Trigonometry Mastery Bank | Equation Station SA

How to use this page

Work in order. Start with familiar L2 methods, then combine ideas in L3 questions.
Write full working on paper first — never scroll to the answer first.
Draw first. If the question suggests a triangle, CAST wheel or 3D sketch, put that on paper before you simplify.
For graph questions, reset to the parent graph. Sketch \(y=\sin x\), \(y=\cos x\) or \(y=\tan x\) first, then apply the parameter changes one at a time.
Open the worked solution only after you finish, then compare every line.
Use marks carefully. Until a card links to its official memo, the marks shown are suggested practice marks.
Redo any question you dropped a mark on. Mastery comes from correcting, not just attempting.

Source note: Cards with paper-and-memo links are memo-confirmed. All other cards are clearly labelled Equation Station practice with suggested practice marks.

L2 - Routine procedures9 Qs L3 - Complex procedures22 Qs L4 - Problem solving2 Qs

Work it out, then check your method. Read the givens, sketch or inspect the diagram, choose your rule, and show your working before opening the solution. Keep full calculator precision until your final answer. Level tags are editorial guidance, not a claim that this bank reproduces the NSC assessment weighting. This collection currently covers L2 and L3 and L4. The L4 cards are extended problems: attempt them only after you can complete the L2 and L3 cards without prompts.

Find a skill, source or questionLevelSource

Routine procedures

9 questions

Q1L2: Routine procedures5 memo marks

Simplify Using Three Identity Families

Without using a calculator, simplify completely:

\[\cos(-\theta)\,\sin(90^\circ-\theta)\left(1+\tan^2\theta\right)\]

Work where the original expression is defined, so \(\cos\theta\ne0\).

Cape Winelands September 2024 Mathematics P2, Q5.1. Question, answer and marking points checked line by line against the supplied September 2024 marking guideline.

Show worked solution

Memo-aligned working
Working and reasons	Marks
Negative angle: \(\cos(-\theta)=\cos\theta\).	1
Co-ratio: \(\sin(90^\circ-\theta)=\cos\theta\).	1
Write \(\tan^2\theta=\dfrac{\sin^2\theta}{\cos^2\theta}\).	1
\(\cos^2\theta\left(1+\dfrac{\sin^2\theta}{\cos^2\theta}\right)=\cos^2\theta+\sin^2\theta\).	1
Therefore the expression is \(\boxed{1}\), for \(\cos\theta\ne0\).	1
Total	5

Q2L2: Routine procedures5 practice marks

CAST Sketch then Double Angle

Given: \(\cos\alpha = -\dfrac{3}{5}\), where \(0°\leq\alpha\leq 180°\). With the aid of a sketch, determine the value of \(\sin 2\alpha\).

Equation Station practice, not an official exam question.

Show worked solution

Solution sketch: constructed from the given ratio and quadrant.

Worked solution
Working and reasons	Practice marks
Since \(\cos\alpha < 0\) and \(0°\leq\alpha\leq 180°\), angle \(\alpha\) lies in Quadrant II.	1
Sketch: horizontal = \(-3\), hypotenuse = \(5\), so vertical \(= +4\) (positive in Q2).	1
\(\sin\alpha = \dfrac{4}{5}\)	1
Apply double angle formula: \(\sin 2\alpha = 2\sin\alpha\cos\alpha\)	1
\(\sin 2\alpha = 2\cdot\dfrac{4}{5}\cdot\left(-\dfrac{3}{5}\right) = \boxed{-\dfrac{24}{25}}\)	1
Total	5

Suggested practice allocation only. These marks are not copied from an official memo.

Q4L2: Routine procedures5 practice marks

Evaluate sin 75° Exactly

Determine, without a calculator, the exact value of \(\sin 75°\).

Hint: Write \(75° = 45° + 30°\)

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
\(\sin75° = \sin(45°+30°)\)	1
Apply compound formula: \(= \sin45°\cos30°+\cos45°\sin30°\)	1
\(= \dfrac{\sqrt{2}}{2}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot\dfrac{1}{2}\)	1
\(= \dfrac{\sqrt{6}}{4}+\dfrac{\sqrt{2}}{4}\)	1
\(= \boxed{\dfrac{\sqrt{6}+\sqrt{2}}{4}}\)	1
Total	5

Suggested practice allocation only. These marks are not copied from an official memo.

Q5L2: Routine procedures5 practice marks

Simplify to a Single Ratio

Simplify fully, without a calculator:

\[\frac{\sin(180°+x)\cdot\cos(360°+x)}{\cos(90°-x)\cdot\cos(-x)}\]

Domain: \(\sin x\ne0\text{ and }\cos x\ne0\). Keep this restriction after simplifying.

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
\(\sin(180°+x)=-\sin x\)	1
\(\cos(360°+x)=\cos x\)	1
\(\cos(90°-x)=\sin x\) (co-ratio)	1
\(\cos(-x)=\cos x\)	1
\(=\dfrac{(-\sin x)(\cos x)}{(\sin x)(\cos x)}=\dfrac{-\sin x\cos x}{\sin x\cos x}=\boxed{-1}\)	1
Total	5

Suggested practice allocation only. These marks are not copied from an official memo.

Q8L2: Routine procedures4 practice marks

Exact Value of cos 105° Without a Calculator

Determine, without a calculator, the exact value of \(\cos 105°\).

