Trigonometry — Mathematics Grade 12 | Learn on Stanmorephysics

11 exam-style question sets on Trigonometry (Paper 2), each with a hint and a fully worked answer. The app holds 55 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Exact values, identities, compound-angle deduction & triple anglesSine & cosine graphs: range on an interval, graphical inequalities & transformations3-D problems in terms of parameters: area, sine & cosine rulesCompound & double angles: terminal arm and expressions in terms of pIdentities, reduction formulae & trigonometric equationsTrigonometric graphs & general solutionsSine rule in 2-D and 3-D problemsExpressions in terms of k, reduction formulae & square-root domainsIdentities with reductions & trig arithmetic sequences (general solutions)Trigonometric graphs & inequalitiesCosine rule in 2-D and 3-D problems

Practise this section in the app →

Question 1

Exact values, identities, compound-angle deduction & triple anglesRoutine31 marksPaper 2

Given: $sin β = \frac{2}{5}$ , where $β \in [9 0^{\circ}; 27 0^{\circ}]$ .

Answer (a) to (c) WITHOUT using a calculator, leaving answers in simplest surd form where necessary. Show ALL your working in every part of this question.

aWith the aid of a sketch, determine the value of $cos β$ . (3)
bDetermine the value of $sin 2 β$ . (3)
cDetermine the value of $cos (99 0^{\circ} - β)$ . (3)
dProve the identity: $\frac{sin ^{4} x + sin ^{2} x cos ^{2} x}{1 - cos x} = 1 + cos x$ (4)
eDetermine ALL the values of $x$ in the interval $[0^{\circ}; 36 0^{\circ}]$ for which the left-hand side of the identity in (d) is undefined. (2)
fHence, determine the minimum value of $\frac{sin ^{4} x + sin ^{2} x cos ^{2} x}{1 - cos x}$ . (2)
gUse the identity $cos (A - B) = cos A cos B + sin A sin B$ to deduce the expansion of $sin (A + B)$ . (3)
hHence, determine the general solution of $sin x cos 1 5^{\circ} + cos x sin 1 5^{\circ} = cos 4 x$ . (5)
iSimplify $\frac{cos 3 x - cos x}{cos 2 x - 1}$ to a single trigonometric ratio. (6)

Hint

Decide in which part of $[9 0^{\circ}; 27 0^{\circ}]$ the sine is positive, then use $x^{2} + y^{2} = r^{2}$ with the correct sign for $x$ . In (c), first remove whole revolutions of $36 0^{\circ}$ . In (d), take out the common factor $sin^{2} x$ and look for a difference of two squares. In (g), start from $sin (A + B) = cos [9 0^{\circ} - (A + B)]$ ; in (h), write $cos 4 x$ as the sine of a related angle; in (i), expand $cos 3 x$ as $cos (2 x + x)$ .

Worked answer

a) $sin β = \frac{2}{5} > 0$ and $β \in [9 0^{\circ}; 27 0^{\circ}]$ , so $β$ lies in the second quadrant, with $y = 2$ and $r = 5$ ✓
$x^{2} + 2^{2} = 5^{2}$ , so $x^{2} = 21$ and $x = - 21$ (second quadrant) ✓
$cos β = \frac{x}{r} = - \frac{21}{5}$ ✓

b) $sin 2 β = 2 sin β cos β$ ✓
$= 2 (\frac{2}{5}) (- \frac{21}{5})$ ✓
$= - \frac{4 21}{25}$ ✓

c) $cos (99 0^{\circ} - β) = cos (72 0^{\circ} + (27 0^{\circ} - β)) = cos (27 0^{\circ} - β)$ ✓
$= - sin β$ ✓
$= - \frac{2}{5}$ ✓

d) LHS $= \frac{sin ^{2} x ( sin ^{2} x + cos ^{2} x )}{1 - cos x}$ ✓ (common factor $sin^{2} x$ )
$= \frac{sin ^{2} x}{1 - cos x}$ ✓ ( $sin^{2} x + cos^{2} x = 1$ )
$= \frac{1 - cos ^{2} x}{1 - cos x} = \frac{( 1 - cos x ) ( 1 + cos x )}{1 - cos x}$ ✓ (difference of squares)
$= 1 + cos x =$ RHS ✓

e) The left-hand side is undefined when $1 - cos x = 0$ , i.e. $cos x = 1$ ✓
$x = 0^{\circ}$ or $x = 36 0^{\circ}$ ✓

f) Wherever it is defined, the expression equals $1 + cos x$ , and $- 1 \leq cos x \leq 1$ ✓
Minimum value $= 1 + (- 1) = 0$ , reached at $x = 18 0^{\circ}$ , where $1 - cos x = 2 \neq = 0$ , so this value is allowed ✓

g) $sin (A + B) = cos [9 0^{\circ} - (A + B)] = cos [(9 0^{\circ} - A) - B]$ ✓
$= cos (9 0^{\circ} - A) cos B + sin (9 0^{\circ} - A) sin B$ ✓
$= sin A cos B + cos A sin B$ ✓

