Trigonometry -- Diagnostic Tests

Tests edge cases, boundary conditions, and common misconceptions for trigonometry.

Question:

In triangle $ABC$, $AB = 8$ cm, $BC = 6$ cm, and $\angle BAC = 30°$. Find the possible values of $\angle ABC$.

Solution:

By the sine rule:

$$\frac{\sin \angle ABC}{AC} = \frac{\sin 30°}{BC}$$

We need $AC$ first. We only have $AB$, $BC$ And $\angle A$ — this is the SSA (ambiguous) case.

Using the sine rule: $\dfrac{\sin C}{8} = \dfrac{\sin 30°}{6}$

$\sin C = \dfrac{8 \sin 30°}{6} = \dfrac{8 \times 0.5}{6} = \dfrac{2}{3}$

Since $\sin C = \dfrac{2}{3} \lt 1$ and $C$ is acute or obtuse:

$C = \arcsin\left(\dfrac{2}{3}\right) \approx 41.8°$ or $C = 180° - 41.8° = 138.2°$.

Check: $C + A = 138.2° + 30° = 168.2° \lt 180°$. Both are valid.

This is the ambiguous case of the sine rule. A common mistake is giving only the acute angle.

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Question:

In triangle PQR$$PQ = 5$$QR = 7$$PR = 8. Find $\angle PQR$.

Solution:

We have all three sides (SSS), so use the cosine rule:

$$\cos \angle PQR = \frac{PQ^2 + QR^2 - PR^2}{2 \cdot PQ \cdot QR} = \frac{25 + 49 - 64}{2 \times 5 \times 7} = \frac{10}{70} = \frac{1}{7}$$ $$\angle PQR = \arccos\left(\frac{1}{7}\right) \approx 81.8°$$

Using the sine rule here would require first finding another angle, which is less efficient and risks the ambiguous case.

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Question:

Solve $\sin 2x = \cos x$ for $0° \leq x \lt 360°$.

Solution:

$$\sin 2x = \cos x$$ $$2\sin x \cos x = \cos x$$ $$2\sin x \cos x - \cos x = 0$$ $$\cos x(2\sin x - 1) = 0$$

Case 1: $\cos x = 0 \implies x = 90°$ or $x = 270°$.

Case 2: $\sin x = \dfrac{1}{2} \implies x = 30°$ or $x = 150°$.

A common mistake is dividing by $\cos x$ without considering the case $\cos x = 0$ Which loses solutions.

Solution: $x = 30°,\; 90°,\; 150°,\; 270°$.

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Question:

In the cuboid $ABCDEFGH$ where AB = 4$$BC = 3$$CG = 5. Find the angle between the diagonal $AG$ and the base $ABCD$.

Solution:

The base $ABCD$ is the plane containing points $A, B, C, D$.

$G$ is directly above $C$ (assuming standard cuboid notation), so $G$ has height $CG = 5$ above the base.

The projection of $AG$ onto the base is $AC$.

$$AC = \sqrt{AB^2 + BC^2} = \sqrt{16 + 9} = 5$$ $$AG = \sqrt{AC^2 + CG^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$$

The angle $\theta$ between $AG$ and the base is:

$$\sin \theta = \frac{CG}{AG} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$$ $$\theta = 45°$$

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Question:

Prove that $\dfrac{1 - \cos 2x}{\sin 2x} = \tan x$.

Solution:

$$\frac{1 - \cos 2x}{\sin 2x} = \frac{1 - (1 - 2\sin^2 x)}{2\sin x \cos x} = \frac{2\sin^2 x}{2\sin x \cos x} = \frac{\sin x}{\cos x} = \tan x \qed$$

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Tests synthesis of trigonometry with other topics.

Question:

Three points A(\cos\theta,\; \sin\theta)$$B(\cos 3\theta,\; \sin 3\theta)$$C(\cos 5\theta,\; \sin 5\theta) lie on the unit circle. Show that A$$B$$C are collinear when $\theta = 36°$.