Hint: Write \(105°=60°+45°\)

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
\(\cos 105°=\cos(60°+45°)=\cos 60°\cos 45°-\sin 60°\sin 45°\)	1
\(=\dfrac{1}{2}\cdot\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{4}-\dfrac{\sqrt{6}}{4}\)	1
\(\cos 105°=\boxed{\dfrac{\sqrt{2}-\sqrt{6}}{4}}\)	1
Check: cos105° should be negative (Q2) since √2 < √6.	1
Total	4

Suggested practice allocation only. These marks are not copied from an official memo.

Q9L2: Routine procedures4 practice marks

Classic Identity: Tangent From Double Angle

Prove that:

\[\frac{1-\cos 2x}{\sin 2x} = \tan x\]

Note: always start with the more complex side (LHS).

Domain: \(\sin x\ne0\text{ and }\cos x\ne0\). Keep this restriction after simplifying.

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
\(\text{LHS} = \dfrac{1-\cos 2x}{\sin 2x}\)	1
Substitute \(\cos 2x = 1-2\sin^2 x\): \(= \dfrac{1-(1-2\sin^2 x)}{2\sin x\cos x}\)	1
\(= \dfrac{2\sin^2 x}{2\sin x\cos x}\)	1
\(= \dfrac{\sin x}{\cos x} = \tan x = \text{RHS}\quad\square\)	1
Total	4

Suggested practice allocation only. These marks are not copied from an official memo.

Q17L2: Routine procedures4 practice marks

General Solution: Linear Form

Determine the general solution of:

\[2\sin x - 1 = 0\]

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
\(\sin x = \dfrac{1}{2}\)	1
Reference angle: \(x = 30°\)	1
\(x = 30° + k\cdot 360°\), \(k\in\mathbb{Z}\) (Q1)	1
\(x = 150° + k\cdot 360°\), \(k\in\mathbb{Z}\) (Q2, since sine is positive there)	1
Total	4

Suggested practice allocation only. These marks are not copied from an official memo.

Q21L2: Routine procedures5 practice marks

General Solution: Reduce to tan Equation

Determine the general solution of:

\[3\sin x = \cos x\]

Hint: first test whether cos x = 0 can satisfy the equation. Only then decide whether division by cos x is safe.

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
First check \(\cos x=0\): then \(\sin x=\pm1\), so the equation \(3\sin x=\cos x\) cannot hold. Division by cos x is therefore safe for solutions of this equation.	1
Divide both sides by \(\cos x\) (valid since \(\cos x\ne 0\)): \(\tan x=\dfrac{1}{3}\)	1
Reference angle: \(\tan^{-1}\!\left(\dfrac{1}{3}\right)\approx 18{,}43°\)	1
General solution: \(x=\tan^{-1}\left(\dfrac13\right)+180^\circ k\), \(k\in\mathbb Z\). Numerically, \(x\approx18{,}43^\circ+180^\circ k\).	1
First few solutions: \(x\approx 18{,}43°;\;198{,}43°;\;-161{,}57°;\ldots\)	1
Total	5

Suggested practice allocation only. These marks are not copied from an official memo.

Q25L2: Routine procedures5 practice marks

Triangle: Cosine Rule then Area

Diagram for this practice question; use the given values, not measurements from the screen.

In \(\triangle ABC\): \(AB = 7\text{ m}\), \(BC = 10\text{ m}\) and \(\hat{B} = 65°\).

Calculate \(AC\). (3)
Calculate the area of \(\triangle ABC\). (2)

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Part 1. Cosine rule: \(AC^2=AB^2+BC^2-2(AB)(BC)\cos B\).	1
\(AC^2=149-140\cos65^\circ\).	1
\(AC=\sqrt{149-140\cos65^\circ}\approx\boxed{9{,}48\text{ m}}\).	1
Part 2. \(\text{Area}=\dfrac12(7)(10)\sin65^\circ\).	1
\(\text{Area}\approx\boxed{31{,}72\text{ m}^2}\).	1
Total	5

Suggested practice allocation only. These marks are not copied from an official memo.

Complex procedures

22 questions

Q3L3: Complex procedures4 practice marks

Find a Compound-Angle Value from Two Quadrants

Given \(\cos\alpha=-\dfrac{3}{5}\), where \(\alpha\in(90^\circ;180^\circ)\), and \(\cos\beta=\dfrac{12}{13}\), where \(\beta\in(0^\circ;90^\circ)\), determine \(\sin(\alpha+\beta)\).

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
\(\alpha\) is in Quadrant II, so \(\sin\alpha>0\). Therefore \(\sin\alpha=\sqrt{1-\cos^2\alpha}=\sqrt{1-\dfrac{9}{25}}=\dfrac45\).	1
\(\beta\) is in Quadrant I, so \(\sin\beta>0\). Therefore \(\sin\beta=\sqrt{1-\cos^2\beta}=\sqrt{1-\dfrac{144}{169}}=\dfrac5{13}\).	1
Use \(\sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta\).	1
\[\begin{aligned}\sin(\alpha+\beta)&=\left(\dfrac45\right)\left(\dfrac{12}{13}\right)+\left(-\dfrac35\right)\left(\dfrac5{13}\right)\\&=\dfrac{48}{65}-\dfrac{15}{65}\\&=\boxed{\dfrac{33}{65}}.\end{aligned}\]	1
Total	4

Suggested practice allocation only. These marks are not copied from an official memo.

Q6L3: Complex procedures5 practice marks

CAST Sketch — Given tan θ, Find Two Values

Given \(\tan\theta = -\dfrac{3}{4}\) and \(\theta\in(90°;\,180°)\). With the aid of a sketch in the correct quadrant, determine without a calculator:

The value of \(\sin\theta+\cos\theta\). (4)
The value of \(\cos 2\theta\). (3)

Equation Station practice, not an official exam question.

