Chapter 4: Trigonometry

Answer: B

Explanation

1. Label the triangle from angle $\theta$'s perspective. The side opposite $\theta$ is 5, the side adjacent to $\theta$ is 12, and the hypotenuse (opposite the right angle) is 13.

2. Apply the definition: $\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5}{12}$.

3. A quick check with the Pythagorean theorem: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$ — the triangle is valid.

Why the distractors are wrong:

A) $5/13$ is $\sin \theta$ (opposite over hypotenuse), not tangent. Many students confuse the "O" in SOHCAHTOA — SOH gives sin, not tan.

C) $12/13$ is $\cos \theta$ (adjacent over hypotenuse). Another common mix-up when the three ratios aren't firmly memorised.

D) $12/5$ is $\cot \theta$, the reciprocal of tangent. Students who recall "tangent involves 12 and 5" sometimes write them in the wrong order.

Takeaway: Anchor SOHCAHTOA firmly: Sin = Opp/Hyp, Cos = Adj/Hyp, Tan = Opp/Adj. For tangent, the hypotenuse never appears.

Answer: A

Explanation

1. From $\cos \theta = 7/25$, label the right triangle: adjacent $= 7$, hypotenuse $= 25$.

2. Use the Pythagorean theorem to find the opposite side: $\text{opp}^2 = 25^2 - 7^2 = 625 - 49 = 576$, so $\text{opp} = 24$.

3. Therefore $\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{24}{25}$.

Why the distractors are wrong:

B) $7/24$ equals $\tan \theta = \text{adj}/\text{opp}$. The student found the missing side (24) but then formed the ratio adjacent/opposite rather than opposite/hypotenuse.

C) $25/24$ is $\csc \theta$, the reciprocal of $\sin \theta$. Flipping numerator and denominator is a frequent slip once the opposite side is correctly found.

D) $7/25$ simply copies the given value of $\cos \theta$. This happens when a student forgets that a different ratio is required.

Takeaway: When given one trig ratio, draw a right triangle, label two sides, use Pythagoras to find the third, then read off whichever ratio the question asks for.

Answer: D

Explanation

1. By definition, $\cot \theta$ is the reciprocal of $\tan \theta$.

2. Since $\tan \theta = \dfrac{\sin \theta}{\cos \theta}$, we have $\cot \theta = \dfrac{\cos \theta}{\sin \theta}$.

3. In triangle terms: $\tan \theta = \text{opp}/\text{adj}$, so $\cot \theta = \text{adj}/\text{opp}$.

Why the distractors are wrong:

A) $\sin\theta / \cos\theta$ is the definition of $\tan \theta$, not $\cot \theta$.

B) opp/adj is $\tan \theta$ written in triangle form — again the reciprocal of what's needed.

C) hyp/adj equals $\sec \theta$. The hypotenuse appears in $\sec$ and $\csc$; cotangent involves only the two legs.

Takeaway: The "co-" prefix signals a reciprocal: $\cot = 1/\tan$, $\csc = 1/\sin$, $\sec = 1/\cos$. Learning all six functions as three reciprocal pairs removes a great deal of memorisation.

Answer: C

Explanation

1. We need the side opposite the 60° angle. Use $\sin 60° = \text{opp}/\text{hyp}$.

2. $\sin 60° = \dfrac{\sqrt{3}}{2}$, so $\text{opp} = 20 \times \dfrac{\sqrt{3}}{2} = 10\sqrt{3}$ cm.

Why the distractors are wrong:

A) 10 comes from using $\sin 30° = 1/2$ instead of $\sin 60°$. This gives the side adjacent to 60° (i.e., opposite the 30° angle).

B) $10\sqrt{2}$ uses $\sin 45° = \sqrt{2}/2$. The student substituted the wrong special angle.

D) $20/\sqrt{3}$ arises from writing $\text{opp} = \text{hyp}/\tan 60°$. Since $\tan 60° = \sqrt{3}$, this gives $20/\sqrt{3} \approx 11.5$, which is neither the opposite nor the adjacent side for this triangle.

Takeaway: For a 30-60-90 triangle, the sides are in ratio $1 : \sqrt{3} : 2$. The side opposite 30° is half the hypotenuse; the side opposite 60° is $(\sqrt{3}/2) \times \text{hyp}$.

Answer: D

Explanation

Find the third side. $AC^2 = 169 - 25 = 144$, so $AC = 12$.

Read the ratios from angle $A$ (opposite $= 5$, adjacent $= 12$, hypotenuse $= 13$): $$\sin A = \tfrac{5}{13}, \qquad \cos A = \tfrac{12}{13}, \qquad \sin A + \cos A = \tfrac{17}{13}$$

Why the others are wrong:

• B (1) confuses this with $\sin^2 A + \cos^2 A = 1$. That identity sums the squares.
• C ($\frac{7}{13}$) subtracted instead of adding.
• A ($\frac{60}{169}$) multiplied the two ratios.

Takeaway: $\sin^2\theta+\cos^2\theta = 1$ and $\sin\theta+\cos\theta$ are completely different expressions. The identity only applies to the squares.

Answer: B

Explanation

Use the identity $1 + \cot^2\theta = \csc^2\theta$ directly. Since $\csc\theta = \frac53$: $$1 + \cot^2\theta = \csc^2\theta = \frac{25}{9}$$

Check with a triangle: $\sin\theta = \frac35$, so opp 3, hyp 5, adj 4. Then $\cot^2\theta = \frac{16}{9}$ and $1 + \frac{16}{9} = \frac{25}{9}$ ✓

Why the others are wrong:

• A ($\frac{16}{9}$) is $\cot^2\theta$ alone — forgot to add 1.
• C (1) borrowed from $\sin^2+\cos^2 = 1$. This identity does not equal 1.
• D ($\frac53$) just repeats the given value.

Takeaway: Three Pythagorean identities: $\sin^2+\cos^2=1$, $1+\tan^2=\sec^2$, $1+\cot^2=\csc^2$. The last two come from dividing the first by $\cos^2\theta$ and $\sin^2\theta$.

Answer: A

Explanation

1. Apply the Pythagorean identities to each factor: - $1 - \cos^2\theta = \sin^2\theta$ - $1 + \cot^2\theta = \csc^2\theta = \dfrac{1}{\sin^2\theta}$

2. Multiply: $\sin^2\theta \times \dfrac{1}{\sin^2\theta} = 1$.

Why the distractors are wrong:

C) $\cos^2\theta$ results from simplifying only the first factor and ignoring the second.

B) $\csc^2\theta$ results from simplifying only the second factor and ignoring the first.

D) $\tan^2\theta$ arises from confusing $1 + \cot^2\theta$ with $1 + \tan^2\theta = \sec^2\theta$, then computing $\sin^2\theta \times \sec^2\theta = \sin^2\theta/\cos^2\theta = \tan^2\theta$. Using the wrong identity out of the three is the root error.

Takeaway: When an expression has two factors, simplify each independently using known identities, then combine. Products of the form $\sin^k\theta \cdot \csc^k\theta$ always collapse to 1.

Answer: C

Explanation

1. From $\tan\alpha = 3/4$: opp $= 3$, adj $= 4$, hyp $= 5$.

2. $\sin\alpha = 3/5$, $\cos\alpha = 4/5$.

3. Numerator: $3/5 - 4/5 = -1/5$.

4. Denominator: $3/5 + 4/5 = 7/5$.

5. Ratio: $\dfrac{-1/5}{7/5} = -\dfrac{1}{7}$.

Why the distractors are wrong:

A) $1/7$ is the correct magnitude but the wrong sign. The numerator is negative because $\cos\alpha > \sin\alpha$ when $\tan\alpha < 1$.