h) By (g), the left-hand side is $sin (x + 1 5^{\circ})$ ✓
$sin (x + 1 5^{\circ}) = cos 4 x = sin (9 0^{\circ} - 4 x)$ ✓
$x + 1 5^{\circ} = 9 0^{\circ} - 4 x + k \cdot 36 0^{\circ}$ , so $5 x = 7 5^{\circ} + k \cdot 36 0^{\circ}$ and $x = 1 5^{\circ} + k \cdot 7 2^{\circ}$ ✓
or $x + 1 5^{\circ} = 18 0^{\circ} - (9 0^{\circ} - 4 x) + k \cdot 36 0^{\circ}$ ✓
$x + 1 5^{\circ} = 9 0^{\circ} + 4 x + k \cdot 36 0^{\circ}$ , so $- 3 x = 7 5^{\circ} + k \cdot 36 0^{\circ}$ and $x = - 2 5^{\circ} + k \cdot 12 0^{\circ}$ (replacing $k$ by $- k$ ); $k \in Z$ ✓

i) Numerator: $cos 3 x - cos x = cos (2 x + x) - cos x$ ✓
$= cos 2 x cos x - sin 2 x sin x - cos x$ ✓
$= cos x (cos 2 x - 1) - 2 sin^{2} x cos x$ ✓
$= cos x (- 2 sin^{2} x) - 2 sin^{2} x cos x = - 4 sin^{2} x cos x$ ✓
Denominator: $cos 2 x - 1 = (1 - 2 sin^{2} x) - 1 = - 2 sin^{2} x$ ✓
$\frac{- 4 sin ^{2} x cos x}{- 2 sin ^{2} x} = 2 cos x$ ✓

Modelled on Nov 2023 Paper 2, Question 5 — same skills, new scenario.

Question 2

Sine & cosine graphs: range on an interval, graphical inequalities & transformationsComplex12 marksPaper 2

The diagram shows the graphs of $f (x) = - 4 cos 2 x$ and $g (x) = sin (x + 3 0^{\circ})$ for $0^{\circ} \leq x \leq 18 0^{\circ}$ . The point $A (3 0^{\circ}; - 2)$ lies on the graph of $f$ , the graph with the greater amplitude.

aWrite down the period of $f$ . (1)
bDetermine the range of $g$ for $x \in [0^{\circ}; 18 0^{\circ}]$ . (2)
cUse the graphs to find the interval(s) in which $f (x) \cdot g (x) > 0$ . (2)
dFor which values of $x$ in the interval is $f (x) + 2 \leq 0$ ? (2)
eThe graph of $f$ is shifted $1 5^{\circ}$ to the LEFT to form the graph of $p$ , so that $p (x) = f (x + 1 5^{\circ})$ for $x \in [0^{\circ}; 18 0^{\circ}]$ . The point $B (t; - 2)$ lies on the graph of $p$ . WITHOUT using a calculator, use the point $A$ and the symmetry of the graph of $f$ to determine the possible values of $t$ . (3)
fThe graph of $g$ is translated $6 0^{\circ}$ to the LEFT to form the graph of $h$ . Write down the equation of $h$ in its simplest form. (2)

This question has a diagram, shown in the app.

Hint

Find where each graph cuts the $x$ -axis and compare their signs region by region: a product is positive where both factors have the same sign, so test the endpoints of the interval as well. For (d), the graph of $f$ is symmetric about the vertical line through its maximum. For (e), $p (t) = f (t + 1 5^{\circ})$ , so you need the angles between $1 5^{\circ}$ and $19 5^{\circ}$ at which $f$ equals $- 2$ .

Worked answer

a) Period of $f = \frac{36 0 ^{\circ}}{2} = 18 0^{\circ}$ ✓

b) $g (0^{\circ}) = sin 3 0^{\circ} = \frac{1}{2}$ ; $g$ rises to its maximum $g (6 0^{\circ}) = sin 9 0^{\circ} = 1$ ✓ and then falls to $g (18 0^{\circ}) = sin 21 0^{\circ} = - \frac{1}{2}$ at the right-hand endpoint ✓, never reaching $- 1$ on this interval.
Range: $- \frac{1}{2} \leq y \leq 1$ , i.e. $y \in [- \frac{1}{2}; 1]$

c) $f (x) = 0$ when $cos 2 x = 0$ : $2 x = 9 0^{\circ}$ or $27 0^{\circ}$ , i.e. $x = 4 5^{\circ}$ or $x = 13 5^{\circ}$ ; $g (x) = 0$ when $x + 3 0^{\circ} = 18 0^{\circ}$ , i.e. $x = 15 0^{\circ}$ .
On $[0^{\circ}; 4 5^{\circ})$ , $f < 0$ while $g > 0$ ; on $(4 5^{\circ}; 13 5^{\circ})$ both graphs lie above the $x$ -axis; on $(13 5^{\circ}; 15 0^{\circ})$ , $f < 0$ while $g > 0$ ; on $(15 0^{\circ}; 18 0^{\circ}]$ both lie below the $x$ -axis, and at the endpoint $f (18 0^{\circ}) \cdot g (18 0^{\circ}) = (- 4) (- \frac{1}{2}) = 2 > 0$ .
$4 5^{\circ} < x < 13 5^{\circ}$ ✓ or $15 0^{\circ} < x \leq 18 0^{\circ}$ ✓