Show worked solution

Solution sketch: constructed from the given ratio and quadrant.

Worked solution
Working and reasons	Practice marks
θ ∈ (90°; 180°) → Quadrant II. In Q2: x is negative, y is positive. Since tanθ = y/x = −3/4, let x = −4, y = 3.	1
By Pythagoras: \(r = \sqrt{(-4)^2+3^2} = \sqrt{16+9} = 5\)	1
\(\sin\theta = \dfrac{3}{5},\quad\cos\theta = -\dfrac{4}{5}\)	1
6.1 \(\sin\theta+\cos\theta = \dfrac{3}{5}+\left(-\dfrac{4}{5}\right) = \boxed{-\dfrac{1}{5}}\)	1
6.2 \(\cos 2\theta = 1-2\sin^2\theta = 1-2\cdot\dfrac{9}{25} = 1-\dfrac{18}{25} = \boxed{\dfrac{7}{25}}\)	1
Total	5

Suggested practice allocation only. These marks are not copied from an official memo.

Q7L3: Complex procedures9 memo marks

Express Three Ratios in Terms of k

Given that \(\sin10^\circ=\sqrt{k}\), write each expression in terms of \(k\), without using a calculator:

\(\sin190^\circ\) (2)
\(\cos20^\circ\) (3)
\(\cos50^\circ\) (4)

Cape Winelands September 2024 Mathematics P2, Q5.3. Checked against marking-guideline page 8, first method. In 5.3.3 one mark is for a sketch; draw it, rather than quoting only the ratio.

Show worked solution

Memo-aligned working
Working and reasons	Marks
5.3.1 \(\sin190^\circ=\sin(180^\circ+10^\circ)=-\sin10^\circ\).	1
\(\boxed{\sin190^\circ=-\sqrt{k}}\).	1
5.3.2 \(\cos20^\circ=\cos(2\cdot10^\circ)\).	1
Use \(\cos2A=1-2\sin^2A\): \(\cos20^\circ=1-2\sin^210^\circ\).	1
\(\boxed{\cos20^\circ=1-2k}\).	1
5.3.3 \(\cos50^\circ=\cos(60^\circ-10^\circ)\) and apply the compound-angle formula.	1
Sketch a right triangle with angle \(10^\circ\), hypotenuse \(1\), opposite side \(\sqrt{k}\) and adjacent side \(\sqrt{1-k}\). Hence \(\cos10^\circ=\sqrt{1-k}\). Draw and label the triangle to show the sketch required for this memo mark.	1
Substitute the special angles: \(\cos50^\circ=\dfrac12\sqrt{1-k}+\dfrac{\sqrt3}{2}\sqrt{k}\).	1
\(\boxed{\cos50^\circ=\dfrac{\sqrt{1-k}+\sqrt{3k}}{2}}\).	1
Total	9

Q10L3: Complex procedures8 memo marks

Simplify, Then Find Where the Expression Is Undefined

Consider:

\[\frac{1}{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}-\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}\]

Simplify the expression to a single trigonometric ratio. (5)
Determine the general solution for the values of \(\theta\) at which the original expression is undefined. (3)

Cape Winelands September 2024 Mathematics P2, Q5.4. Question, answer and marking points checked line by line against the supplied September 2024 marking guideline.

Show worked solution

Memo-aligned working
Working and reasons	Marks
5.4.1 Use a common denominator: \(\dfrac{1-(\cos\theta+\sin\theta)^2}{(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)}\).	1
Expand the numerator and use \(\sin^2\theta+\cos^2\theta=1\).	1
The numerator becomes \(-2\sin\theta\cos\theta=-\sin2\theta\).	1
The denominator is \(\cos^2\theta-\sin^2\theta=\cos2\theta\).	1
Therefore \(\boxed{-\tan2\theta}\).	1
5.4.2 The original denominators are zero when \(\cos\theta=\pm\sin\theta\), so \(\tan\theta=\pm1\).	1
The principal values are \(45^\circ\) and \(135^\circ\).	1
\(\boxed{\theta=45^\circ+90^\circ k,\ k\in\mathbb Z}\).	1
Total	8

Q11L3: Complex procedures4 memo marks

Derive the Sine Difference Formula

Given \(\cos(A+B)=\cos A\cos B-\sin A\sin B\), use this formula to derive a formula for \(\sin(A-B)\). (4)

Cape Winelands September 2024 Mathematics P2, Q5.2.1. Question, answer and marking points checked line by line against the supplied September 2024 marking guideline.

Show worked solution

Memo-aligned working
Working and reasons	Marks
Use a co-ratio: \(\sin(A-B)=\cos[90^\circ-(A-B)]\).	1
Regroup the angle: \(=\cos[(90^\circ-A)+B]\).	1
Apply the given formula: \(=\cos(90^\circ-A)\cos B-\sin(90^\circ-A)\sin B\).	1
Use co-ratios: \(\boxed{\sin(A-B)=\sin A\cos B-\cos A\sin B}\).	1
Total	4

Q12L3: Complex procedures5 memo marks

Prove an Exact Compound-Angle Identity

Without using a calculator, show that:

\[\sin(x+63^\circ)\cos(x+378^\circ)+\cos(x+63^\circ)\cos(x+108^\circ)=\frac{1}{\sqrt2}\]

Cape Winelands September 2024 Mathematics P2, Q5.2.2. Question, answer and marking points checked line by line against the supplied September 2024 marking guideline.