B) $7$ inverts the fraction ($7/5 \div 1/5 = 7$), suggesting the student placed numerator and denominator the wrong way round.

D) $-7$ both inverts and keeps the negative sign, combining the two errors above.

Takeaway: When $\tan\alpha < 1$ (i.e., $\alpha < 45°$), we have $\cos\alpha > \sin\alpha$, so $\sin\alpha - \cos\alpha$ is negative. Always assign a sign before simplifying fractions.

Section 4.2 — Trigonometry in Right-Angled Triangles

Answer: C

Explanation

1. The ladder is the hypotenuse (8 m) and the base distance is the adjacent side (4 m).

2. $\cos\theta = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{4}{8} = 0.5$.

3. $\theta = \cos^{-1}(0.5) = 60°$.

Why the distractors are wrong:

B) 30° is the angle the ladder makes with the wall (the complement of the angle with the ground). Mixing up the reference direction — measuring from the wall instead of the ground — produces this error.

A) 45° is a common guess when students feel uncertain; it has no basis in the given measurements.

D) 90° would mean the ladder is horizontal — impossible if it's leaning against a wall.

Takeaway: Identify which side is adjacent and which is the hypotenuse relative to the angle you are finding, then choose the appropriate inverse trig function.

Answer: D

Explanation

1. The rise (2 m) is opposite the angle, and the horizontal run (5 m) is adjacent.

2. $\tan\theta = \dfrac{2}{5} = 0.4$.

3. $\theta = \tan^{-1}(0.4) \approx 21.8°$.

Why the distractors are wrong:

A) 68.2° comes from inverting the fraction: $\tan^{-1}(5/2) = \tan^{-1}(2.5) \approx 68.2°$. This is the complement of the correct angle.

C) 23.6° uses $\sin^{-1}(2/5) = \sin^{-1}(0.4) \approx 23.6°$. The student used sine when the hypotenuse is unknown.

B) 24.0° is a rounded estimate without a principled calculation — a guess close to C.

Takeaway: When both legs of a right triangle are given, tangent is the correct ratio (no hypotenuse needed). Use $\theta = \tan^{-1}(\text{rise}/\text{run})$ for inclination problems.

Answer: A

Explanation

The angle of depression from the cliff equals the angle of elevation from the boat (alternate angles).

In the triangle, the 40 m cliff is opposite the 35° angle and the distance $x$ is adjacent: $$\tan 35° = \frac{40}{x} \quad\Rightarrow\quad x = \frac{40}{\tan 35°} \approx 57.1 \text{ m}$$

Why the others are wrong — each multiplies where it should divide:

• C (28.0) used $40 \times \tan 35°$.
• B (22.9) used $40 \times \sin 35°$, which would need 40 to be the hypotenuse.
• D (32.8) used $40 \times \cos 35°$.

Takeaway: Write the equation before rearranging. With the known side opposite the angle, the adjacent side is $\frac{\text{opp}}{\tan\theta}$ — a division, which is easy to forget.

Answer: B

Explanation

Two angles, two right triangles, one shared 50 m base.

Below eye level (depression 15°) gives the shorter building's height: $$h = 50\tan 15° \approx 13.4 \text{ m}$$

Above eye level (elevation 25°) gives the extra height: $$d = 50\tan 25° \approx 23.3 \text{ m}$$

Total: $13.4 + 23.3 \approx 36.7$ m

Why the others are wrong:

• A (23.3) is only the part above eye level.
• D (13.4) is only the part below it.
• C (50.0) is the horizontal distance — not a height at all.

Takeaway: One elevation and one depression describe two triangles sharing a base. Add the two vertical pieces.

Answer: D

Explanation

1. The right angle is at $R$, so $PQ$ is the hypotenuse.

2. $PQ^2 = PR^2 + QR^2 = 9^2 + 12^2 = 81 + 144 = 225$.

3. $PQ = \sqrt{225} = 15$ cm. (Recognise the 3-4-5 triple scaled by 3.)

Why the distractors are wrong:

B) 21 adds the two legs: $9 + 12 = 21$. Adding sides never gives the hypotenuse except by coincidence.

C) $\sqrt{63}$ computes $\sqrt{12^2 - 9^2} = \sqrt{144 - 81} = \sqrt{63}$. This subtracts instead of adds, which would find a leg if 12 were the hypotenuse — but the hypotenuse is $PQ$, not $QR$.

A) 7.5 is $\frac{9}{12} = 0.75$ multiplied by 10 or some other unconnected arithmetic.

Takeaway: The Pythagorean theorem adds the squares of the two legs to get the square of the hypotenuse. Only subtract if you are finding a leg and the hypotenuse is given.

Answer: A

Explanation

Let $h$ be the tower's height and $x$ the distance from the closer position.

Two equations, one for each position: $$h = x\tan 68° \qquad h = (x+80)\tan 42°$$

Equate and solve for $x$: $$x(\tan 68° - \tan 42°) = 80\tan 42° \implies x = \frac{72.03}{1.5747} \approx 45.7 \text{ m}$$

Then find $h$: $\ h = 45.7 \times \tan 68° \approx 113.2$ m

Why the others are wrong:

• B (72.0) is the numerator alone — stopped before solving for $x$.
• C (161.5) swapped the angles, putting 68° at the far position.
• D (52.6) never completed the final multiplication.

Takeaway: Two positions give two equations in $h$ and $x$. Equate them, solve for $x$ first, then substitute back for $h$.

Answer: C

Explanation

1. The diagonal $AC$ is the hypotenuse of right triangle $ABC$ (right angle at $B$ since $ABCD$ is a rectangle).

2. Side $BC$ is opposite angle $\angle CAB = 67°$.

3. $BC = AC \times \sin 67° = 26 \times 0.9205 \approx 23.9$ cm.

Why the distractors are wrong:

A) 10.2 m uses $\cos 67° \approx 0.3907$: $26 \times 0.3907 \approx 10.2$. This gives $AB$ (the adjacent side), not $BC$.

B) 28.2 cm comes from $26/\sin 67° \approx 28.2$ — dividing instead of multiplying, as if 26 were a leg rather than the hypotenuse.

D) 24.2 cm uses $\cos 23° \approx 0.921$, making the same computational choice but from the wrong angle.

Takeaway: In any rectangle, the diagonal creates a right triangle with the sides. Identify which side is opposite and which is adjacent to the given angle, then apply SOH or CAH accordingly.

Section 4.3 — Special Angles (30°, 45°, 60°)

Answer: B

Explanation

1. $\sin 30° = \dfrac{1}{2}$ and $\cos 60° = \dfrac{1}{2}$.

2. $\sin 30° + \cos 60° = \dfrac{1}{2} + \dfrac{1}{2} = 1$.

Why the distractors are wrong:

A) $\sqrt{3}$ results from computing $\sin 60° + \cos 30°$ instead: both equal $\sqrt{3}/2$, so their sum is $\sqrt{3}$. The question uses 30° with sin and 60° with cos, which both equal $1/2$.

D) $\sqrt{3}/2$ is the value of a single function such as $\sin 60°$ or $\cos 30°$ — not the sum of two terms.

C) $\sqrt{2}$ is a plausible guess based on mixing 45° values; $2 \times (\sqrt{2}/2) = \sqrt{2}$, but that applies to $\sin 45° + \cos 45°$, not this expression.