d) $f (x) + 2 \leq 0 ⟺ f (x) \leq - 2$ . From $A$ , $f (3 0^{\circ}) = - 2$ ; the graph of $f$ is symmetric about its maximum at $x = 9 0^{\circ}$ , so $f (15 0^{\circ}) = - 2$ as well. The graph lies on or below the line $y = - 2$ from the left-hand endpoint, where $f (0^{\circ}) = - 4$ , up to $A$ , and again from $x = 15 0^{\circ}$ to the right-hand endpoint.
$0^{\circ} \leq x \leq 3 0^{\circ}$ ✓ or $15 0^{\circ} \leq x \leq 18 0^{\circ}$ ✓

e) $p (t) = - 2 \Rightarrow f (t + 1 5^{\circ}) = - 2$ ✓, where $1 5^{\circ} \leq t + 1 5^{\circ} \leq 19 5^{\circ}$ because $0^{\circ} \leq t \leq 18 0^{\circ}$ .
From (d), $f = - 2$ at $x = 3 0^{\circ}$ (the point $A$ ) and at $x = 15 0^{\circ}$ (symmetry about the maximum). Every other solution differs from one of these by a multiple of the period $18 0^{\circ}$ (for example $21 0^{\circ}$ or $- 3 0^{\circ}$ ) and falls outside this range, so $t + 1 5^{\circ} = 3 0^{\circ}$ or $t + 1 5^{\circ} = 15 0^{\circ}$ ✓
$t = 1 5^{\circ}$ or $t = 13 5^{\circ}$ ✓
(Check: $p (x) = - 4 cos (2 x + 3 0^{\circ})$ , and $- 4 cos 6 0^{\circ} = - 4 cos 30 0^{\circ} = - 2$ .)

f) $h (x) = g (x + 6 0^{\circ}) = sin (x + 6 0^{\circ} + 3 0^{\circ}) = sin (x + 9 0^{\circ})$ ✓
$h (x) = cos x$ ✓

Modelled on Nov 2023 Paper 2, Question 6 — same skills, new scenario.

Question 3

3-D problems in terms of parameters: area, sine & cosine rulesRoutine7 marksPaper 2

A tower crane is being put up on a building site in Umhlanga, KwaZulu-Natal. The crane's mast is represented by the vertical line $M N$ : its foot $M$ is on the level ground and $N$ is the top of the mast. Reinforcing steel is stored in a triangular yard $M G H$ , one corner of which is at the foot $M$ of the mast. The points $M$ , $G$ and $H$ lie in the same horizontal plane, and $N \hat{M} G = N \hat{M} H = 9 0^{\circ}$ . The site plan gives the following:

the side $M G$ of the yard is $d$ metres long, and $G \hat{M} H = θ$
the yard covers an area of $k$ m²
from the corner $H$ , the top of the mast is seen at an angle of elevation of $x$ , that is $N \hat{H} M = x$

aUse the area of the yard to express $M H$ in terms of $k$ , $d$ and $θ$ . (2)
bHence show that the height of the mast is $M N = \frac{2 k tan x}{d sin θ}$ . (2)
cThe mast is $52$ m tall. On site, the foreman measures $d = 30$ m, $θ = 6 4^{\circ}$ and $x = 5 4^{\circ}$ . Calculate $k$ , the area of the yard, correct to TWO decimal places. (3)

Hint

One corner of the yard is at $M$ , so use the area rule with the two sides that meet at $M$ and the angle between them. The mast is vertical, which makes $△ N M H$ right-angled at $M$ , with the angle $x$ at $H$ . In (c) the unknown is the area: substitute the site values into the formula from (b) and make $k$ the subject.

Worked answer

a) Area rule in $△ M G H$ : $k = \frac{1}{2} \cdot M G \cdot M H \cdot sin (G \hat{M} H) = \frac{1}{2} d \cdot M H \cdot sin θ$ ✓
$\Rightarrow M H = \frac{2 k}{d sin θ}$ ✓

b) In $△ N M H$ : $N \hat{M} H = 9 0^{\circ}$ and $N \hat{H} M = x$ , so $tan x = \frac{M N}{M H}$ ✓
$\Rightarrow M N = M H tan x = \frac{2 k}{d sin θ} \cdot tan x = \frac{2 k tan x}{d sin θ}$ ✓

c) $52 = \frac{2 k tan 5 4 ^{\circ}}{30 sin 6 4 ^{\circ}}$ ✓
$k = \frac{52 \times 30 sin 6 4 ^{\circ}}{2 tan 5 4 ^{\circ}}$ ✓
$k = 509, 35$ m² ✓

Modelled on Nov 2023 Paper 2, Question 7 — same skills, new scenario.