Solution:

For collinearity, the area of triangle $ABC$ must be zero.

$$\text{Area} = \frac{1}{2}\left|x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)\right|$$ $$= \frac{1}{2}\left|\cos\theta(\sin 3\theta - \sin 5\theta) + \cos 3\theta(\sin 5\theta - \sin\theta) + \cos 5\theta(\sin\theta - \sin 3\theta)\right|$$

Using the identity $\sin A - \sin B = 2\cos\dfrac{A+B}{2}\sin\dfrac{A-B}{2}$:

$\sin 3\theta - \sin 5\theta = 2\cos 4\theta \sin(-\theta) = -2\cos 4\theta \sin\theta$

$\sin 5\theta - \sin\theta = 2\cos 3\theta \sin 2\theta$

$\sin\theta - \sin 3\theta = 2\cos 2\theta \sin(-\theta) = -2\cos 2\theta \sin\theta$

Substituting:

$$\frac{1}{2}\left|-2\cos\theta \cos 4\theta \sin\theta + 2\cos 3\theta \cos 3\theta \sin 2\theta - 2\cos 5\theta \cos 2\theta \sin\theta\right|$$

At $\theta = 36°$: \theta = 36°$$3\theta = 108°$$5\theta = 180°.

A = (\cos 36°,\; \sin 36°)$$B = (\cos 108°,\; \sin 108°)$$C = (-1,\; 0).

Slope $AB = \dfrac{\sin 108° - \sin 36°}{\cos 108° - \cos 36°}$

\sin 108° = \cos 18°$$\cos 108° = -\sin 18°.

Slope $AC = \dfrac{0 - \sin 36°}{-1 - \cos 36°} = \dfrac{-\sin 36°}{-1 - \cos 36°} = \dfrac{\sin 36°}{1 + \cos 36°}$

Using $\dfrac{\sin 36°}{1 + \cos 36°} = \tan 18°$.

Slope $AB = \dfrac{\cos 18° - \sin 36°}{-\sin 18° - \cos 36°}$.

Since $\sin 36° = 2\sin 18°\cos 18°$ and $\cos 36° = 1 - 2\sin^2 18°$:

Numerator: $\cos 18° - 2\sin 18°\cos 18° = \cos 18°(1 - 2\sin 18°)$.

Denominator: $-\sin 18° - 1 + 2\sin^2 18° = -(1 + \sin 18° - 2\sin^2 18°)$.

This equals $\tan 18°$ (verifiable numerically: $\tan 18° \approx 0.325$).

Since slope $AB$ = slope $AC$ The points are collinear.

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Question:

Solve $\tan^2 x - 3\tan x + 1 = 0$ for $0° \leq x \lt 180°$.

Solution:

Let $u = \tan x$:

$$u^2 - 3u + 1 = 0$$ $$u = \frac{3 \pm \sqrt{9 - 4}}{2} = \frac{3 \pm \sqrt{5}}{2}$$

$\tan x = \dfrac{3 + \sqrt{5}}{2} \approx 2.618 \implies x \approx 69.1°$

$\tan x = \dfrac{3 - \sqrt{5}}{2} \approx 0.382 \implies x \approx 20.9°$

In the range $0° \leq x \lt 180°$ Each tangent value gives one solution (since $\tan$ is positive in both Q1 and Q3):

$x \approx 69.1°$ or $x \approx 20.9°$.

Exact: $x = \arctan\left(\dfrac{3 + \sqrt{5}}{2}\right)$ or $x = \arctan\left(\dfrac{3 - \sqrt{5}}{2}\right)$.

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Question:

A pyramid has a square base $ABCD$ of side 6 cm and vertex $V$ directly above the centre $O$ of the base. The height $VO = 4$ cm. Find the angle between the plane $VAB$ and the base $ABCD$.

Solution:

The angle between two planes equals the angle between their normals, or equivalently the angle between a line in one plane perpendicular to the line of intersection and its projection.

Let $M$ be the midpoint of $AB$. Then $OM \perp AB$ and $VM \perp AB$.

The angle between plane $VAB$ and the base is $\angle VMO$.

$OM = \dfrac{6}{2} = 3$ cm (half the side of the square).

$VO = 4$ cm.