Takeaway: Memorise the special-angle table: $\sin 30° = \cos 60° = 1/2$, $\sin 45° = \cos 45° = \sqrt{2}/2$, $\sin 60° = \cos 30° = \sqrt{3}/2$. Note that $\sin\theta = \cos(90°-\theta)$.

Answer: A

Explanation

1. In a 45-45-90 triangle the two legs are equal; if each leg $= 1$, the hypotenuse $= \sqrt{2}$.

2. $\tan 45° = \dfrac{\text{opp}}{\text{adj}} = \dfrac{1}{1} = 1$.

Why the distractors are wrong:

B) $\sqrt{2}$ is $\sec 45° = 1/\cos 45° = \sqrt{2}$. Confusing tangent with secant produces this.

C) $\sqrt{3}$ is $\tan 60°$.

D) $1/\sqrt{2}$ equals $\sin 45° = \cos 45°$. The student may know the 45° values but apply them to the wrong function.

Takeaway: $\tan 45° = 1$ is perhaps the most useful special-angle value to know instantly, since it signals equal opposite and adjacent sides and arises constantly in symmetry arguments.

Answer: C

Explanation

Both special values are $\frac{\sqrt3}{2}$, so both squares are $\frac34$:

$$\sin^2 60° + \cos^2 30° - 1 = \tfrac34 + \tfrac34 - 1 = \tfrac32 - 1 = \tfrac12$$

Why the others are wrong:

• A (0) treats this as $\sin^2 60° + \cos^2 \mathbf{60°} - 1$. The question uses $\cos 30°$, which equals $\sin 60°$ — so both terms are $\frac34$, not $\frac34$ and $\frac14$.
• B ($\frac32$) added the squares but never subtracted the 1.
• D ($\frac14$) used $\sin^2 60° = \frac12$, which is wrong.

Takeaway: $\sin 60° = \cos 30°$ — a co-function pair, so they are not the complementary pair the Pythagorean identity needs. Read the angles carefully, then finish the arithmetic.

Answer: B

Explanation

1. $\tan 30° = \dfrac{1}{\sqrt{3}}$, so $\tan^2 30° = \dfrac{1}{3}$.

2. Numerator: $2 \times \dfrac{1}{\sqrt{3}} = \dfrac{2}{\sqrt{3}}$.

3. Denominator: $1 - \dfrac{1}{3} = \dfrac{2}{3}$.

4. $\dfrac{2/\sqrt{3}}{2/3} = \dfrac{2}{\sqrt{3}} \times \dfrac{3}{2} = \dfrac{3}{\sqrt{3}} = \sqrt{3}$.

Recognition: This is exactly the double-angle formula $\tan(2 \times 30°) = \tan 60° = \sqrt{3}$.

Why the distractors are wrong:

A) $2/\sqrt{3}$ is just the numerator before dividing by the denominator — the student stopped halfway.

C) $2$ arises from incorrectly simplifying: perhaps dividing $2/\sqrt{3}$ by $1/3$ and getting $6/\sqrt{3} = 2\sqrt{3}$, then rounding.

D) $1/\sqrt{3}$ is $\tan 30°$ itself — the student evaluated the input rather than the expression.

Takeaway: The formula $\dfrac{2\tan\theta}{1-\tan^2\theta} = \tan(2\theta)$ is a powerful identity. Recognising its form allows you to evaluate it without tedious arithmetic.

Answer: D

Explanation

1. $\sin\theta = \cos\theta$ implies $\dfrac{\sin\theta}{\cos\theta} = 1$, i.e., $\tan\theta = 1$.

2. The unique solution in $[0°, 90°]$ is $\theta = 45°$, so $\sin\theta = \cos\theta = \dfrac{1}{\sqrt{2}}$ and $\tan\theta = 1$.

Why the distractors are wrong:

A) $\theta = 30°$ is wrong: $\sin 30° = 0.5 \neq \cos 30° = \sqrt{3}/2 \approx 0.866$.

B) $\sin\theta = 1$ would require $\theta = 90°$, at which point $\cos 90° = 0 \neq 1$, contradicting the condition.

C) $\cos\theta = 0$ would require $\theta = 90°$ — same contradiction as B.

Takeaway: $\sin\theta = \cos\theta$ is equivalent to $\tan\theta = 1$. Dividing both sides of a trig equation by a function is a valid algebraic step and often simplifies the equation dramatically.

Answer: B

Explanation

1. Write $75° = 45° + 30°$ and apply the compound angle formula:

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

2. $\sin 75° = \sin 45°\cos 30° + \cos 45°\sin 30°$

$= \dfrac{\sqrt{2}}{2} \cdot \dfrac{\sqrt{3}}{2} + \dfrac{\sqrt{2}}{2} \cdot \dfrac{1}{2}$

$= \dfrac{\sqrt{6}}{4} + \dfrac{\sqrt{2}}{4} = \dfrac{\sqrt{6} + \sqrt{2}}{4}$.

Why the distractors are wrong:

A) $(\sqrt{6}-\sqrt{2})/4$ is $\sin 15° = \sin(45° - 30°)$. The student used subtraction instead of addition in the compound angle.

C) $(\sqrt{2}+1)/2$ is an incorrect expansion, perhaps treating the square roots incorrectly.

D) $\sqrt{3}/2$ is $\sin 60°$, off by 15°.

Takeaway: To evaluate $\sin$ or $\cos$ of a non-standard angle such as 75°, decompose it into a sum or difference of special angles (30°, 45°, 60°) and apply the compound angle formula.

Section 4.4 — Unit Circle and Reference Angles

Answer: C

Explanation

1. 240° lies in the third quadrant (between 180° and 270°).

2. Reference angle $= 240° - 180° = 60°$.

Why the distractors are wrong:

A) 30° would be the reference angle for 210° (= 180° + 30°). Off by one common multiple of 30°.

B) 120° is $360° - 240°$, which gives the supplementary distance from 240° to the full circle — not the reference angle (which is always measured from the nearest x-axis).

D) 300° is $360° - 60°$, a valid angle in Q4 that has the same reference angle (60°) but is not the reference angle itself.

Takeaway: The reference angle is always the acute angle between the terminal side and the x-axis. For Q3: subtract 180°. For Q2: subtract from 180°. For Q4: subtract from 360°.

Answer: D

Explanation

1. The quadrants: Q1 (0°–90°), Q2 (90°–180°), Q3 (180°–270°), Q4 (270°–360°).

2. 310° is between 270° and 360° → Quadrant IV.

Why the distractors are wrong:

A) Q1 applies to angles less than 90°. 310° has completed most of the circle.

B) Q2 (90° to 180°) is on the opposite side; 310° is past 270°.

C) Q3 (180° to 270°): 310° > 270°, so it has passed Q3.

Takeaway: A quick check: does the angle exceed 270°? If so, it is in Q4. Locate by comparing to the quadrant boundaries 0°, 90°, 180°, 270°, 360°.

Answer: A

Explanation

Using the CAST rule (All positive in Q1, Sin positive in Q2, Tan positive in Q3, Cos positive in Q4):

• Q1: both sin and cos positive.
• Q2: sin positive, cos negative.
• Q3: both sin and cos negative. ✓
• Q4: cos positive, sin negative.

Both functions negative $\Rightarrow$ Quadrant III.

Why the distractors are wrong:

D) Q1 has both functions positive.

B) Q2 has sin positive but cos negative — only one is negative here.