Show worked solution

Memo-aligned working
Working and reasons	Marks
Reduce \(\cos(x+378^\circ)=\cos(x+18^\circ)\).	1
Regroup the other cosine: \(\cos(x+108^\circ)=\cos[(x+18^\circ)+90^\circ]\).	1
Use the co-ratio: \(\cos[(x+18^\circ)+90^\circ]=-\sin(x+18^\circ)\).	1
The expression is now \(\sin(x+63^\circ)\cos(x+18^\circ)-\cos(x+63^\circ)\sin(x+18^\circ)=\sin[(x+63^\circ)-(x+18^\circ)]\).	1
\(=\sin45^\circ=\boxed{\dfrac{1}{\sqrt2}}\), so LHS = RHS.	1
Total	5

Q13L3: Complex procedures5 practice marks

Prove, Then Use the Result

Prove that \((\sin x-\cos x)^2=1-\sin 2x\).

Hence, without a calculator, determine the value of \((\sin 75°-\cos 75°)^2\).

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
LHS: expand the bracket: \(\sin^2 x-2\sin x\cos x+\cos^2 x\)	1
Apply Pythagorean identity: \(=1-2\sin x\cos x\)	1
Apply double angle: \(=1-\sin 2x=\text{RHS}\quad\square\)	1
Hence: \((\sin 75°-\cos 75°)^2 = 1-\sin(2\times 75°)=1-\sin 150°\)	1
\(=1-\dfrac{1}{2}=\boxed{\dfrac{1}{2}}\)	1
Total	5

Suggested practice allocation only. These marks are not copied from an official memo.

Q14L3: Complex procedures5 practice marks

Identity with Restrictions

Prove that:

\[\frac{1-\sin 2A}{\cos 2A}=\frac{\cos A-\sin A}{\cos A+\sin A}\]

Also state the values of \(A\in[0°;\,180°]\) for which the identity is undefined.

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Work on LHS. Use \(\sin 2A=2\sin A\cos A\) and \(\cos 2A=\cos^2\!A-\sin^2\!A=(\cos A-\sin A)(\cos A+\sin A)\):	1
\(=\dfrac{\sin^2\!A+\cos^2\!A-2\sin A\cos A}{(\cos A-\sin A)(\cos A+\sin A)}=\dfrac{(\cos A-\sin A)^2}{(\cos A-\sin A)(\cos A+\sin A)}\)	1
\(=\dfrac{\cos A-\sin A}{\cos A+\sin A}=\text{RHS}\). This holds on the original domain: both \(\cos A-\sin A\ne0\) and \(\cos A+\sin A\ne0\).	1
The original denominator is zero when \(\cos2A=0\), so \(2A=90^\circ+180^\circ n\), \(n\in\mathbb Z\).	1
On \([0^\circ;180^\circ]\), exclude \(\boxed{A=45^\circ\text{ or }135^\circ}\). Cancellation does not restore either excluded value.	1
Total	5

Suggested practice allocation only. These marks are not copied from an official memo.

Q15L3: Complex procedures4 practice marks

Product Identity Using 30° Angles

Prove that:

\[\cos(A-30°)\cdot\cos(A+30°)=\cos^2\!A-\tfrac{1}{4}\]

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Expand both factors using compound angle formula: \(\cos(A-30°)=\cos A\cos30°+\sin A\sin30°=\frac{\sqrt{3}}{2}\cos A+\frac{1}{2}\sin A\) \(\cos(A+30°)=\cos A\cos30°-\sin A\sin30°=\frac{\sqrt{3}}{2}\cos A-\frac{1}{2}\sin A\)	1
Use difference of squares \((p+q)(p-q)=p^2-q^2\): \(=\left(\frac{\sqrt{3}}{2}\cos A\right)^2-\left(\frac{1}{2}\sin A\right)^2=\frac{3}{4}\cos^2\!A-\frac{1}{4}\sin^2\!A\)	1
Replace \(\sin^2\!A=1-\cos^2\!A\):	1
\(=\frac{3}{4}\cos^2\!A-\frac{1}{4}(1-\cos^2\!A)=\frac{3}{4}\cos^2\!A-\frac{1}{4}+\frac{1}{4}\cos^2\!A=\cos^2\!A-\frac{1}{4}=\text{RHS}\quad\square\)	1
Total	4

Suggested practice allocation only. These marks are not copied from an official memo.

Q16L3: Complex procedures4 practice marks

Ratio Identity (Level 2/3)

Prove that, for \(\cos x\ne 0\) and \(\cos y\ne 0\) and \(\sin(x-y)\ne 0\):

\[\frac{\sin(x+y)}{\sin(x-y)}=\frac{\tan x+\tan y}{\tan x-\tan y}\]

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Work on RHS. Write \(\tan x=\frac{\sin x}{\cos x}\) and \(\tan y=\frac{\sin y}{\cos y}\):	1
\(\text{RHS}=\dfrac{\dfrac{\sin x}{\cos x}+\dfrac{\sin y}{\cos y}}{\dfrac{\sin x}{\cos x}-\dfrac{\sin y}{\cos y}}=\dfrac{\sin x\cos y+\cos x\sin y}{\sin x\cos y-\cos x\sin y}\) (multiply by \(\cos x\cos y\))	1
Recognise: numerator = \(\sin(x+y)\), denominator = \(\sin(x-y)\)	1
\(=\dfrac{\sin(x+y)}{\sin(x-y)}=\text{LHS}\quad\square\)	1
Total	4

Suggested practice allocation only. These marks are not copied from an official memo.