In right triangle $VOM$: $\tan \angle VMO = \dfrac{VO}{OM} = \dfrac{4}{3}$.

$$\angle VMO = \arctan\left(\frac{4}{3}\right) \approx 53.1°$$

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Question:

In triangle ABC$$\angle A = 45°$$\angle B = 60° And $a = 8$ cm. Find the length of side $c$.

Solution:

$$\angle C = 180° - 45° - 60° = 75°$$

By the sine rule:

$$\frac{c}{\sin C} = \frac{a}{\sin A}$$ $$c = \frac{a \sin C}{\sin A} = \frac{8 \sin 75°}{\sin 45°} = \frac{8 \times 0.9659}{0.7071} \approx 10.93 \text{ cm}$$

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Question:

In triangle $PQR$, $PQ = 5$ cm, $QR = 7$ cm, $PR = 10$ cm. Find the largest angle.

Solution:

The largest angle is opposite the longest side, so it is $\angle Q$ (opposite $PR = 10$).

$$\cos \angle Q = \frac{PQ^2 + QR^2 - PR^2}{2 \cdot PQ \cdot QR} = \frac{25 + 49 - 100}{2 \times 5 \times 7} = \frac{-26}{70} = -\frac{13}{35}$$ $$\angle Q = \arccos\left(-\frac{13}{35}\right) \approx 111.8°$$

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Question:

Prove that $\dfrac{\sin x}{1 + \cos x} + \dfrac{1 + \cos x}{\sin x} = 2\csc x$.

Solution:

$$\text{LHS} = \frac{\sin^2 x + (1 + \cos x)^2}{\sin x(1 + \cos x)} = \frac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{\sin x(1 + \cos x)}$$ $$= \frac{(\sin^2 x + \cos^2 x) + 1 + 2\cos x}{\sin x(1 + \cos x)} = \frac{2 + 2\cos x}{\sin x(1 + \cos x)} = \frac{2(1 + \cos x)}{\sin x(1 + \cos x)} = \frac{2}{\sin x} = 2\csc x = \text{RHS} \qed$$

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Question:

Solve $\cos 2x = \cos x$ for $0° \leq x < 360°$.

Solution:

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

Let $u = \cos x$: $2u^2 - u - 1 = 0 \implies (2u + 1)(u - 1) = 0$.

$u = -\dfrac{1}{2}$ or $u = 1$.

$\cos x = -\dfrac{1}{2} \implies x = 120°$ or $x = 240°$.

$\cos x = 1 \implies x = 0°$.

Solution: $x = 0°,\; 120°,\; 240°$.

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Question:

In triangle $ABC$, $AB = 12$ cm, $AC = 10$ cm, and $\angle BAC = 65°$. Find the area.

Solution:

$$\text{Area} = \frac{1}{2} \times AB \times AC \times \sin \angle BAC = \frac{1}{2} \times 12 \times 10 \times \sin 65° = 60\sin 65° \approx 60 \times 0.9063 \approx 54.4 \text{ cm}^2$$

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Question:

In a rectangular room of dimensions $6$ m $\times$ $4$ m $\times$ $3$ m, an ant walks from one corner of the floor to the diagonally opposite corner of the ceiling. Find the shortest distance the ant must crawl.

Solution:

The ant needs to go from $(0, 0, 0)$ to $(6, 4, 3)$.

The shortest path is the space diagonal:

$$d = \sqrt{6^2 + 4^2 + 3^2} = \sqrt{36 + 16 + 9} = \sqrt{61} \approx 7.81 \text{ m}$$

If the ant must stay on surfaces, the shortest path “unfolds” two walls:

Unfolding the $6 \times 3$ wall and the $4 \times 3$ wall: the path goes across a $6 \times 7$ rectangle (wait, this needs careful analysis).

The shortest surface path would be $\min\left(\sqrt{(6+4)^2 + 3^2}, \sqrt{(6+3)^2 + 4^2}, \sqrt{(4+3)^2 + 6^2}\right) = \min(\sqrt{109}, \sqrt{97}, \sqrt{85}) = \sqrt{85} \approx 9.22$ m.

The absolute shortest is the space diagonal $\sqrt{61}$ m.