C) Q4 has cos positive but sin negative — the reverse of Q2.

Takeaway: The CAST mnemonic places "All" in Q1, "Sin" in Q2, "Tan" in Q3, "Cos" in Q4 to indicate which primary function is positive. The sign of both sin and cos being negative uniquely identifies Q3.

Answer: D

Explanation

1. $\sin\theta < 0$ and $\cos\theta > 0$ places $\theta$ in Quadrant IV.

2. With opp $= 5$, hyp $= 13$: adj $= \sqrt{169 - 25} = 12$.

3. In Q4: $\cos\theta = +12/13$, $\sin\theta = -5/13$.

4. $\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{-5/13}{12/13} = -\dfrac{5}{12}$.

Why the distractors are wrong:

B) $5/12$ is the magnitude of $\tan\theta$ but with the wrong sign. In Q4 tangent is negative.

C) $-12/5$ inverts the fraction. The student may have confused opposite and adjacent.

A) $-5/13$ copies the given sine value — the student didn't compute tangent at all.

Takeaway: In Q4, sin is negative, cos is positive, so tan (= sin/cos) is negative. Always determine the quadrant before assigning signs to the reciprocal/quotient functions.

Answer: A

Explanation

Using the compound angle formula:

$\cos(180° + \theta) = \cos 180°\cos\theta - \sin 180°\sin\theta = (-1)\cos\theta - (0)\sin\theta = -\cos\theta$.

Why the distractors are wrong:

B) $\cos\theta$ forgets the sign change. Moving 180° into Q3 (where cosine is negative) must change the sign.

C) $\sin\theta$ would be the result of $\cos(90° - \theta)$, not $\cos(180° + \theta)$.

D) $-\sin\theta$ is $\cos(90° + \theta)$, a different reduction formula.

Takeaway: The four key reduction formulas: $\cos(180°\pm\theta) = -\cos\theta$, $\sin(180° - \theta) = \sin\theta$, $\sin(180° + \theta) = -\sin\theta$, and $\cos(360° - \theta) = \cos\theta$. Each reflects the CAST sign pattern in the relevant quadrant.

Answer: C

Explanation

1. The reference angle: $|\tan\theta| = \sqrt{3}$, so the reference angle is 60°.

2. With $\theta$ in Q2 ($90° < \theta < 180°$), we have $\theta = 180° - 60° = 120°$.

3. $\sin 120° = \sin(180° - 60°) = \sin 60° = \dfrac{\sqrt{3}}{2}$.

(In Q2, sin is positive.)

Why the distractors are wrong:

A) $-\sqrt{3}/2$ places $\theta$ in Q3 where sin is also negative. Forgetting that tan is negative in Q2 (not just Q3) causes this error.

B) $1/2$ is $\sin 30°$, from confusing the reference angle: $\tan 60° = \sqrt{3}$ gives reference 60°, not 30°.

D) $2/\sqrt{3}$ is an algebraic error — perhaps $1/\sin\theta$ or inversion of the triangle sides.

Takeaway: Tan is negative in both Q2 and Q4. A given interval for $\theta$ (here Q2) removes the ambiguity. In Q2 sin is always positive, so the result is $+\sqrt{3}/2$.

Answer: B

Explanation

1. $\sin(360° - \theta) = \sin(-\theta) = -\sin\theta$. (In Q4, sin is negative and the co-angle of $\theta$.)

2. $\cos(180° + \theta) = -\cos\theta$. (From Q26 above.)

3. Sum $= -\sin\theta + (-\cos\theta) = -(\sin\theta + \cos\theta)$.

Why the distractors are wrong:

A) $\sin\theta + \cos\theta$ gets both signs wrong: the student forgot that both reduction formulas introduce a negative sign.

D) $\sin\theta - \cos\theta$ gets the first sign wrong and the second right.

C) $-\sin\theta + \cos\theta$ gets the first right but forgets the sign on the cosine term.

Takeaway: Apply each reduction formula independently before combining. Write out each step explicitly: $\sin(360°-\theta) = -\sin\theta$, then $\cos(180°+\theta) = -\cos\theta$, then add.

Section 4.5 — Trigonometric Identities

Answer: A

Explanation

From the fundamental identity $\sin^2\theta + \cos^2\theta = 1$:

$1 - \sin^2\theta = \cos^2\theta$.

Why the distractors are wrong:

B) $1 + \cos^2\theta$ adds rather than recognising the rearrangement. The result would exceed 1 for most $\theta$.

C) $-\cos^2\theta$ introduces a spurious negative sign.

D) $\sin^2\theta$ gives back part of what was subtracted, which would imply $1 - \sin^2\theta = \sin^2\theta$, i.e., $\sin^2\theta = 1/2$ — true only for $\theta = 45°$, not generally.

Takeaway: The Pythagorean identity is best learned in all three rearrangements: $\sin^2+\cos^2=1$, $\cos^2 = 1 - \sin^2$, $\sin^2 = 1 - \cos^2$. Recognising which form to use is half the battle.

Answer: D

Explanation

1. Rewrite the numerator: $\sin^2\theta - 1 = -(1 - \sin^2\theta) = -\cos^2\theta$.

2. $\dfrac{-\cos^2\theta}{\cos\theta} = -\cos\theta$.

Why the distractors are wrong:

A) $\cos\theta$ drops the negative sign introduced in step 1.

C) $-\sin^2\theta$ results from cancelling $\theta$-values incorrectly without factoring the numerator first.

B) $\tan\theta$ would require the numerator to be $\sin^2\theta/\cos\theta \cdot \cos\theta = \sin^2\theta$, not $\sin^2\theta - 1$.

Takeaway: $\sin^2\theta - 1 = -\cos^2\theta$ is a one-step Pythagorean rearrangement that frequently appears in simplification. Spotting it quickly is a mark of identity fluency.

Answer: B

Explanation

1. $\tan\theta + \cot\theta = \dfrac{\sin\theta}{\cos\theta} + \dfrac{\cos\theta}{\sin\theta} = \dfrac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \dfrac{1}{\sin\theta\cos\theta}$.

2. Multiply by $\sin\theta\cos\theta$: $\dfrac{1}{\sin\theta\cos\theta} \times \sin\theta\cos\theta = 1$.

Why the distractors are wrong:

A) $\sin 2\theta$ comes from recognising $2\sin\theta\cos\theta = \sin 2\theta$ but not completing the simplification of the bracket first.

C) $\tan\theta$ is a partial simplification — perhaps only $\tan\theta$ was factored from the bracket.

D) $2$ may arise from mistaking $\tan\theta + \cot\theta = 2$ (which is false in general).

Takeaway: Combine $\tan + \cot$ by placing over a common denominator first. The Pythagorean identity then collapses the numerator to 1, and the $\sin\theta\cos\theta$ cancels cleanly.

Answer: C

Explanation

1. Recognise $1 - \cos^2\theta = \sin^2\theta$.

2. $\dfrac{\sin^2\theta}{\sin\theta} = \sin\theta = \dfrac{4}{5}$.

Why the distractors are wrong:

A) $16/25$ is $\sin^2\theta = (4/5)^2$, i.e., the student simplified the numerator but forgot to divide by $\sin\theta$.

B) $3/5$ is $\cos\theta$ (since $\cos\theta = \sqrt{1 - 16/25} = 3/5$). A student who computed the cosine and then confused numerator and denominator would land here.

D) $5/4$ is $1/\sin\theta = \csc\theta$, suggesting the student inverted the expression.