Question 4

Compound & double angles: terminal arm and expressions in terms of pComplex12 marksPaper 2

$O$ is the origin and $T (- 8; - 15)$ is a point in the third quadrant. $O T$ is the terminal arm of angle $A$ , which is measured anticlockwise from the positive $x$ -axis. Answer this question WITHOUT using a calculator.

aDetermine the value of $cos A$ . (2)
bDetermine the value of $cos 2 A$ . (2)
cIf $sin B = \frac{3}{5}$ , where $9 0^{\circ} < B < 18 0^{\circ}$ , determine the value of $sin (A - B)$ . (4)
dGiven that $cos α = p$ , express $sin (\frac{α}{2} - 4 5^{\circ}) \cdot cos (\frac{α}{2} - 4 5^{\circ})$ in terms of $p$ . (4)

Hint

First find the length of $O T$ with the theorem of Pythagoras and keep the quadrant signs in mind; in part (d), double the product so that the sine of a double angle appears.

Worked answer

a) $O T^{2} = (- 8)^{2} + (- 15)^{2} = 289$ , so $r = O T = 17$ ✓
$cos A = \frac{x}{r} = - \frac{8}{17}$ ✓

b) $cos 2 A = 2 cos^{2} A - 1 = 2 (- \frac{8}{17})^{2} - 1$ ✓ $= \frac{128}{289} - \frac{289}{289} = - \frac{161}{289}$ ✓

c) $B$ lies in the second quadrant, so $cos B = - \frac{4}{5}$ ✓ (3-4-5 triple). From the point $T$ : $sin A = \frac{y}{r} = - \frac{15}{17}$ ✓
$sin (A - B) = sin A cos B - cos A sin B$ ✓
$= (- \frac{15}{17}) (- \frac{4}{5}) - (- \frac{8}{17}) (\frac{3}{5}) = \frac{60}{85} + \frac{24}{85} = \frac{84}{85}$ ✓

d) $sin (\frac{α}{2} - 4 5^{\circ}) cos (\frac{α}{2} - 4 5^{\circ}) = \frac{1}{2} \cdot 2 sin (\frac{α}{2} - 4 5^{\circ}) cos (\frac{α}{2} - 4 5^{\circ})$ ✓
$= \frac{1}{2} sin (2 (\frac{α}{2} - 4 5^{\circ})) = \frac{1}{2} sin (α - 9 0^{\circ})$ ✓
$= - \frac{1}{2} sin (9 0^{\circ} - α)$ ✓ $= - \frac{1}{2} cos α = - \frac{p}{2}$ ✓

Modelled on Nov 2024 Paper 2, Question 5 — same skills, new scenario.

Question 5

Identities, reduction formulae & trigonometric equationsComplex18 marksPaper 2

In this question, you may assume that $cos (x - y) = cos x cos y + sin x sin y$ . Show ALL working.

aUse the given identity to derive a formula for $cos (x + y)$ . (2)
bProve the identity: $\frac{sin ( 18 0 ^{\circ} - x ) cos ( - y ) + cos ( 36 0 ^{\circ} - x ) sin ( 18 0 ^{\circ} - y )}{sin ( 9 0 ^{\circ} + x ) cos ( 36 0 ^{\circ} - y ) + cos ( 9 0 ^{\circ} - x ) sin ( - y )} = tan (x + y)$ (6)
cDetermine the values of $x$ for which $2 sin^{2} x - cos (9 0^{\circ} + x) + 3 = 2$ , where $x \in (0^{\circ}; 36 0^{\circ})$ . (6)
dDetermine the maximum value of $f (x) = 4 sin^{2} x - 1$ . (3)
eWrite down the smallest value of $x$ , $x \in [0^{\circ}; 36 0^{\circ}]$ , at which $f$ attains this maximum. (1)

Hint

For the identity, reduce each factor on its own before comparing with the expansions of $sin (x + y)$ and $cos (x + y)$ ; in the equation, square first and use $cos (9 0^{\circ} + x) = - sin x$ .

Worked answer

a) $cos (x + y) = cos (x - (- y))$ ✓
$= cos x cos (- y) + sin x sin (- y) = cos x cos y - sin x sin y$ ✓ (since $cos (- y) = cos y$ and $sin (- y) = - sin y$ )

b) Numerator: $sin (18 0^{\circ} - x) = sin x$ and $cos (- y) = cos y$ ✓; $cos (36 0^{\circ} - x) = cos x$ and $sin (18 0^{\circ} - y) = sin y$ ✓
Numerator $= sin x cos y + cos x sin y = sin (x + y)$ ✓
Denominator: $sin (9 0^{\circ} + x) = cos x$ and $cos (36 0^{\circ} - y) = cos y$ ✓; $cos (9 0^{\circ} - x) = sin x$ and $sin (- y) = - sin y$ ✓
Denominator $= cos x cos y - sin x sin y = cos (x + y)$ , so LHS $= \frac{sin ( x + y )}{cos ( x + y )} = tan (x + y)$ ✓