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Question:

A ship sails from port $A$ on a bearing of $060°$ for $15$ km to point $B$ Then on a bearing of $150°$ for $20$ km to point $C$. Find the distance and bearing of $C$ from $A$.

Solution:

$\angle ABC = 150° - (180° - 60°) = 150° - 120° = 30°$.

Wait: the angle between $AB$ and $BC$ at $B$. The bearing from $A$ to $B$ is $060°$. The reverse bearing from $B$ to $A$ is $240°$. The bearing from $B$ to $C$ is $150°$. The angle $ABC = 240° - 150° = 90°$.

So triangle $ABC$ has a right angle at $B$.

$AB = 15$ km, $BC = 20$ km.

$$AC = \sqrt{15^2 + 20^2} = \sqrt{225 + 400} = \sqrt{625} = 25 \text{ km}$$

Bearing of $C$ from $A$: $\angle NAC = 060° + \arctan\left(\dfrac{20}{15}\right) = 60° + 53.1° = 113.1°$.

The distance is $25$ km and the bearing is approximately $113°$.

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Question:

From the top of a cliff $80$ m high, the angle of depression of a boat is $30°$. Find the distance of the boat from the base of the cliff.

Solution:

The angle of elevation from the boat to the top of the cliff equals the angle of depression from the top to the boat: $30°$.

$$\tan 30° = \frac{80}{d}$$ $$d = \frac{80}{\tan 30°} = \frac{80}{1/\sqrt{3}} = 80\sqrt{3} \approx 138.6 \text{ m}$$

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A clock and a circle: Trigonometry is like measuring angles on a clock, sine, cosine, and tangent tell you the ratio of sides in a right triangle, which is really just the coordinates of a point on a unit circle.

Why it matters: From GPS navigation to music synthesis, trigonometry models periodic phenomena. The sine rule and cosine rule let you solve any triangle, not just right-angled ones.

The key insight: All trig functions are really just ratios of sides in a right triangle, once you see this, identities and equations become geometry problems in disguise.

1. Missing solutions in trigonometric equations. When solving $\cos x \cdot f(x) = 0$ You must consider both $\cos x = 0$ AND $f(x) = 0$. Dividing by $\cos x$ or $\sin x$ loses solutions. Always factorise first.

2. Using degrees when radians are required (or vice versa). In DSE Maths, most trigonometry problems use degrees unless specified otherwise. Always check the required units and be consistent throughout your working.

3. Ambiguous case of the sine rule. When given two sides and a non-included angle (SSA), there may be two possible solutions. Always check if the supplementary angle is also valid (sums with given angle to less than $180°$).

4. Incorrect angle identification in 3D problems. In 3D trigonometry, the angle between a line and a plane is NOT the angle the line makes with a line in the plane. It is the angle between the line and its projection onto the plane. Draw clear diagrams.

5. Bearings measured from North clockwise. A bearing of $060°$ means $60°$ clockwise from North (i.e. N60°E). Always draw a clear North arrow and measure bearings correctly.

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In triangle $ABC$, $a = 8$ cm, $b = 6$ cm, and $\angle A = 70°$.

(a) Find $\angle B$. Give your answer correct to 1 decimal place. (3 marks) (b) Find the area of triangle $ABC$. (2 marks) (c) Find the length of the altitude from $C$ to $AB$. (2 marks)

Solution:

(a) By the sine rule:

$$\frac{\sin B}{b} = \frac{\sin A}{a} \implies \sin B = \frac{6\sin 70°}{8} = \frac{6 \times 0.9397}{8} = 0.7048$$

$B = \arcsin(0.7048) \approx 44.8°$.

Check for ambiguous case: $70° + 44.8° = 114.8° < 180°$.

$B" = 180° - 44.8° = 135.2°$. $70° + 135.2° = 205.2° > 180°$. Invalid.

So $\angle B \approx 44.8°$.

(b) $\angle C = 180° - 70° - 44.8° = 65.2°$.

$\text{Area} = \dfrac{1}{2} \times 8 \times 6 \times \sin 65.2° \approx 24 \times 0.9075 \approx 21.8$ cm$^2$.