Takeaway: Before substituting numbers, simplify the algebraic form entirely. Here the expression equals $\sin\theta$ regardless of the specific value given — recognising that saves computation.

Answer: B

Explanation

1. Factor the numerator as a difference of squares: $\cos^2\theta - \sin^2\theta = (\cos\theta + \sin\theta)(\cos\theta - \sin\theta)$.

2. Cancel $(\cos\theta - \sin\theta)$ (provided $\cos\theta \neq \sin\theta$):

$\dfrac{(\cos\theta + \sin\theta)(\cos\theta - \sin\theta)}{\cos\theta - \sin\theta} = \cos\theta + \sin\theta$.

Why the distractors are wrong:

A) $\cos\theta - \sin\theta$ results from cancelling $(\cos\theta + \sin\theta)$ instead of $(\cos\theta - \sin\theta)$ — picking the wrong factor.

C) $1$ confuses this with the Pythagorean identity: $\cos^2\theta + \sin^2\theta = 1$, but the numerator here has a minus sign.

D) $\cos^2\theta + \sin^2\theta$ is 1, the Pythagorean identity — same error as C written differently.

Takeaway: $a^2 - b^2 = (a+b)(a-b)$ with trig functions: $\cos^2\theta - \sin^2\theta = (\cos\theta+\sin\theta)(\cos\theta-\sin\theta)$. This factoring appears regularly in identity simplifications.

Answer: D

Explanation

Start from $\sin^2\theta + \cos^2\theta = 1$. Divide both sides by $\cos^2\theta$:

$\tan^2\theta + 1 = \sec^2\theta$ ✓

This is valid for all $\theta$ where $\cos\theta \neq 0$.

Checking the others:

• A: $\sin 2\theta = 2\sin\theta\cos\theta \neq 2\sin\theta$ (missing $\cos\theta$).
• C: $\cos 2\theta = 1 - 2\sin^2\theta \neq 1 - \sin^2\theta$.
• B: $\tan\theta + \cot\theta = 2/\sin 2\theta$ exactly (no $-1$).

Why the distractors are wrong:

A) Drops the $\cos\theta$ factor from the double angle formula.

C) Uses only one $\sin^2$ instead of two in the correct formula $\cos 2\theta = 1 - 2\sin^2\theta$.

B) Is almost right but the $-1$ is incorrect: $\tan\theta + \cot\theta = 1/(\sin\theta\cos\theta) = 2/(2\sin\theta\cos\theta) = 2/\sin 2\theta$ exactly.

Takeaway: Checking an identity means testing whether both sides are equal for every $\theta$ — substituting one specific value like $\theta = 30°$ quickly eliminates false options.

Answer: A

Explanation

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

2. $\dfrac{(1-\cos\theta)(1+\cos\theta)}{1-\cos\theta} = 1 + \cos\theta$, provided $\cos\theta \neq 1$.

Why the distractors are wrong:

C) $1 - \cos\theta$ cancels the wrong factor: the student divided by $(1+\cos\theta)$ rather than cancelling it.

B) $\sin\theta + \cos\theta$ has no correct derivation from this expression; it may reflect a guess based on familiar-looking terms.

D) $1/\sin\theta = \csc\theta$ comes from incorrectly treating the numerator as $\sin\theta$ (first power) and cancelling, or from inverting.

Takeaway: $\sin^2\theta = (1-\cos\theta)(1+\cos\theta)$ is the factored Pythagorean identity. When the denominator is $(1 \pm \cos\theta)$, this factoring creates an immediate cancellation.

Section 4.6 — Solving Trigonometric Equations

Answer: C

Explanation

1. $\sin\theta = \sqrt{2}/2 > 0$, so $\theta$ lies in Q1 or Q2.

2. Reference angle: $\sin^{-1}(\sqrt{2}/2) = 45°$.

3. Q1: $\theta = 45°$. Q2: $\theta = 180° - 45° = 135°$.

4. Solutions: $\theta = 45°$ and $\theta = 135°$.

Why the distractors are wrong:

A) 45° only misses the Q2 solution, a very common error — students stop as soon as they find one answer.

B) 45° and 225° gives Q3 instead of Q2 for the second solution. In Q3, $\sin\theta < 0$, contradicting the equation.

D) 135° only finds the Q2 angle but misses the Q1 angle.

Takeaway: When $\sin\theta = k > 0$, there are always two solutions in $[0°, 360°]$: one in Q1 and one in Q2 ($\theta = 180° - \text{ref}$). A positive sine value is never satisfied in Q3 or Q4.

Answer: C

Explanation

1. $\cos\theta = -1/2 < 0$: solutions are in Q2 and Q3.

2. Reference angle: $\cos^{-1}(1/2) = 60°$.

3. Q2: $\theta = 180° - 60° = 120°$. Q3: $\theta = 180° + 60° = 240°$.

Why the distractors are wrong:

B) 60° and 300° are solutions for $\cos\theta = +1/2$ (Q1 and Q4), not $-1/2$.

A) 60° only uses the reference angle directly without considering the sign change or finding both solutions.

D) 120° only finds Q2 but misses Q3.

Takeaway: Cosine is negative in Q2 and Q3. For Q2 use $180° - \text{ref}$; for Q3 use $180° + \text{ref}$.

Answer: A

Explanation

1. Isolate: $\sin\theta = -\dfrac{1}{2}$.

2. Negative sine: Q3 and Q4. Reference angle: $\sin^{-1}(1/2) = 30°$.

3. Q3: $180° + 30° = 210°$. Q4: $360° - 30° = 330°$.

Why the distractors are wrong:

B) 30° and 150° solves $\sin\theta = +1/2$ (forgot the negative sign from the equation).

C) 30° only uses the raw reference angle without adjusting for the negative value or finding both solutions.

D) 210° only finds Q3 but misses Q4.

Takeaway: Always isolate the trig function first, then determine the sign and the appropriate quadrants. Negative sine → Q3 and Q4; positive sine → Q1 and Q2.

Answer: D

Explanation

1. $\tan\theta = \pm\sqrt{3}$.

2. $\tan\theta = +\sqrt{3}$: reference $= 60°$. Tan positive in Q1 and Q3: $\theta = 60°, 240°$.

3. $\tan\theta = -\sqrt{3}$: reference $= 60°$. Tan negative in Q2 and Q4: $\theta = 120°, 300°$.

4. All four solutions: 60°, 120°, 240°, 300°.

Why the distractors are wrong:

A) 60° and 240° includes only the positive root. Taking the square root gives $\pm\sqrt{3}$; both branches must be solved.

B) 60° and 120° gives Q1 for the positive root and Q2 for the negative, but misses Q3 and Q4.

C) 30°, 150°, 210°, 330° uses reference angle 30° instead of 60°. $\tan 30° = 1/\sqrt{3}$, not $\sqrt{3}$.

Takeaway: $\tan^2\theta = k$ always produces four solutions in $[0°, 360°]$ because both $+\sqrt{k}$ and $-\sqrt{k}$ each give two solutions. For tangent, the two solutions for a given sign are separated by 180°.

Answer: B

Explanation

1. Factor: $(2\sin\theta + 1)(\sin\theta - 1) = 0$.

2. $\sin\theta = 1$: $\theta = 90°$.

3. $\sin\theta = -1/2$: reference $= 30°$, solutions in Q3 and Q4: $\theta = 210°, 330°$.

4. Full solution set: $\{90°, 210°, 330°\}$.

Why the distractors are wrong:

A) 90° only solves only $\sin\theta = 1$ and ignores the factor $(2\sin\theta+1)=0$.