Q18L3: Complex procedures6 practice marks

General Solution Using cos 2x Substitution

Determine the general solution of:

\[\cos 2x + \sin x = 0\]

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Substitute \(\cos 2x = 1-2\sin^2 x\): \(1-2\sin^2 x + \sin x = 0\)	1
Rearrange: \(2\sin^2 x - \sin x - 1 = 0\)	1
Factorise: \((2\sin x + 1)(\sin x - 1) = 0\)	1
\(\sin x = -\dfrac{1}{2}\) or \(\sin x = 1\)	1
For \(\sin x = -\dfrac{1}{2}\): \(x = 210° + k\cdot 360°\) or \(x = 330° + k\cdot 360°\), \(k\in\mathbb{Z}\)	1
For \(\sin x = 1\): \(x = 90° + k\cdot 360°\), \(k\in\mathbb{Z}\)	1
Total	6

Suggested practice allocation only. These marks are not copied from an official memo.

Q19L3: Complex procedures5 practice marks

Quadratic in cos — Factorise and Solve

Determine the general solution of:

\[2\cos^2 x - \cos x - 1 = 0\]

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Treat as quadratic in \(\cos x\). Let \(c = \cos x\): \(2c^2 - c - 1 = 0\)	1
Factorise: \((2c+1)(c-1) = 0\)	1
\(\cos x = -\dfrac{1}{2}\) or \(\cos x = 1\)	1
For \(\cos x = -\dfrac{1}{2}\): reference \(= 60°\); \(x = 120° + k\cdot 360°\) or \(x = 240° + k\cdot 360°\), \(k\in\mathbb{Z}\)	1
For \(\cos x = 1\): \(x = 0° + k\cdot 360°\), \(k\in\mathbb{Z}\)	1
Total	5

Suggested practice allocation only. These marks are not copied from an official memo.

Q20L3: Complex procedures6 practice marks

Read, Interpret and Solve from Graphs

Solid blue: f. Dashed red: g. All angles in degrees; domain [0; 270].

The graphs of \(f(x)=\sin 3x\) and \(g(x)=\cos(x-30°)\) are drawn for \(x\in[0°;\;270°]\).

Write down the amplitude of \(g\). (1)
Write down the period of \(f\). (1)
For which values of \(x\) is \(f(x)>g(x)\)? (2)
Use the graphs to solve \(\sin 3x = \cos(x-30°)\) for \(x\in[90°;\;270°]\). (3)

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
\(\text{Amplitude of }g=1;\quad\text{period of }f=360^\circ/3=120^\circ.\)	1
\(\cos(x-30^\circ)=\sin(120^\circ-x)\)	1
\(3x=120^\circ-x+360^\circ k\ \Rightarrow\ x=30^\circ+90^\circ k\) or \(3x=180^\circ-(120^\circ-x)+360^\circ k\ \Rightarrow\ x=30^\circ+180^\circ k\)	1
The second family is contained in the first. On \([0^\circ;270^\circ]\) the intersections are \(30^\circ,120^\circ,210^\circ\).	1
Check the signs between intersections: \(f(x)>g(x)\text{ for }x\in(120^\circ;210^\circ)\cup(210^\circ;270^\circ]\). Exclude 210 degrees because the values are equal. Include 270 degrees because \(f(270^\circ)=1>g(270^\circ)=-1/2\).	1
On the requested smaller interval \([90^\circ;270^\circ]\), equality holds at \(x=120^\circ\text{ or }210^\circ\).	1
Total	6

Suggested practice allocation only. These marks are not copied from an official memo.

Q22L3: Complex procedures4 practice marks

Solve in an Interval — Compound Angle Equation

Solve for \(x\in[-180°;\,180°]\):

\[\sin(x+30°)=\cos x\]

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Write \(\cos x=\sin(90°-x)\), so the equation becomes \(\sin(x+30°)=\sin(90°-x)\)	1
Case 1: \(x+30°=90°-x+k\cdot 360°\Rightarrow 2x=60°+k\cdot 360°\Rightarrow x=30°+k\cdot 180°\) In \([-180°;180°]\): \(x=30°\) or \(x=30°-180°=-150°\)	1
Case 2: \(x+30°=180°-(90°-x)+k\cdot 360°=90°+x+k\cdot 360°\Rightarrow 30°=90°+k\cdot 360°\) → impossible	1
\[x\in\{-150°;\;30°\}\]	1
Total	4

Suggested practice allocation only. These marks are not copied from an official memo.

Q23L3: Complex procedures5 practice marks

Factorise and Solve — Mixed Equation

Determine the general solution of:

\[\tan x\cdot\sin x+\sin x=0\]

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Factorise common factor sin x: \(\sin x(\tan x+1)=0\)	1
Case 1: \(\sin x=0\Rightarrow x=k\cdot 180°,\quad k\in\mathbb{Z}\)	1
Case 2: \(\tan x=-1\Rightarrow x=-45°+k\cdot 180°,\quad k\in\mathbb{Z}\)	1
Final answer: \(x=k\cdot 180°\) or \(x=-45°+k\cdot 180°,\quad k\in\mathbb{Z}\)	1
Note: the original equation contains tan x, so it is only defined where \(\cos x\ne 0\) (i.e. \(x\ne 90°+k\cdot180°\)). At \(x=k\cdot180°\), \(\cos x=\pm1\ne0\), so \(\tan x=0\) is perfectly defined there — tan x is undefined where \(\cos x=0\), not where \(\sin x=0\). So the \(x=k\cdot180°\) solutions are valid.	1
Total	5

Suggested practice allocation only. These marks are not copied from an official memo.

