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: $\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5}{12}\mathclose{.}$

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 $\cot \theta\mathclose{,}$ 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: C

Explanation

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

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

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

Why the distractors are wrong:

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

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

A) $7/25$ simply copies the given value of $\cos \theta\mathclose{.}$ 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 $\tan \theta\mathclose{.}$

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

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

Why the distractors are wrong:

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

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

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

Takeaway: The "co-" prefix signals a reciprocal: $\cot = 1/\tan\mathclose{,}$ $\csc = 1/\sin\mathclose{,}$ $\sec = 1/\cos\mathclose{.}$ 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 $\sin 60° = \text{opp}/\text{hyp}\mathclose{.}$

2. $\sin 60° = \dfrac{\sqrt{3}}{2}\mathclose{,}$ 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 $\sin 60°\mathclose{.}$ This gives the side adjacent to 60° (i.e., opposite the 30° angle).

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

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

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

Answer: D

Explanation

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

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

Why the others are wrong:

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

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: D

Explanation

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

Check with a triangle: $\sin\theta = \frac35\mathclose{,}$ 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 (1) borrowed from $\sin^2+\cos^2 = 1\mathclose{.}$ This identity does not equal 1.
• B $(\frac53)$ just repeats the given value.
• C $(\frac{16}{9})$ is $\cot^2\theta$ alone — forgot to add 1.

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

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

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 $1 + \tan^2\theta = \sec^2\theta\mathclose{,}$ then computing $\sin^2\theta \times \sec^2\theta = \sin^2\theta/\cos^2\theta = \tan^2\theta\mathclose{.}$ 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 $\tan\alpha = 3/4\mathclose{:}$ opp $= 3\mathclose{,}$ adj $= 4\mathclose{,}$ hyp $= 5\mathclose{.}$

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

3. Numerator: $3/5 - 4/5 = -1/5\mathclose{.}$

4. Denominator: $3/5 + 4/5 = 7/5\mathclose{.}$

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

Why the distractors are wrong:

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

A) $7$ inverts the fraction $(7/5 \div 1/5 = 7)\mathclose{,}$ 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., $\alpha < 45°)\mathclose{,}$ we have $\cos\alpha > \sin\alpha\mathclose{,}$ so $\sin\alpha - \cos\alpha$ is negative. Always assign a sign before simplifying fractions.

Section 4.2 — Trigonometry in Right-Angled Triangles

Answer: D

Explanation

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

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

3. $\theta = \cos^{-1}(0.5) = 60°\mathclose{.}$

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.

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

A) 27° uses $\tan^{-1}\left(\tfrac{4}{8}\right) \approx 26.6°\mathclose{.}$ The 4 m and the 8 m are the side next to the angle and the hypotenuse, so the ratio is cos, not tan.

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

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

Why the distractors are wrong:

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

C) 23.6° uses $\sin^{-1}(2/5) = \sin^{-1}(0.4) \approx 23.6°\mathclose{.}$ 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: D

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:

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

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: C

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 (13.4) is only the part below it.
• B (23.3) is only the part above eye level.
• D (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 $R\mathclose{,}$ so $PQ$ is the hypotenuse.

2. $PQ^2 = PR^2 + QR^2 = 9^2 + 12^2 = 81 + 144 = 225\mathclose{.}$

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: $9 + 12 = 21\mathclose{.}$ Adding sides never gives the hypotenuse except by coincidence.

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

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: D

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

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

Why the others are wrong:

• A (39.0) uses the difference of the two angles: $80 \times \tan(68° - 42°)\mathclose{.}$ Angles of elevation measured from two different points cannot be subtracted like that.
• B (52.8) adds the two horizontal distances, $\frac{h}{\tan 42°} + \frac{h}{\tan 68°} = 80\mathclose{.}$ Both sightings are from the same side of the tower, so the 80 m is their difference.
• C (72.0) is the numerator alone — stopped before solving for $x\mathclose{.}$

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

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 $\angle CAB = 67°\mathclose{.}$

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

Why the distractors are wrong:

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

D) 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.

B) 23.4 cm rounds $\sin 67°$ to $0.9$ before multiplying: $26 \times 0.9 = 23.4\mathclose{.}$ Round at the end, never in the middle — those extra decimals in the sine are worth half a centimetre here.

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 $\cos 60° = \dfrac{1}{2}\mathclose{.}$

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

Why the distractors are wrong:

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

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; $2 \times (\sqrt{2}/2) = \sqrt{2}\mathclose{,}$ but that applies to $\sin 45° + \cos 45°\mathclose{,}$ not this expression.

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

Answer: A

Explanation

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

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

Why the distractors are wrong:

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

C) $\sqrt{3}$ is $\tan 60°\mathclose{.}$

D) $1/\sqrt{2}$ equals $\sin 45° = \cos 45°\mathclose{.}$ 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 $\frac{\sqrt3}{2}\mathclose{,}$ so both squares are $\frac34\mathclose{:}$

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

Why the others are wrong:

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

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

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

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

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

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

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 $6/\sqrt{3} = 2\sqrt{3}\mathclose{,}$ 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 $\dfrac{\sin\theta}{\cos\theta} = 1\mathclose{,}$ i.e., $\tan\theta = 1\mathclose{.}$

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

Why the distractors are wrong:

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

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

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

Takeaway: $\sin\theta = \cos\theta$ is equivalent to $\tan\theta = 1\mathclose{.}$ 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. $\sin 75° = \sin 45°\cos 30° + \cos 45°\sin 30°$

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

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

Why the distractors are wrong:

A) $(\sqrt{6}-\sqrt{2})/4$ is $\sin 15° = \sin(45° - 30°)\mathclose{.}$ 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 $\sin 60°\mathclose{,}$ 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: B

Explanation

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

2. Reference angle $= 240° - 180° = 60°\mathclose{.}$

Why the distractors are wrong:

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

C) 120° is $360° - 240°\mathclose{,}$ 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 $360° - 60°\mathclose{,}$ 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: B

Explanation

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

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

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

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

Why the distractors are wrong:

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

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

C) $-5/13$ 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:

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

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 $\cos(90° - \theta)\mathclose{,}$ not $\cos(180° + \theta)\mathclose{.}$

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

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

Answer: C

Explanation

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

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

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

(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 $\sin 30°\mathclose{,}$ 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 $+\sqrt{3}/2\mathclose{.}$

Answer: B

Explanation

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

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

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

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: $\sin(360°-\theta) = -\sin\theta\mathclose{,}$ then $\cos(180°+\theta) = -\cos\theta\mathclose{,}$ then add.

Section 4.5 — Trigonometric Identities

Answer: A

Explanation

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

$1 - \sin^2\theta = \cos^2\theta\mathclose{.}$

Why the distractors are wrong:

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

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

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

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

Answer: D

Explanation

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

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

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 $\sin^2\theta/\cos\theta \cdot \cos\theta = \sin^2\theta\mathclose{,}$ not $\sin^2\theta - 1\mathclose{.}$

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

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

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 $1 - \cos^2\theta = \sin^2\theta\mathclose{.}$

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

Why the distractors are wrong:

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

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

D) $5/4$ is $1/\sin\theta = \csc\theta\mathclose{,}$ 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.