D) 30°, 150°, 210°, 330° solves $\sin\theta = -1/2$ correctly (Q3 and Q4: 210° and 330°) but replaces the $\sin\theta = 1$ solution with $\sin\theta = 1/2$ solutions (30° and 150°).

C) 90° and 270° confuses $\sin\theta = 1$ (gives 90°) with $\sin\theta = -1$ (gives 270°); the actual second factor gives $-1/2$, not $-1$.

Takeaway: Quadratics in $\sin\theta$ or $\cos\theta$ factor just like ordinary quadratics. Treat $\sin\theta$ as a single variable $u$, factor, solve for $u$, then solve each trig equation separately.

Answer: C

Explanation

1. Divide both sides by $\cos\theta$ (valid when $\cos\theta \neq 0$): $\tan\theta = 1$.

2. Reference angle: 45°. Tan positive in Q1 and Q3: $\theta = 45°, 225°$.

Why the distractors are wrong:

B) 45° only finds Q1 but misses the Q3 solution at 225°.

A) 135° and 315° solves $\tan\theta = -1$, which would arise from $\sin\theta = -\cos\theta$.

D) 0° and 180° are solutions to $\sin\theta = 0$, a completely different equation.

Takeaway: $\sin\theta = \cos\theta$ is most efficiently solved by dividing to get $\tan\theta = 1$. This avoids squaring (which can introduce spurious solutions) and immediately reveals the quadrant structure.

Answer: B

Explanation

1. Substitute $\cos 2\theta = 2\cos^2\theta - 1$:

$2\cos^2\theta - 1 = \cos\theta$

$2\cos^2\theta - \cos\theta - 1 = 0$

$(2\cos\theta + 1)(\cos\theta - 1) = 0$.

2. $\cos\theta = 1$: $\theta = 0°$ (or 360°, same point).

3. $\cos\theta = -1/2$: Q2: $120°$, Q3: $240°$.

4. Solutions in $[0°, 360°)$: $\{0°, 120°, 240°\}$.

Why the distractors are wrong:

A) 0°, 120°, 240°, 360° lists 360° as a separate solution. Since $[0°, 360°]$ is a closed interval, 360° is technically valid — but most conventions treat $[0°, 360°)$ as the standard period, giving three distinct solutions. The question specifies $[0°, 360°]$, making this partially defensible, but the standard answer is B.

C) 60°, 180°, 300° uses $\cos 2\theta = 2\cos^2\theta$ (missing the $-1$), leading to a different factoring.

D) 120° and 240° forgets the $\cos\theta = 1$ solution at $0°$.

Takeaway: When a double-angle equation appears, always rewrite in terms of a single angle using a double-angle identity, then factor.

Answer: A

Explanation

1. $\sin\theta = \sqrt{3}/2$: principal value $60°$ (Q1) and supplementary value $120°$ (Q2).

2. Both repeat every $360°$, giving the general solution:

$\theta = 60° + 360°n$ or $\theta = 120° + 360°n$, $n \in \mathbb{Z}$.

Why the distractors are wrong:

C) Gives only the Q1 family, omitting the Q2 family entirely.

B) $60° + 180°n$ generates 60°, 240°, 420°, ... but $\sin 240° = -\sqrt{3}/2 \neq \sqrt{3}/2$. The period for a fixed positive sine is 360°, not 180°.

D) $n \times 60°$ generates 0°, 60°, 120°, 180°, ..., many of which give incorrect sine values.

Takeaway: For $\sin\theta = k$ (with $k > 0$), the general solution has two families: $\theta = \alpha + 360°n$ (Q1) and $\theta = (180°-\alpha) + 360°n$ (Q2). For $\cos\theta = k$: $\theta = \pm\alpha + 360°n$.

Answer: D

Explanation

1. Factor: $(2\cos\theta - 1)(\cos\theta + 2) = 0$.

2. $\cos\theta = 1/2$: valid. Solutions: $\theta = 60°, 300°$.

3. $\cos\theta = -2$: impossible since $|\cos\theta| \leq 1$.

4. Two solutions.

Why the distractors are wrong:

A) 4 assumes both roots of the quadratic give two trig solutions each. The second root $\cos\theta = -2$ is outside $[-1, 1]$ and must be discarded.

B) 0 mistakenly believes no solutions exist — perhaps the student solved for the quadratic roots but misread the discriminant.

C) 3 has no clear origin; possibly the student counted 60°, 300°, and one of the spurious $\cos\theta = -2$ angles.

Takeaway: Always check that the value produced by a quadratic falls within $[-1, 1]$ before computing angles. Roots outside this range give no real solutions and must be discarded.

Section 4.7 — Trigonometric Graphs

Answer: D

Explanation

The amplitude is the magnitude of the coefficient of the trig function: $|-3| = 3$. Amplitude is always non-negative.

Why the distractors are wrong:

B) $-3$ includes the sign. Amplitude is a distance and is therefore always taken as an absolute value.

C) 2 is the period-affecting coefficient (it compresses the period), not the amplitude.

A) 6 multiplies the two coefficients: $2 \times 3 = 6$. The 2 affects period, not amplitude.

Takeaway: For $y = a\sin(bx) + d$, amplitude $= |a|$, period $= 360°/b$, vertical shift $= d$. Each parameter affects exactly one feature of the graph.

Answer: C

Explanation

Period $= \dfrac{360°}{b} = \dfrac{360°}{3} = 120°$.

Why the distractors are wrong:

A) 360° is the period of the basic $y = \sin x$; forgetting to divide by the frequency multiplier.

B) 3° confuses the coefficient with the period value.

D) 180° would be correct for $b = 2$ (e.g., $\sin 2x$). Off by one common step.

Takeaway: The coefficient $b$ in $\sin(bx)$ compresses the graph horizontally by a factor of $b$, reducing the period from 360° to $360°/b$.

Answer: B

Explanation

In $y = \cos(x - d)$, the graph shifts $d$ units to the right (the peak that was at $x = 0$ now occurs at $x = d$).

Here $d = 30°$, so the graph moves 30° to the right.

Why the distractors are wrong:

A) 30° left is the shift for $y = \cos(x + 30°)$. The sign inside the argument determines the direction: $-30°$ shifts right, $+30°$ shifts left.

C) 30° upward would require $y = \cos x + 30°$ (a vertical shift outside the function).

D) 60° right doubles the shift without justification.

Takeaway: Phase shift direction is counter-intuitive: $y = f(x - d)$ shifts right by $d$ (positive $d$). A way to remember it: the "zero point" occurs when the argument equals zero, i.e., when $x - d = 0$, i.e., $x = d$.

Answer: A

Explanation

Adding a constant outside the trig function translates the entire graph vertically. $y = \sin x + 2$ shifts every point 2 units upward; the range changes from $[-1, 1]$ to $[1, 3]$.

Why the distractors are wrong:

D) The amplitude is still 1; $+2$ is a vertical translation, not a stretch.

B) The period is unchanged at 360°; only the coefficient inside the argument affects the period.

C) A horizontal shift requires $+2$ inside the argument: $y = \sin(x + 2°)$, not outside.

Takeaway: The four graph transformations and their locations: amplitude (coefficient in front), period (coefficient inside), phase shift (constant inside), vertical shift (constant outside/added after).

Answer: A

Explanation

The maximum of $\sin x$ is 1, so the maximum of $2\sin x$ is $2 \times 1 = 2$.