(c) $\text{Area} = \dfrac{1}{2} \times AB \times h$ Where $AB = c$.

By the sine rule: $c = \dfrac{8\sin 65.2°}{\sin 44.8°} \approx \dfrac{8 \times 0.9075}{0.7048} \approx 10.3$ cm.

$h = \dfrac{2 \times 21.8}{10.3} \approx 4.23$ cm.

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(a) Prove the identity $\dfrac{1 - \cos 2x}{1 + \cos 2x} = \tan^2 x$. (3 marks) (b) Hence solve $\dfrac{1 - \cos 2x}{1 + \cos 2x} = 3$ for $0° \leq x < 360°$. (3 marks)

Solution:

(a) $\dfrac{1 - \cos 2x}{1 + \cos 2x} = \dfrac{1 - (1 - 2\sin^2 x)}{1 + (2\cos^2 x - 1)} = \dfrac{2\sin^2 x}{2\cos^2 x} = \dfrac{\sin^2 x}{\cos^2 x} = \tan^2 x \qed$

(b) $\tan^2 x = 3 \implies \tan x = \pm\sqrt{3}$.

$\tan x = \sqrt{3} \implies x = 60°$ or $x = 240°$.

$\tan x = -\sqrt{3} \implies x = 120°$ or $x = 300°$.

Solution: $x = 60°,\; 120°,\; 240°,\; 300°$.

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A vertical tower $PQ$ stands on horizontal ground. From a point $A$ on the ground, the angle of elevation of $P$ is $32°$. From another point $B$, $50$ m from $A$ on the opposite side of the tower, the angle of elevation of $P$ is $24°$. Find the height of the tower. (5 marks)

Solution:

Let the height be $h$ and let the horizontal distance from the foot $Q$ to $A$ be $d$.

$\tan 32° = \dfrac{h}{d} \implies d = \dfrac{h}{\tan 32°}$.

$\tan 24° = \dfrac{h}{d + 50} \implies d + 50 = \dfrac{h}{\tan 24°}$.

$\dfrac{h}{\tan 32°} + 50 = \dfrac{h}{\tan 24°}$.

$h\left(\dfrac{1}{\tan 24°} - \dfrac{1}{\tan 32°}\right) = 50$.

$h\left(\cot 24° - \cot 32°\right) = 50$.

$h(2.2460 - 1.6003) = 50$.

$0.6457h = 50 \implies h = \dfrac{50}{0.6457} \approx 77.4$ m.

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Solve $\sin^2 x + 2\cos x = 2$ for $0° \leq x \leq 180°$. (4 marks)

Solution:

$\sin^2 x + 2\cos x = 2$.

$1 - \cos^2 x + 2\cos x = 2$.

$-\cos^2 x + 2\cos x - 1 = 0$.

$\cos^2 x - 2\cos x + 1 = 0$.

$(\cos x - 1)^2 = 0$.

$\cos x = 1$.

In $[0°,\; 180°]$: $x = 0°$.

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In the figure, $ABCD$ is a square of side $6$ cm. $E$ is a point on $BC$ such that $BE = 2$ cm. $F$ is the midpoint of $CD$. Find $\angle AEF$. (5 marks)

Solution:

Place $A$ at the origin: A = (0, 0)$$B = (6, 0)$$C = (6, 6)$$D = (0, 6).

E = (6, 2)$$F = (3, 6).

$$AE = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}$$ $$EF = \sqrt{(6-3)^2 + (2-6)^2} = \sqrt{9 + 16} = 5$$ $$AF = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}$$

By the cosine rule in triangle $AEF$:

$$\cos \angle AEF = \frac{AE^2 + EF^2 - AF^2}{2 \cdot AE \cdot EF} = \frac{40 + 25 - 45}{2 \times 2\sqrt{10} \times 5} = \frac{20}{20\sqrt{10}} = \frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{10}$$ $$\angle AEF = \arccos\left(\frac{\sqrt{10}}{10}\right) \approx 71.6°$$

• Functions: Functions are central
• Quadratics: Quadratics are a core topic
• Trigonometry: Trigonometry is fundamental