Grade 11 Maths · Trigonometry · Paper 2

Identities Grade 11

Real Identities past paper questions and answers from 2023 to 2025, with each memo split into the ticks examiners give. Try one, mark yourself, and see exactly which skill cost you marks.

past paper questions: 25 past paper questions
skills: 3 skills
videos: 3 videos

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Question 3.5 · June 2024 · Paper 2 · Provincial · 2 marks

In the diagram below, the graphs of

f (x) = a cos x

and

g (x) = sin (x + p)

are drawn for

x \in [- 18 0^{\circ}; 18 0^{\circ}]

3.5 If

k (x) = cos (2 x + t)

, determine the value of

t

k (x) = g (2 x)

.(2)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

Identities skills

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More Identities Grade 11 questions and answers

Try each one on paper before you open the memo.

4. Question 3.4.1 · March 2025 · Provincial · 5 marks
3.4 Given: $(1 - cos^{2} x) (tan x + \frac{c o s x}{s i n x})$
3.4.1 Prove that $(1 - cos^{2} x) (tan x + \frac{c o s x}{s i n x}) = tan x$ (5)
Show the memo (5 ticks)Memo
1. ✓
  Identify square identity $1 - cos^{2} x = sin^{2} x$
  1 mark · Use the square identity
2. ✓
  Replace $tan x$ by $\frac{s i n x}{c o s x}$ $sin^{2} x (\frac{s i n x}{c o s x} + \frac{c o s x}{s i n x})$
  1 mark · Write tan as sin over cos
3. ✓
  LCD $(cos x sin x)$ $= sin^{2} x (\frac{s i n ^{2} x + c o s ^{2} x}{c o s x s i n x})$
  1 mark · Add or subtract algebraic fractions
4. ✓
  $sin^{2} x + cos^{2} x = 1$ $= sin^{2} x (\frac{1}{c o s x s i n x})$
  1 mark · Use the square identity
5. ✓
  Simplification to $\frac{s i n x}{c o s x}$ $= \frac{s i n x}{c o s x} = tan x = R H S$
  1 mark · Simplify an algebraic fraction
5. Question 4.3.1 · November 2025 · Paper 2 · Provincial · 6 marks
4.3 Given: $\frac{c o s x . t a n x}{2 [ s i n x . c o s ( x - 90° ) + c o s x . c o s ( - x ) ]}$
4.3.1 Simplify the above expression to a single trigonometric ratio of $x$ .(6)
Show the memo (6 ticks)Memo
1. ✓
  $\frac{s i n x}{c o s x}$ $= \frac{c o s x . \frac{s i n x}{c o s x}}{2 [ s i n x . ( \dots ) + c o s x ( \dots ) ]}$
  1 mark · Write tan as sin over cos
2. ✓
  $sin x$ $cos (x - 90°) = cos (90° - x) = sin x$
  1 mark · Use a reduction formula
3. ✓
  $cos x$ $cos (- x) = cos x$
  $= \frac{c o s x . \frac{s i n x}{c o s x}}{2 [ s i n x . ( s i n x ) + c o s x ( c o s x ) ]}$
  1 mark · Use a reduction formula
4. ✓
  Simplification $= \frac{s i n x}{2 [ s i n ^{2} x + c o s ^{2} x ]}$
  1 mark · Simplify an algebraic fraction
5. ✓
  $sin^{2} x + cos^{2} x = 1$ $= \frac{s i n x}{2 ( 1 )}$
  1 mark · Use the square identity
6. ✓
  Answer $= \frac{s i n x}{2}$
  1 mark · Use the square identity
6. Question 2.1.3 · June 2024 · Paper 2 · Provincial · 3 marks
2.1 Given: $sin 2 0^{\circ} = p$ . Determine the following in terms of $p$ :
2.1.3 $1 - sin^{2} 7 0^{\circ}$ (3)
Show the memo (3 ticks)Memo
1. ✓
  Use identity $1 - sin^{2} 7 0^{\circ} = cos^{2} 7 0^{\circ}$