c) Both sides are non-negative, so squaring: $2 sin^{2} x - cos (9 0^{\circ} + x) + 3 = 4$ ✓
$cos (9 0^{\circ} + x) = - sin x$ ✓, so $2 sin^{2} x + sin x - 1 = 0$ ✓
$(2 sin x - 1) (sin x + 1) = 0$ ✓
$sin x = \frac{1}{2}$ : $x = 3 0^{\circ}$ or $x = 15 0^{\circ}$ ✓
$sin x = - 1$ : $x = 27 0^{\circ}$ ✓

d) $f (x) = 4 (\frac{1 - cos 2 x}{2}) - 1$ ✓ $= 1 - 2 cos 2 x$ ✓
Since $- 1 \leq cos 2 x \leq 1$ , the maximum occurs when $cos 2 x = - 1$ : maximum $= 1 - 2 (- 1) = 3$ ✓

e) $cos 2 x = - 1 \Rightarrow 2 x = 18 0^{\circ}$ , so $x = 9 0^{\circ}$ ✓

Modelled on Nov 2024 Paper 2, Question 6 — same skills, new scenario.

Question 6

Trigonometric graphs & general solutionsRoutine11 marksPaper 2

In the diagram, the graph of $f (x) = tan x$ is drawn for $x \in [- 18 0^{\circ}; 18 0^{\circ}]$ . The dashed vertical lines are the asymptotes of $f$ . The graph of $g (x) = 2 cos x + 1$ is NOT drawn.

aWrite down the equation of the asymptote of $f$ for $x \in (0^{\circ}; 18 0^{\circ})$ . (1)
bFor which values of $x$ , where $x \in (- 18 0^{\circ}; 0^{\circ}]$ , is $f (x) \leq 0$ ? (2)
cWrite down the period of $g$ . (1)
dDraw the graph of $g$ for $x \in [- 18 0^{\circ}; 18 0^{\circ}]$ on the grid in your ANSWER BOOK. Clearly show ALL intercepts with the axes and the turning points of the graph. (3)
eUse the graphs of $f$ and $g$ to determine the general solution of $2 sin x + tan x = 0$ . (4)

This question has a diagram, shown in the app.

Hint

For the last part, write $2 sin x$ as $tan x \cdot 2 cos x$ so that the equation factorises as $f (x) \cdot g (x) = 0$ ; the solutions are the $x$ -intercepts of the two drawn graphs.

Worked answer

a) $x = 9 0^{\circ}$ ✓

b) On $(- 18 0^{\circ}; 0^{\circ}]$ the graph of $f$ lies on or below the $x$ -axis only between its asymptote at $x = - 9 0^{\circ}$ and the origin ✓, so $- 9 0^{\circ} < x \leq 0^{\circ}$ ✓

c) $36 0^{\circ}$ ✓

d) $g$ is a cosine curve with amplitude 2, shifted 1 unit up, completing one full period on the interval ✓. Maximum turning point $(0^{\circ}; 3)$ ; minimum turning points $(- 18 0^{\circ}; - 1)$ and $(18 0^{\circ}; - 1)$ ✓; $y$ -intercept $(0^{\circ}; 3)$ ; $x$ -intercepts $(- 12 0^{\circ}; 0)$ and $(12 0^{\circ}; 0)$ ✓

e) $2 sin x + tan x = tan x \cdot 2 cos x + tan x$ ✓ $= tan x (2 cos x + 1) = f (x) \cdot g (x) = 0$ ✓
So $f (x) = 0$ or $g (x) = 0$ : the solutions are the $x$ -intercepts of the two drawn graphs.
From the graph of $f$ : $x = k \cdot 18 0^{\circ}$ ✓. From the graph of $g$ : $cos x = - \frac{1}{2}$ , so $x = \pm 12 0^{\circ} + k \cdot 36 0^{\circ}$ ✓
General solution: $x = k \cdot 18 0^{\circ}$ or $x = \pm 12 0^{\circ} + k \cdot 36 0^{\circ}$ , $k \in Z$

Modelled on Nov 2024 Paper 2, Question 7 — same skills, new scenario.

Question 7

Sine rule in 2-D and 3-D problemsProblem-solving9 marksPaper 2

A vertical communications mast $D C$ stands on level horizontal ground, with its foot at $C$ and its top at $D$ . $A$ , $B$ and $C$ are points on the ground in the same horizontal plane, so that $D \hat{C} A = D \hat{C} B = 9 0^{\circ}$ . A maintenance platform $E$ is fitted to the mast between $C$ and $D$ . The following measurements are known:

$D C = 36$ m
the angle of elevation of $D$ from $A$ is $4 2^{\circ}$
$B \hat{A} C = 5 8^{\circ}$ and $A \hat{C} B = 7 5^{\circ}$
at the platform, the angle between the sightline $E B$ and the mast section $E D$ is $B \hat{E} D = 107, 9 3^{\circ}$

aCalculate $A C$ , the distance from $A$ to the foot of the mast, correct to TWO decimal places. (2)
bCalculate $E D$ , the length of the mast section above the platform, correct to TWO decimal places. (7)

Hint

Use tan in right triangle $A C D$ first; then find the third angle of the ground triangle, apply the sine rule for $B C$ , use right triangle $B C D$ for $B D$ and $B \hat{D} C$ , and finish with the sine rule in triangle $B E D$ .