Q24L3: Complex procedures12 memo marks

Tangent and Sine Graph Analysis

Original question-paper graph of f of x equals tangent of x plus p, with asymptotes at negative 45 degrees and 135 degrees. — Original question-paper graph with the full interval, axes and asymptotes retained.

The graph of \(f(x)=\tan(x+p^\circ)\) is shown for \(x\in[-135^\circ;180^\circ]\), with asymptotes at \(x=-45^\circ\) and \(x=135^\circ\).

Write down \(p\). (1)
Draw \(g(x)=\sin2x\) on the same interval, showing all intercepts and turning points. (3)
Write down the period of \(g\). (1)
Shift \(g\) left by \(45^\circ\) to form \(h\). Give the simplest equation of \(h\). (2)
For \(x\in[-135^\circ;0^\circ]\), determine where \(f(x)\le-1\). (2)
On the same interval, solve \(\sin x\cos x+2<2\). (3)

Cape Winelands September 2024 Mathematics P2, Q6. Checked against marking-guideline page 10. Its left-endpoint label has a sign typo: sin(-270 degrees) = +1, so the correct point is (-135 degrees; 1).

Show worked solution

Memo-aligned working
Working and reasons	Marks
6.1 \(\boxed{p=-45^\circ}\).	1
6.2 Plot the x-intercepts \((-90^\circ,0)\), \((0^\circ,0)\), \((90^\circ,0)\) and \((180^\circ,0)\).	1
Plot maxima at \((-135^\circ,1)\), \((45^\circ,1)\) and minima at \((-45^\circ,-1)\), \((135^\circ,-1)\).	1
Join the points with the smooth sine shape for \(g(x)=\sin2x\).	1
6.3 \(\boxed{\text{Period of }g=180^\circ}\).	1
6.4 \(h(x)=\sin[2(x+45^\circ)]=\sin(2x+90^\circ)\).	1
Therefore \(\boxed{h(x)=\cos2x}\).	1
6.5.1 Use the critical value \(f(0^\circ)=-1\) and the asymptote at \(x=-45^\circ\).	1
\(\boxed{-45^\circ	1
6.5.2 \(\sin x\cos x<0\), so \(\sin2x<0\).	1
The critical values in the interval are \(-90^\circ\) and \(0^\circ\).	1
\(\boxed{-90^\circ	1
Total	12

Q26L3: Complex procedures9 practice marks

3D Flagpole: Prove the Length Formula

Horizontal triangle EGF with EF 12 metres and the two given angles. — First solve the horizontal triangle with the sine rule.

Vertical right triangle DEG with angle of elevation alpha at G. — Then use the horizontal length EG in the right triangle.

In the diagram, \(DE\) is a vertical flagpole. Points \(E\), \(F\) and \(G\) lie on the same horizontal plane. The angle of elevation of \(D\) (top of pole) from \(G\) is \(\alpha\).

\(\hat{GEF} = 30°\), \(\hat{EFG} = 150°-\alpha\), and \(EF = 12\text{ m}\).

Show that \(EG = \dfrac{6\cos\alpha + 6\sqrt{3}\sin\alpha}{\sin\alpha}\). (5)
Hence write down an expression for the height \(DE\). (4)

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Part 1. In \(\triangle EGF\): \(\hat{FGE} = 180°-30°-(150°-\alpha) = \alpha\)	1
Apply sine rule: \(\dfrac{EG}{\sin\hat{EFG}} = \dfrac{EF}{\sin\hat{FGE}}\)	1
\(EG = \dfrac{12\sin(150°-\alpha)}{\sin\alpha}\)	1
Expand: \(\sin(150°-\alpha)=\sin150°\cos\alpha-\cos150°\sin\alpha = \tfrac{1}{2}\cos\alpha+\tfrac{\sqrt{3}}{2}\sin\alpha\)	1
\(EG = \dfrac{12\left(\tfrac{1}{2}\cos\alpha+\tfrac{\sqrt{3}}{2}\sin\alpha\right)}{\sin\alpha} = \dfrac{6\cos\alpha+6\sqrt{3}\sin\alpha}{\sin\alpha}\quad\square\)	1
Part 2. In right-angled \(\triangle DEG\): \(\tan\alpha = \dfrac{DE}{EG}\), so \(DE = EG\cdot\tan\alpha\)	1
\(DE = \dfrac{6\cos\alpha+6\sqrt{3}\sin\alpha}{\sin\alpha}\cdot\dfrac{\sin\alpha}{\cos\alpha}\)	1
\(DE = \dfrac{6\cos\alpha+6\sqrt{3}\sin\alpha}{\cos\alpha} = 6+6\sqrt{3}\tan\alpha\)	2
Total	9

Suggested practice allocation only. These marks are not copied from an official memo.

Q28L3: Complex procedures8 practice marks

3D Height Problem — Fully Worked

Horizontal triangle ABC with AB 20 metres and given angles 50 and 70 degrees. — Step 1 finds AC without using the vertical point.

Vertical right triangle APC with angle of elevation 35 degrees. — Step 2 uses the unrounded AC value to find PC.

Points \(A\), \(B\) and \(C\) lie in the same horizontal plane. \(AB = 20\text{ m}\), \(\hat{ABC}=70°\) and \(\hat{BAC}=50°\). Point \(P\) is directly above \(C\). The angle of elevation of \(P\) from \(A\) is \(35°\).