  1 mark · Use the square identity
2. ✓
  Co-function $cos^{2} 7 0^{\circ} = sin^{2} (9 0^{\circ} - 7 0^{\circ}) = sin^{2} 2 0^{\circ}$
  1 mark · Use co-ratio identities
3. ✓
  Substitute p $p^{2}$
  1 mark · Use co-ratio identities
7. Question 4.2.1 · March 2024 · Paper 1 · Provincial · 3 marks
4.2 Given: $\frac{s i n θ}{1 + c o s θ}$
4.2.1 Prove that $\frac{s i n θ}{1 + c o s θ} = \frac{1 - c o s θ}{s i n θ}$ (3)
Show the memo (3 ticks)Memo
1. ✓
  Multiply LHS by conjugate $\frac{s i n θ}{1 + c o s θ} \times \frac{1 - c o s θ}{1 - c o s θ}$
  1 mark · Simplify an algebraic fraction
2. ✓
  Simplify denominator using identity $\frac{s i n θ ( 1 - c o s θ )}{1 - c o s ^{2} θ} = \frac{s i n θ ( 1 - c o s θ )}{s i n ^{2} θ}$
  1 mark · Use the square identity
3. ✓
  Simplify fraction $\frac{1 - c o s θ}{s i n θ}$
  1 mark · Simplify an algebraic fraction
8. Question 5.2.2 · November 2024 · Paper 2 · Provincial · 2 marks
5.2 If $cos 49° = k$ , determine the values of the following in terms of $k$ .
5.2.2 $1 - cos^{2} 41°$ (2)
Show the memo (2 ticks)Memo
1. ✓
  METHOD 1: For $sin^{2} 41°$ $1 - cos^{2} 41° = sin^{2} 41°$
  1 mark · Use the square identity
2. ✓
  METHOD 1: Answer $= k^{2}$
  1 mark · Find trig ratios from a sketch
9. Question 4.1 · March 2023 · Paper 1 · Provincial · 5 marks
4.1 Prove the following identity: $\frac{s i n ^{2} ( 18 0 ^{\circ} + θ ) + c o s ( - θ ) \cdot s i n ( 9 0 ^{\circ} - θ )}{s i n θ} + \frac{1}{t a n θ} = \frac{1 + c o s θ}{s i n θ}$ (5)
Show the memo (3 ticks)Memo
1. ✓
  Reduce LHS termsNumerator: $(- sin θ)^{2} + cos θ \cdot cos θ = sin^{2} θ + cos^{2} θ$
  Term 2: $\frac{1}{t a n θ} = \frac{c o s θ}{s i n θ}$
  3 marks · Use a reduction formula
2. ✓
  Apply identity $\frac{1}{s i n θ} + \frac{c o s θ}{s i n θ}$
  1 mark · Use the square identity
3. ✓
  Combine fractions $\frac{1 + c o s θ}{s i n θ} = RHS$
  1 mark · Add or subtract algebraic fractions
10. Question 3.3 · March 2023 · Paper 1 · Provincial · 4 marks
3.3 Prove the identity: $tan θ - \frac{c o s θ}{s i n θ} = \frac{1 - 2 c o s ^{2} θ}{s i n θ c o s θ}$ (4)
Show the memo (4 ticks)Memo
1. ✓
  $\frac{s i n θ}{c o s θ}$ LHS $= \frac{s i n θ}{c o s θ} - \frac{c o s θ}{s i n θ}$
  1 mark · Write tan as sin over cos
2. ✓
  Numerator $sin^{2} θ - cos^{2} θ$ $= \frac{s i n ^{2} θ - c o s ^{2} θ}{s i n θ c o s θ}$
  1 mark · Add or subtract algebraic fractions
3. ✓
  LCD: $sin θ cos θ$ Common denominator $sin θ cos θ$
  1 mark · Add or subtract algebraic fractions
4. ✓
  $1 - cos^{2} θ$ $= \frac{1 - c o s ^{2} θ - c o s ^{2} θ}{s i n θ c o s θ} = \frac{1 - 2 c o s ^{2} θ}{s i n θ c o s θ}$ = RHS
  1 mark · Use the square identity

Identities Grade 11 videos

Short lessons from South African teachers first. Videos load only when you tap play, to save your data.

Grade 11 Trigonometry Identities IntroductionMiss Martins Maths and Science

Grade 11 Trig - Topic 6 Restrictions on IdentitiesJessica Ballantyne

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