Worked answer

a) In right $△ A C D$ : $tan 4 2^{\circ} = \frac{36}{A C}$ ✓ $\Rightarrow A C = \frac{36}{tan 4 2 ^{\circ}} = 39, 98$ m ✓

b) $A \hat{B} C = 18 0^{\circ} - 5 8^{\circ} - 7 5^{\circ} = 4 7^{\circ}$ (∠ sum of $△ A B C$ ) ✓
Sine rule in $△ A B C$ : $\frac{B C}{sin 5 8 ^{\circ}} = \frac{A C}{sin 4 7 ^{\circ}}$ ✓ $\Rightarrow B C = \frac{39 , 98 sin 5 8 ^{\circ}}{sin 4 7 ^{\circ}} = 46, 36$ m ✓
In right $△ B C D$ : $B D = 46, 3 6^{2} + 3 6^{2} = 58, 70$ m ✓ and $tan B \hat{D} C = \frac{46 , 36}{36}$ , so $B \hat{D} C = 52, 1 7^{\circ}$ ✓. Since $E$ lies on $D C$ , $B \hat{D} E = 52, 1 7^{\circ}$ .
In $△ B E D$ : $D \hat{B} E = 18 0^{\circ} - 107, 9 3^{\circ} - 52, 1 7^{\circ} = 19, 9 0^{\circ}$ (∠ sum of $△ B E D$ ) ✓
Sine rule: $\frac{E D}{sin D B ^ E} = \frac{B D}{sin B E ^ D}$ $\Rightarrow E D = \frac{58 , 70 sin 19 , 9 0 ^{\circ}}{sin 107 , 9 3 ^{\circ}} = 21, 00$ m ✓

Modelled on Nov 2024 Paper 2, Question 8 — same skills, new scenario.

Question 8

Expressions in terms of k, reduction formulae & square-root domainsComplex17 marksPaper 2

In this question, do NOT use a calculator. It is given that $tan 6 5^{\circ} = k$ . Answer (a) to (c) in terms of $k$ , showing ALL working.

aExpress $cos 2 5^{\circ}$ in terms of $k$ . (2)
bDetermine, in terms of $k$ , the value of $\frac{2 sin 32 , 5 ^{\circ} \cdot cos 32 , 5 ^{\circ}}{2 - 4 sin ^{2} 32 , 5 ^{\circ}}$ (5)
cDetermine $sin 4 0^{\circ}$ in terms of $k$ . (4)
dSimplify $\frac{cos ( 9 0 ^{\circ} + x ) \cdot cos ( - x )}{sin ( 18 0 ^{\circ} - x )}$ to a SINGLE trigonometric ratio of $x$ . (4)
eHence determine ALL values of $x$ in the interval $[0^{\circ}; 36 0^{\circ}]$ for which the square root of the answer to (d) will be real. (2)

Hint

Sketch a right triangle with opposite side $k$ and adjacent side $1$ for $6 5^{\circ}$ ; in (b) write the denominator as $2 (1 - 2 sin^{2} 32, 5^{\circ})$ , and in (d) reduce each factor separately before you simplify.

Worked answer

a) From $tan 6 5^{\circ} = \frac{k}{1}$ , draw a right triangle: opposite $= k$ , adjacent $= 1$ , hypotenuse $= 1 + k^{2}$ ✓
$cos 2 5^{\circ} = cos (9 0^{\circ} - 6 5^{\circ}) = sin 6 5^{\circ} = \frac{k}{1 + k ^{2}}$ ✓

b) Numerator: $2 sin 32, 5^{\circ} cos 32, 5^{\circ} = sin 6 5^{\circ}$ ✓ (double angle)
Denominator: $2 - 4 sin^{2} 32, 5^{\circ} = 2 (1 - 2 sin^{2} 32, 5^{\circ})$ ✓ $= 2 cos 6 5^{\circ}$ ✓
So the expression $= \frac{sin 6 5 ^{\circ}}{2 cos 6 5 ^{\circ}} = \frac{1}{2} tan 6 5^{\circ}$ ✓ $= \frac{k}{2}$ ✓

c) $sin 4 0^{\circ} = sin (6 5^{\circ} - 2 5^{\circ})$ ✓ $= sin 6 5^{\circ} cos 2 5^{\circ} - cos 6 5^{\circ} sin 2 5^{\circ}$ ✓
Since $cos 2 5^{\circ} = sin 6 5^{\circ} = \frac{k}{1 + k ^{2}}$ and $sin 2 5^{\circ} = cos 6 5^{\circ} = \frac{1}{1 + k ^{2}}$ ✓:
$sin 4 0^{\circ} = \frac{k ^{2}}{1 + k ^{2}} - \frac{1}{1 + k ^{2}} = \frac{k ^{2} - 1}{1 + k ^{2}}$ ✓