Why the distractors are wrong:

B) 1 is the maximum of the unscaled $\sin x$. Forgetting the amplitude factor produces this.

C) 4 doubles the amplitude again: perhaps the student doubled a value they thought was already 2.

D) $\pi$ confuses radian measure with function values.

Takeaway: Maximum value $=$ amplitude $=$ coefficient of the trig function (assuming no vertical shift). For $y = a\sin x + d$, the maximum is $a + d$ and minimum is $-a + d$.

Answer: C

Explanation

• Amplitude $= 4$: coefficient in front is 4.
• Period $= 180°$: $360°/b = 180° \Rightarrow b = 2$.
• Equation: $y = 4\sin(2x)$.

Why the distractors are wrong:

A) $4\sin(x/2)$ has period $360° \div (1/2) = 720°$ — too long.

B) $2\sin(4x)$ has amplitude 2 and period 90° — both wrong.

D) $4\sin x$ has amplitude 4 but period 360° — period not halved.

Takeaway: Build the equation in two steps: write the amplitude first (coefficient in front), then determine $b$ from $b = 360°/\text{period}$.

Answer: D

Explanation

The maximum of $\sin$ occurs when the argument $= 90°$:

$2x = 90° \Rightarrow x = 45°$.

Why the distractors are wrong:

A) 180° is where $y = \sin x$ (uncompressed) reaches its next zero, not its maximum.

B) 90° is where $y = \sin x$ first reaches its maximum. Forgetting to adjust for the factor of 2 inside the argument.

C) 30° has no standard derivation here.

Takeaway: To find the $x$-coordinate of a maximum for $y = \sin(bx)$, set $bx = 90°$ and solve. For minima, set $bx = 270°$; for zeros, set $bx = 0°$ or $180°$.

Answer: B

Explanation

1. Range of $\sin x$: $[-1, 1]$.

2. Multiply by 3: $[-3, 3]$.

3. Subtract 2: $[-3 - 2,\; 3 - 2] = [-5, 1]$.

Why the distractors are wrong:

A) $[-3, 3]$ is the range after multiplying by 3, before subtracting 2.

D) $[-2, 2]$ ignores the amplitude entirely and focuses only on the shift.

C) $[0, 1]$ has no basis in the function.

Takeaway: Find range by transforming the known range $[-1, 1]$ step by step: first apply the amplitude (multiply), then the vertical shift (add or subtract). The endpoints transform exactly like any real number under those operations.

Section 4.8 — Sine Rule and Cosine Rule

Answer: B

Explanation

Cosine rule: $c^2 = a^2 + b^2 - 2ab\cos C$

$c^2 = 64 + 25 - 2(8)(5)\cos 60° = 89 - 80 \times \tfrac{1}{2} = 89 - 40 = 49$

$c = 7$.

Why the distractors are wrong:

A) $\sqrt{89}$ uses $c^2 = a^2 + b^2$ only, dropping the $2ab\cos C$ correction term — as if applying Pythagoras to a non-right triangle.

C) 3 subtracts the sides: $|8 - 5| = 3$. This has no trig basis.

D) 13 adds the sides: $8 + 5 = 13$. The triangle inequality says $c < 13$, but equality only occurs when $C = 180°$.

Takeaway: The cosine rule modifies the Pythagorean theorem with a correction factor $-2ab\cos C$. When $C = 90°$, $\cos C = 0$ and the formula reduces to Pythagoras. When $C < 90°$, the correction is positive (subtracted), giving $c < \sqrt{a^2+b^2}$.

Answer: D

Explanation

Sine rule: $\dfrac{b}{\sin B} = \dfrac{a}{\sin A}$

$b = 12 \times \dfrac{\sin 75°}{\sin 35°} = 12 \times \dfrac{0.9659}{0.5736} \approx 12 \times 1.684 \approx 20.2$.

Why the distractors are wrong:

A) 7.1 inverts the ratio: $12 \times \dfrac{\sin 35°}{\sin 75°} \approx 7.1$. The side opposite the larger angle should be the longer side, not the shorter.

C) 14.4 and B) 10.2 arise from incorrect arithmetic or mixing up the angles in the ratio.

Takeaway: The sine rule states that the side over the sine of its opposite angle is constant throughout a triangle: $a/\sin A = b/\sin B = c/\sin C$. The larger angle is always opposite the longer side.

Answer: C

Explanation

$\cos R = \dfrac{p^2 + q^2 - r^2}{2pq} = \dfrac{49 + 81 - 121}{2 \times 7 \times 9} = \dfrac{9}{126} = \dfrac{1}{14} \approx 0.0714$

$R = \cos^{-1}(0.0714) \approx 85.9°$.

Why the distractors are wrong:

A) 15.1° uses $\cos^{-1}(13/14)$, arising from adding instead of subtracting in the numerator: $49 + 81 + 121 = 251$ ... or some other sign error.

B) 74.1° comes from computing $\cos R = (p^2 + r^2 - q^2)/(2pr)$ — placing $q$ where $r$ should be in the formula.

D) 94.1° is $180° - 85.9°$: the supplement. Taking the supplement is tempting if the calculator result seems too small.

Takeaway: To find angle $R$ using the cosine rule, put $r^2$ on the numerator "receiving side" (after the minus sign): $\cos R = (p^2 + q^2 - r^2)/(2pq)$. The angle is always opposite the side that is subtracted.

Answer: A

Explanation

$\sin Y = \dfrac{y \sin X}{x} = \dfrac{8 \times \sin 50°}{10} = \dfrac{8 \times 0.7660}{10} = 0.6128$

$Y = \sin^{-1}(0.6128) \approx 37.8°$.

Since $y = 8 < x = 10$, we have $Y < X = 50°$: only one solution. ✓

Why the distractors are wrong:

D) 52.2° exceeds $X = 50°$, which is impossible if $y < x$.

B) 73.2° inverts the ratio: $\sin Y = 10\sin 50°/8 \approx 0.9575$, giving $\arcsin(0.9575) \approx 73.2°$. This is the result of writing the formula upside down.

C) 42.8° comes from an arithmetic error in the computation.

Takeaway: When using the sine rule to find an angle, check whether two solutions are geometrically possible by comparing the given sides. If the side opposite the unknown angle is shorter, that angle must be acute and unique.

Answer: D

Explanation

Area $= \dfrac{1}{2}ab\sin C = \dfrac{1}{2} \times 6 \times 9 \times \sin 30° = \dfrac{1}{2} \times 54 \times \dfrac{1}{2} = 13.5$.

Why the distractors are wrong:

B) 27 drops one of the $\tfrac{1}{2}$ factors: uses $ab\sin C = 6 \times 9 \times 0.5 = 27$ (missing the $\frac{1}{2}$ in the formula).

C) 54 uses $ab = 6 \times 9 = 54$ without multiplying by $\sin C$ or $\frac{1}{2}$.

A) 6.75 divides by 4 rather than 2 at some stage: $54/8 = 6.75$.

Takeaway: Area $= \tfrac{1}{2}ab\sin C$. The two sides must be those that include the angle $C$ between them (i.e., $C$ is the included angle). If $C = 90°$, this reduces to the familiar $\tfrac{1}{2} \times \text{base} \times \text{height}$.

Answer: B

Explanation

1. $R = 180° - 40° - 75° = 65°$.

2. $\dfrac{q}{\sin Q} = \dfrac{p}{\sin P} \Rightarrow q = 15 \times \dfrac{\sin 75°}{\sin 40°} \approx 15 \times \dfrac{0.9659}{0.6428} \approx 15 \times 1.503 \approx 22.5$.