Calculate \(AC\). (4)
Hence calculate the height \(PC\). (4)

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Part 1. In \(\triangle ABC\): \(\hat{ACB} = 180°-70°-50° = 60°\)	1
Apply sine rule: \(\dfrac{AC}{\sin\hat{ABC}} = \dfrac{AB}{\sin\hat{ACB}}\)	1
\(\dfrac{AC}{\sin70°} = \dfrac{20}{\sin60°}\)	1
\(AC = \dfrac{20\sin70°}{\sin60°} \approx 21{,}70\text{ m}\)	1
Part 2. In \(\triangle APC\) (right-angled at \(C\)): \(\tan 35° = \dfrac{PC}{AC}\)	2
Use the unrounded value of \(AC\): \(PC = AC\tan35° \approx 15{,}20\text{ m}\)	2
Total	8

Suggested practice allocation only. These marks are not copied from an official memo.

Q29L3: Complex procedures8 practice marks

Triangle KLM — Diagram Given

Diagram for this practice question; use the given values, not measurements from the screen.

In the diagram, \(\triangle KLM\) has \(KL=15\text{ cm}\), \(LM=10\text{ cm}\) and \(\hat{L}=55°\).

Calculate \(KM\). (3)
Calculate \(\hat{M}\). (3)
Calculate the area of \(\triangle KLM\). (2)

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
\(KM=\sqrt{15^2+10^2-2(15)(10)\cos55^\circ}\approx12.37\text{ cm}\)	3
Retain the unrounded value of KM. Use the cosine rule to decide the angle without the inverse-sine ambiguity: \(\cos M=\frac{10^2+KM^2-15^2}{2(10)(KM)}\)	1
\(M\approx83.52^\circ\). A fixed SAS triangle does not, by itself, justify choosing the acute inverse-sine answer.	2
\(\text{Area}=\frac12(15)(10)\sin55^\circ\approx61.44\text{ cm}^2\)	2
Total	8

Suggested practice allocation only. These marks are not copied from an official memo.

Q30L3: Complex procedures9 practice marks

D on BC — Two Triangles Sharing a Side

Triangles ABD and ACD share AD while B, D and C lie on one straight line. — B, D and C lie on a straight line. The shared side AD links the two triangles.

In the diagram, D is a point on BC. \(AB=9\text{ m}\), \(\hat{ABD}=52°\), \(\hat{ADB}=67°\), \(\hat{ACD}=44°\).

Find \(\hat{DAB}\). (1)
Calculate \(AD\). (3)
Find \(\hat{DAC}\). (2)
Calculate \(AC\). (3)

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
\(\angle DAB=180^\circ-52^\circ-67^\circ=61^\circ\)	1
\(AD=\frac{9\sin52^\circ}{\sin67^\circ}\approx7.70\text{ m}\)	3
\(\angle ADC=113^\circ,\quad\angle DAC=23^\circ\)	2
\(AC=\frac{AD\sin113^\circ}{\sin44^\circ}\approx10.21\text{ m}\) (use unrounded AD).	3
Total	9

Suggested practice allocation only. These marks are not copied from an official memo.

Q31L3: Complex procedures7 practice marks

3D — Vertical Point Above Horizontal Triangle

Horizontal triangle ABC with BC 20 metres and given angles 53 and 72 degrees. — Step 1 uses the sine rule to find AB.

Vertical right triangle BAT with angle of elevation 40 degrees. — Step 2 uses the unrounded AB value to find AT.

In the diagram, A, B and C are points on the same horizontal plane. T is a point directly above A. \(\hat{BAC}=53°\), \(\hat{ABC}=72°\), \(BC=20\text{ m}\). The angle of elevation of T from B is 40°.

Calculate \(AB\). (3)
Hence calculate the height \(AT\). (4)

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
\(\angle ACB=55^\circ\)	1
\(AB=\frac{20\sin55^\circ}{\sin53^\circ}\approx20.51\text{ m}\)	2
\(AT=\frac{20\sin55^\circ}{\sin53^\circ}\tan40^\circ\approx17.21\text{ m}\) (do not round AB first).	4
Total	7

Suggested practice allocation only. These marks are not copied from an official memo.

Q32L3: Complex procedures10 practice marks

3D Tower — Prove the Height Formula

Right triangle ABC with BC equal to p, AB equal to h and angle alpha at C. — Use this elevation triangle for parts 1 and 3.

Ground right triangle BCD with BC equal to p and angle beta at D. — Use this horizontal triangle to find CD and BD.

Right triangle ABD with AB equal to h and angle gamma at D. — Use this final elevation triangle for part 4.

In the diagram, AB is a vertical tower of height \(h\) metres at B on level ground. B, C, D are on the ground with \(BC=p\text{ m}\), \(\hat{BCD}=90°\), \(\hat{BDC}=\beta\). The angle of elevation of A from C is \(\alpha\).

Express \(BC\) in terms of \(h\) and \(\alpha\). (1)
Express \(CD\) in terms of \(p\) and \(\beta\). (2)
Hence show that: \(h = p\cdot\tan\alpha\). (2)
The angle of elevation of A from D is \(\gamma\). Show that \(\tan\gamma=\dfrac{p\tan\alpha}{\sqrt{p^2+CD^2}}\). (5)

Equation Station practice, not an official exam question.