d) $cos (9 0^{\circ} + x) = - sin x$ ✓; $cos (- x) = cos x$ ✓; $sin (18 0^{\circ} - x) = sin x$ ✓
Expression $= \frac{( - sin x ) ( cos x )}{sin x} = - cos x$ ✓

e) $- cos x$ is real when $- cos x \geq 0$ , i.e. $cos x \leq 0$ ✓
$x \in [9 0^{\circ}; 27 0^{\circ}]$ ✓ (the zeros at $9 0^{\circ}$ and $27 0^{\circ}$ are included)

Modelled on Nov 2025 Paper 2, Question 5 — same skills, new scenario.

Question 9

Identities with reductions & trig arithmetic sequences (general solutions)Problem-solving13 marksPaper 2

Show ALL your working in this question.

aProve the identity: $\frac{tan ( 18 0 ^{\circ} - x ) \cdot cos x + 1}{cos ^{2} x} = \frac{1}{1 + sin x}$ (6)
bThe expressions $sin^{2} x$ ; $sin x \cdot cos x$ and $- 3 cos^{2} x$ are the first THREE terms of an arithmetic sequence whose common difference is NOT zero. Determine the general solution for $x$ . (7)

Hint

Replace $tan (18 0^{\circ} - x)$ with $- tan x$ and factorise $1 - sin^{2} x$ as a difference of squares; in (b) apply $2 T_{2} = T_{1} + T_{3}$ and divide by $cos^{2} x$ to obtain a quadratic in $tan x$ .

Worked answer

a) LHS $= \frac{( - tan x ) cos x + 1}{cos ^{2} x}$ ✓ ( $tan (18 0^{\circ} - x) = - tan x$ )
$= \frac{1 - sin x}{cos ^{2} x}$ ✓ ( $tan x cos x = sin x$ )
$= \frac{1 - sin x}{1 - sin ^{2} x}$ ✓
$= \frac{1 - sin x}{( 1 - sin x ) ( 1 + sin x )}$ ✓ (difference of squares)
$= \frac{1}{1 + sin x} =$ RHS ✓✓
Restriction: $cos x \neq = 0$ , i.e. $x \neq = 9 0^{\circ} + k \cdot 18 0^{\circ}$ , $k \in Z$ (this also ensures $sin x \neq = \pm 1$ )

b) Arithmetic sequence: $2 T_{2} = T_{1} + T_{3}$ , so $2 sin x cos x = sin^{2} x - 3 cos^{2} x$ ✓
If $cos x = 0$ , the equation gives $0 = sin^{2} x = 1$ , which is impossible, so $cos x \neq = 0$ ; divide by $cos^{2} x$ ✓
$2 tan x = tan^{2} x - 3$ ✓
$tan^{2} x - 2 tan x - 3 = 0 \Rightarrow (tan x - 3) (tan x + 1) = 0$ ✓
$tan x = 3$ : $x = 71, 5 7^{\circ} + k \cdot 18 0^{\circ}$ , $k \in Z$ ✓
$tan x = - 1$ : $x = 13 5^{\circ} + k \cdot 18 0^{\circ}$ , $k \in Z$ ✓
Common difference $= sin x cos x - sin^{2} x = sin x (cos x - sin x)$ , which is zero only if $sin x = 0$ or $tan x = 1$ ; neither holds at the solutions, so the difference is NOT zero ✓

Modelled on Nov 2025 Paper 2, Question 6 — same skills, new scenario.

Question 10

Trigonometric graphs & inequalitiesProblem-solving10 marksPaper 2

In the diagram, the graph of $f (x) = sin 2 x$ is drawn for $x \in [- 13 5^{\circ}; 13 5^{\circ}]$ . The dashed horizontal lines through $y = 1$ and $y = - 1$ are shown on the grid. The graph of $g (x) = tan 2 x + 1$ is NOT drawn.

aWrite down the period of $f$ . (1)
bOn the grid provided in your ANSWER BOOK, draw the graph of $g$ for the same interval, $x \in [- 13 5^{\circ}; 13 5^{\circ}]$ . Clearly indicate the asymptotes and ALL intercepts with the axes. (3)
cThe graph of $h$ is obtained when the graph of $f$ is shifted $4 5^{\circ}$ to the RIGHT. Write down the equation of $h$ in its simplest form. (1)
dWrite down the range of $h$ . (1)
eDetermine, with the aid of the two graphs, the values of $x \in [0^{\circ}; 13 5^{\circ}]$ for which $sin 2 x (tan 2 x + 1) \geq 0$ . (4)

This question has a diagram, shown in the app.