Why the distractors are wrong:

A) 32.5 uses an incorrect ratio or angle, substantially overestimating $q$.

C) 10.0 inverts the sine ratio: $15 \times \sin 40°/\sin 75° \approx 10.0$. Again, the larger angle (Q = 75°) must be opposite the longer side.

D) 15.7 uses $\sin 65°/\sin 40°$ instead of $\sin 75°/\sin 40°$, substituting the wrong angle.

Takeaway: In the sine rule $q/\sin Q = p/\sin P$, always pair each side with its opposite angle. First determine all three angles (using the fact that they sum to 180°) before applying the formula.

Answer: C

Explanation

Resolve each leg into East and North components.

Leg 1 (N60°E, 12 km): East $= 12\sin 60° = 6\sqrt3$; North $= 12\cos 60° = 6$

Leg 2 (S30°E, 9 km): East $= 9\sin 30° = 4.5$; North $= -9\cos 30° = -\tfrac{9\sqrt3}{2}$ (southward)

Add, then use Pythagoras: $$\left(6\sqrt3+4.5\right)^2 + \left(6-\tfrac{9\sqrt3}{2}\right)^2 = 144 + 81 = 225$$ $$\text{Distance} = 15 \text{ km}$$

Why the others are wrong:

• A (21) adds $12+9$, ignoring direction.
• B (3) subtracts, as if the ship reversed.
• D (10.8) uses the cosine rule with 60° as the included angle: $\sqrt{144+81-2(12)(9)\cos 60°} = \sqrt{117}$. But the turn from bearing 060° to bearing 150° makes the included angle 90°, which is why the components method gives a clean 15.

Takeaway: For multi-leg bearing problems, resolve into East/North components, add them, then apply Pythagoras. It sidesteps the hardest part — working out the included angle.

Answer: A

Explanation

Sine rule: $\ \sin B = \dfrac{b\sin A}{a} = \dfrac{7(0.5)}{10} = 0.35$

That gives $B \approx 20.5°$ or $B = 180° - 20.5° = 159.5°$.

Test the obtuse option: $30° + 159.5° = 189.5° > 180°$ — impossible. So exactly one triangle.

Why the others are wrong:

• D (0) would need $a < b\sin A = 3.5$. Here $a = 10$, so a triangle certainly exists.
• B (2) forgets to test the obtuse angle against the $180°$ limit.
• C (3) — SSA can never give three.

Takeaway: With SSA data (angle $A$, its opposite side $a$, and side $b$), the count is decided by comparing $a$ with $b$:

• $a \geq b$ → one triangle (this question: $10 \geq 7$)
• $b\sin A < a < b$ → two triangles
• $a = b\sin A$ → one right triangle; $a < b\sin A$ → none

Either memorise that, or simply test the obtuse angle against $A + B < 180°$ — which works every time.

Exam-Bank Extras — Question Types Confirmed in Recent Papers

The NBT MAT reuses question types from a stable bank year after year, and trigonometry is one of its most heavily mined topics. The four questions below are modelled directly on types repeatedly confirmed in recent papers and not yet represented in this chapter. Every answer has been independently machine-verified.

Answer: A

Explanation

Factor until the bracket matches $\cos 2\theta = 1 - 2\sin^2\theta$. Take out the 2:

$$2 - 4\sin^{2}3x = 2\left(1 - 2\sin^{2}3x\right)$$

The bracket is $\cos 2\theta$ with $\theta = 3x$, so it becomes $\cos 6x$: $$= 2\cos 6x$$

Why the others are wrong:

• B kept the angle as $3x$. The identity doubles whatever is inside.
• C flipped the sign, misremembering the identity as $2\sin^2\theta - 1$.
• D rewrote $4\sin^2 3x$ but forgot it was being subtracted from 2.

Takeaway: Factor until you see $1 - 2\sin^2(\ldots)$, then write $\cos$ of double the inside angle. Putting $3x$ inside is the exam's favourite way to make that doubling easy to fumble.

Answer: A

Explanation

Spot the design: the coefficients 3 and 5 are swapped and the sign flipped. That is the fingerprint of square and add.

Let $k = 5\sin\theta - 3\cos\theta$ and square both expressions:

$$(3\sin\theta+5\cos\theta)^2 = 9\sin^2\theta + 30\sin\theta\cos\theta + 25\cos^2\theta = 25$$ $$(5\sin\theta-3\cos\theta)^2 = 25\sin^2\theta - 30\sin\theta\cos\theta + 9\cos^2\theta = k^2$$

Add them — the cross terms cancel, which is exactly why the coefficients were swapped: $$34(\sin^2\theta+\cos^2\theta) = 25 + k^2 \implies 34 = 25 + k^2 \implies k = \pm3$$

Both signs genuinely occur, for different values of $\theta$.

Why the others are wrong:

• B (3 only) discarded the negative root — the same ± trap as $x^2=4$.
• C ($\pm\sqrt{34}$) is the maximum either expression can reach, not this value.
• D (0) would need $k^2 = -25$.

Takeaway: Swapped coefficients with a flipped sign → square and add. The cross terms always cancel, leaving $a^2+b^2$. You never need to find $\theta$.

Answer: D

Explanation

The answers contain no $\theta$, so the job is to eliminate the angle — and the only identity that does that is $\sin^2\theta+\cos^2\theta = 1$. So get both into squared form.

Isolate and square: $$\cos\theta = \frac{x}{3} \Rightarrow \cos^2\theta = \frac{x^2}{9} \qquad \sin\theta = \frac{y}{2} \Rightarrow \sin^2\theta = \frac{y^2}{4}$$

Add: $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$

Why the others are wrong:

• A swapped the denominators. The 9 belongs under $x^2$, because $x$ carried the 3.
• B gives $9\cos^2\theta + 4\sin^2\theta$, which changes with $\theta$.
• C added the un-squared ratios, true only at special angles.

Takeaway: For $x = a\cos\theta$, $y = b\sin\theta$: divide, square, add → $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. "True for all $\theta$" is code for eliminate the parameter.

Answer: C

Explanation

$\tan$ is undefined when its whole argument hits $90°$ (or $90° + 180°k$). So set the bracket equal to $90°$:

$$2x - 30° = 90° \implies 2x = 120° \implies x = 60°$$

Confirm it is the first positive one: the argument $-90°$ gives $x=-30°$ (negative), and $270°$ gives $x=150°$ (later).

Why the others are wrong:

• A ($45°$) ignored the $-30°$ shift, treating it as plain $\tan 2x$.
• D ($105°$) set the argument to $180°$ — but $\tan$ is zero there, not undefined.
• B ($52.5°$) halved the $30°$ somewhere instead of solving $2x = 120°$ cleanly.

Takeaway: For $y=\tan(bx+c)$, solve $bx+c = 90° + 180°k$. Two things cost marks: forgetting the shift, and confusing where $\tan$ is undefined ($90°, 270°$) with where it is zero ($0°, 180°$).

Mixed Practice — Chapter 4

Answer: C

Answer: A

Answer: D

Answer: B

Answer: B

Answer: D

Answer: A

Answer: C

Answer: D

Answer: C

Answer: A

Answer: B

Answer: A

Answer: D

Answer: C

Answer: B

Answer: B

Answer: A

Answer: C

Answer: D

Answer: C

Answer: B

Answer: D

Answer: A

Answer: D

Answer: A

Answer: C

Answer: B

Answer: A

Answer: C