Show worked solution

Worked solution
Working and reasons	Practice marks
Part 1. In right-angled △ABC: \(\tan\alpha=\dfrac{h}{BC}\Rightarrow BC=\dfrac{h}{\tan\alpha}\).	1
Part 2. In right-angled △BCD (\(\hat{BCD}=90°\)): \(\tan\beta=\dfrac{BC}{CD}=\dfrac{p}{CD}\Rightarrow CD=\dfrac{p}{\tan\beta}\)	2
Part 3. From part 1: \(BC=p\Rightarrow\dfrac{h}{\tan\alpha}=p\Rightarrow h=p\tan\alpha\quad\square\)	2
Part 4. \(BD^2=BC^2+CD^2=p^2+CD^2\) (Pythagoras in right-angled △BCD), so \(BD=\sqrt{p^2+CD^2}\)	2
In right-angled △ABD: \(\tan\gamma=\dfrac{AB}{BD}=\dfrac{h}{\sqrt{p^2+CD^2}}\)	1
Substitute \(h=p\tan\alpha\): \(\tan\gamma=\dfrac{p\tan\alpha}{\sqrt{p^2+CD^2}}\quad\square\)	2
Total	10

Suggested practice allocation only. These marks are not copied from an official memo.

Problem solving

2 questions

Q27L4: Problem solving6 memo marks

3D Tower: Prove the Height Formula

Original question-paper diagram of vertical tower AB above horizontal triangle DBC, with p, k, x, y and theta labelled. — Original question-paper diagram with every label retained.

\(AB\) is a vertical tower of \(p\) units high. \(D\) and \(C\) are in the same horizontal plane as \(B\) (the base of the tower). The angle of elevation of \(A\) from \(D\) is \(x\). \(\hat{BDC}=y\), \(\hat{DCB}=\theta\) and \(DC = k\) units.

Express \(p\) in terms of \(DB\) and \(x\). (1)
Hence prove that: \(p = \dfrac{k\sin\theta\tan x}{\sin y\cos\theta+\cos y\sin\theta}\). (5)

Grade 12 Term 2 Geometry and Trigonometry Assignment, Q1.1. Checked against assignment-memo page 3. The memo mislabels the third ground angle: it is DBC, at B, not BDC. The solution below uses the correct vertex.

Show worked solution

Memo-aligned working
Working and reasons	Marks
1.1.1 In \(\triangle ABD\): \(\tan x=\dfrac{p}{DB}\), therefore \(p=DB\tan x\).	1
1.1.2 In \(\triangle BDC\): \(\widehat{DBC}=180^\circ-(y+\theta)\).	1
By the sine rule, \(\dfrac{DB}{\sin\theta}=\dfrac{k}{\sin(180^\circ-(y+\theta))}\).	1
Using the reduction formula, \(\sin(180^\circ-(y+\theta))=\sin(y+\theta)\), so \(DB=\dfrac{k\sin\theta}{\sin(y+\theta)}\).	1
Replace \(DB\) in \(p=DB\tan x\): \(p=\dfrac{k\sin\theta}{\sin(y+\theta)}\tan x\).	1
Expand \(\sin(y+\theta)\): \(\boxed{p=\dfrac{k\sin\theta\tan x}{\sin y\cos\theta+\cos y\sin\theta}}\).	1
Total	6

Q33L4: Problem solving7 memo marks

Vertical Wall and Floor Triangle

Original question-paper diagram of vertical triangular wall ABC on floor triangle BCD, with CA 13, equal sides k, alpha and 2 alpha labelled. — Original question-paper diagram with the complete shape and labels retained.

In the diagram, \(ABC\) is a vertical triangular wall on the horizontal floor \(CBD\). \(CA=13\text{ m}\), \(CD=BD=k\text{ m}\), \(\widehat{ACB}=\alpha\) and \(\widehat{BDC}=2\alpha\).

Show that \(CB=13\cos\alpha\). (1)
Hence show that \(k=\dfrac{13}{2\tan\alpha}\). (4)
Calculate the area of floor \(\triangle BCD\) if \(\alpha=26^\circ\). (2)

Cape Winelands September 2024 Mathematics P2, Q7. Question, final values and marking points checked against the supplied September 2024 marking guideline.

Show worked solution

Memo-aligned working
Working and reasons	Marks
7.1.1 In right-angled \(\triangle ACB\), \(\cos\alpha=\dfrac{CB}{13}\), so \(CB=13\cos\alpha\).	1
7.1.2 Apply the cosine rule and substitute \(CB=13\cos\alpha\): \((13\cos\alpha)^2=k^2+k^2-2k^2\cos2\alpha\).	1
Factorise: \(169\cos^2\alpha=2k^2(1-\cos2\alpha)\).	1
Use \(\cos2\alpha=1-2\sin^2\alpha\): \(169\cos^2\alpha=4k^2\sin^2\alpha\).	1
Lengths are positive, so \(13\cos\alpha=2k\sin\alpha\), hence \(\boxed{k=\dfrac{13}{2\tan\alpha}}\).	1
7.2 \(\text{Area}=\dfrac12 k^2\sin2\alpha=\dfrac12\left(\dfrac{13}{2\tan26^\circ}\right)^2\sin52^\circ\).	1
\(\boxed{\text{Area}=69{,}98\text{ m}^2}\).	1
Total	7

Your final check before opening a solution

Identity

Keep both sides separate. Rewrite one side using valid identities, keep all restrictions, and end at the other side.

Equation

Answer in the form requested: a general solution or every solution in the stated interval. Check endpoints and the original domain.

Diagram

Identify the triangle containing the required value, label the known facts, choose the rule, and round only the final result.