Hint

For part (e), notice that $sin 2 x (tan 2 x + 1) = f (x) \cdot g (x)$ : the product is non-negative wherever the two graphs lie on the same side of the $x$ -axis or one of them touches it, and it does not exist at the asymptotes of $g$ .

Worked answer

a) Period of $f = \frac{36 0 ^{\circ}}{2} = 18 0^{\circ}$ ✓

b) $g$ is a tangent curve of period $9 0^{\circ}$ translated 1 unit up. Dashed vertical asymptotes at $x = \pm 4 5^{\circ}$ and $x = \pm 13 5^{\circ}$ ✓; increasing branches with $y$ -intercept $(0^{\circ}; 1)$ ✓; $x$ -intercepts where $tan 2 x = - 1$ : $(- 112, 5^{\circ}; 0)$ , $(- 22, 5^{\circ}; 0)$ and $(67, 5^{\circ}; 0)$ ✓

c) $h (x) = sin (2 (x - 4 5^{\circ})) = sin (2 x - 9 0^{\circ}) = - cos 2 x$ ✓

d) $- 1 \leq y \leq 1$ ✓

e) The product is non-negative where $f$ and $g$ lie on the same side of the $x$ -axis, or where either graph meets the axis.
From the graphs: on $[0^{\circ}; 4 5^{\circ})$ both $f (x) \geq 0$ and $g (x) > 0$ , so the product is non-negative ✓; on $(4 5^{\circ}; 67, 5^{\circ})$ , $f (x) > 0$ but $g (x) < 0$ , so the product is negative; on $[67, 5^{\circ}; 9 0^{\circ}]$ , $g (x) \geq 0$ and $f (x) \geq 0$ ✓; on $(9 0^{\circ}; 13 5^{\circ})$ , $f (x) < 0$ while $g (x) > 0$ , so the product is negative ✓
$x = 4 5^{\circ}$ and $x = 13 5^{\circ}$ are EXCLUDED because $tan 2 x$ is undefined there.
$∴ 0^{\circ} \leq x < 4 5^{\circ}$ or $67, 5^{\circ} \leq x \leq 9 0^{\circ}$ ✓

Modelled on Nov 2025 Paper 2, Question 7 — same skills, new scenario.

Question 11

Cosine rule in 2-D and 3-D problemsProblem-solving10 marksPaper 2

A rectangular advertising billboard $A B C D$ stands VERTICALLY on a level concrete plaza, which is a horizontal plane. The bottom edge $A B$ of the billboard rests on the plaza, with $D$ directly above $A$ and $C$ directly above $B$ , so that the edges $A D$ and $B C$ are vertical. $E$ is a point on the plaza such that $△ A B E$ lies in the horizontal plane and shares the edge $A B$ with the billboard, with $A \hat{B} E = 12 0^{\circ}$ and $B E = A B$ . The area of $A B C D$ is $32$ m $^{2}$ and $A B = 2 B C$ .

aShow that $B C = 4$ m. (2)
bCalculate the length of $A C$ , a diagonal of the billboard, correct to TWO decimal places. (2)
cCalculate the length of $A E$ , correct to TWO decimal places. (2)
dCalculate the size of $A \hat{C} E$ , correct to TWO decimal places. (4)

Hint

Work with $B C = x$ and $A B = 2 x$ in the area equation; then use Pythagoras in the vertical rectangle, the cosine rule with the obtuse angle in the horizontal triangle, and finally show that $C E = A C$ before applying the cosine rule in $△ A C E$ .

Worked answer

a) Let $B C = x$ . Then $A B = 2 x$ ✓ and area $= A B \times B C = 2 x^{2} = 32$ , so $x^{2} = 16$ and $x = 4$ (length $> 0$ ). $∴ B C = 4$ m ✓

b) $A \hat{B} C = 9 0^{\circ}$ (angle of the rectangle): $A C = A B^{2} + B C^{2} = 8^{2} + 4^{2}$ ✓ $= 80 = 8, 94$ m ✓

c) Cosine rule in $△ A B E$ : $A E^{2} = A B^{2} + B E^{2} - 2 \cdot A B \cdot B E \cdot cos (A \hat{B} E) = 8^{2} + 8^{2} - 2 (8) (8) cos 12 0^{\circ}$ ✓ $= 192$ , so $A E = 192 = 13, 86$ m ✓

d) $B C$ is vertical and $B E$ lies in the horizontal plane, so $C \hat{B} E = 9 0^{\circ}$ and $C E = B E^{2} + B C^{2} = 8^{2} + 4^{2} = 80 = A C$ ✓ — $△ A C E$ is isosceles ✓
Cosine rule in $△ A C E$ : $cos (A \hat{C} E) = \frac{A C ^{2} + C E ^{2} - A E ^{2}}{2 \cdot A C \cdot C E} = \frac{80 + 80 - 192}{2 ( 80 )}$ ✓ $= - 0, 2$ , so $A \hat{C} E = 101, 5 4^{\circ}$ ✓

Modelled on Nov 2025 Paper 2, Question 8 — same skills, new scenario.

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