sin(15\pi/4 - 4\pi) = \sin(-\pi/4) = -\dfrac{\sqrt{2}}{2} = \sin(5\pi/4)$ &check;

Q4 · Exercise 3D Q11 (review, was W4-L3) Challenging

(i) Express $3\cos x + 4\sin x$ in the form $R\cos(x - \alpha)$, where $R > 0$ and $0^{\circ} < \alpha < 90^{\circ}$, stating the exact value of $R$ and giving the value of $\alpha$ correct to 2 decimal places.

(ii) Hence solve the equation $3\cos x + 4\sin x = 4.5$, giving all solutions in the interval $0^{\circ} < x < 360^{\circ}$.

Cambridge International AS & A Level Mathematics 9709 Paper 22 Q6 November 2009

💡 Hint

Part (i): standard R-form. $R = \sqrt{3^2 + 4^2}$ and $\tan\alpha = \dfrac{4}{3}$. Note that $3^2 + 4^2 = 25$ is a perfect square, so $R$ is exact.

Part (ii): divide both sides by $R$ to get $\cos(x - \alpha) = \dfrac{4.5}{R}$, then use the two-quadrant rule for cosine.

👁 Reveal Solution

Part (i) — Find $R$ and $\alpha$

Expand $R\cos(x - \alpha) = R\cos x\cos\alpha + R\sin x\sin\alpha$ and match with $3\cos x + 4\sin x$:

$R\cos\alpha = 3$, $R\sin\alpha = 4$.

$R = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = \boxed{5}$ (exact).

$\tan\alpha = \dfrac{4}{3}$, so $\alpha = \arctan\!\left(\dfrac{4}{3}\right) \approx \boxed{53.13^{\circ}}$ (2 d.p.).

$$3\cos x + 4\sin x = 5\cos(x - 53.13^{\circ})$$

Part (ii) — Solve $3\cos x + 4\sin x = 4.5$

Substitute: $5\cos(x - 53.13^{\circ}) = 4.5$, so $\cos(x - 53.13^{\circ}) = 0.9$.

Let $u = x - 53.13^{\circ}$. Then $\cos u = 0.9$.

Principal value: $u = \arccos(0.9) \approx 25.84^{\circ}$.

Cosine is positive in Q1 and Q4: $u = 25.84^{\circ}$ or $u = 360^{\circ} - 25.84^{\circ} = 334.16^{\circ}$.

So $x = 53.13^{\circ} + 25.84^{\circ} \approx \boxed{78.97^{\circ}}$ (2 d.p.).

Or $x = 53.13^{\circ} + 334.16^{\circ} = 387.29^{\circ} \equiv \boxed{27.29^{\circ}}$ (subtract $360^{\circ}$, 2 d.p.).

Both values lie in $0^{\circ} < x < 360^{\circ}$ &check;

Verification

At $x = 27.29^{\circ}$: $3\cos 27.29^{\circ} + 4\sin 27.29^{\circ} \approx 3(0.8890) + 4(0.4579) \approx 2.667 + 1.832 = 4.499 \approx 4.5$ &check;

At $x = 78.97^{\circ}$: $3\cos 78.97^{\circ} + 4\sin 78.97^{\circ} \approx 3(0.1911) + 4(0.9816) \approx 0.573 + 3.926 = 4.500$ &check;

🎫

Exit Ticket

Before you leave, tick the W3–W4 objectives you feel confident you can achieve. Be honest — this guides what we revisit next week.

I can evaluate $\sec$, $\cosec$, $\cot$ at standard angles ($30^{\circ}$, $45^{\circ}$, $60^{\circ}$, $90^{\circ}$, $135^{\circ}$, $150^{\circ}$, ...) and recall their graphs and domains. I can apply the Pythagorean identities $\sec^2\theta = 1 + \tan^2\theta$ and $\cosec^2\theta = 1 + \cot^2\theta$ in proofs and equation-solving, including the quadrant sign check. I can recall and apply all three compound-angle formulae ($\sin(A\pm B)$, $\cos(A\pm B)$, $\tan(A\pm B)$) and avoid the “distributing sin over a sum” trap. I can choose the right form of $\cos 2A$ ($\cos^2 A - \sin^2 A$, $2\cos^2 A - 1$, or $1 - 2\sin^2 A$) for a given identity proof or equation, and use $\sin 2A = 2\sin A\cos A$. I can express $a\cos\theta + b\sin\theta$ as $R\cos(\theta - \alpha)$ (or $R\sin(\theta + \alpha)$), with $R = \sqrt{a^2+b^2}$ and $\tan\alpha = b/a$, and use it to solve equations and find maxima/minima. I can structure a trig identity proof (start from the more complex side, convert to $\sin$/$\cos$, choose matching $\cos 2A$ form) and solve a mixed trig equation by reducing to a single function and factorising.