Grade 12 Maths · Trigonometry · Paper 2

Trigonometric Identities Grade 12

Real Trigonometric Identities past paper questions and answers from 2021 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: 21 past paper questions
skills: 5 skills
videos: 5 videos

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Question 5.2 · March 2025 · Provincial · 6 marks

5.2 Simplify:

\frac{s i n ( - 18 0 ^{\circ} - θ ) \cdot t a n ( 18 0 ^{\circ} - θ ) \cdot c o s ( - θ )}{c o s ^{2} ( 9 0 ^{\circ} + θ ) + 3 s i n ^{2} θ}

(6)

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

Trigonometric Identities skills

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4. Question 5.1.2 · May/June 2025 · Paper 2 · National (DBE) · 2 marks
5.1 If $cos θ = - \frac{5}{13}$ where $18 0^{\circ} < θ < 36 0^{\circ}$ , determine, without using a calculator, the value of:
5.1.2 $tan (36 0^{\circ} - θ)$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Apply reduction formula $tan (36 0^{\circ} - θ) = - tan θ$
  1 mark · Use a reduction formula
2. ✓
  Substitute values and simplifyFrom 5.1.1, we have $x = - 5$ and $y = - 12$ .
  $= - (\frac{y}{x})$
  $= - (\frac{- 12}{- 5})$
  $= - \frac{12}{5}$
  1 mark · Find trig ratios from a sketch
5. Question 7.1.1 · March 2023 · Provincial · 2 marks
7.1 Given $cos 4 0^{\circ} = t$ .

Without using a calculator, determine each of the following in terms of t.
7.1.1 $tan 4 0^{\circ}$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Draw diagram or use identityDiagram with angle 40, adj t, hyp 1. Opp $1 - t^{2}$ .
  1 mark · Use the theorem of Pythagoras
2. ✓
  Write ratio $tan 4 0^{\circ} = \frac{1 - t ^{2}}{t}$
  1 mark · Find trig ratios from a sketch
6. Question 6.1.1 · March 2022 · Provincial · 2 marks
6.1 Given $cos 2 0^{\circ} = p$ and $sin 1 4^{\circ} = q$ .

Without using a calculator, calculate the value of the following in terms of $p$ , $q$ or $p$ and $q$ .
6.1.1 $sin 2 0^{\circ}$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Method: third side of the sketchRight-angled triangle with angle $2 0^{\circ}$ , adjacent side $p$ and hypotenuse $1$ : opposite side $= 1 - p^{2}$ (Pythagoras)
  1 mark · Use the theorem of Pythagoras
2. ✓
  Answer $sin 2 0^{\circ} = 1 - p^{2}$
  1 mark · Write trig ratios in a triangle
7. Question 7.1.1 · March 2022 · Provincial · 4 marks
7.1 For each of the following:
7.1.1 Prove the following identity: $cos 4 x = 8 cos^{4} x - 8 cos^{2} x + 1$ (4)
Show the memo (4 ticks)Memo
1. ✓
  Double angle expansion $cos 4 x = cos (2 \cdot 2 x) = 2 cos^{2} 2 x - 1$
  1 mark · Use a double angle identity
2. ✓
  Expand cos 2x $= 2 (2 cos^{2} x - 1)^{2} - 1$
  1 mark · Use a double angle identity
3. ✓
  Square the bracket $= 2 (4 cos^{4} x - 4 cos^{2} x + 1) - 1$
  1 mark · Square a binomial
4. ✓
  Simplify $= 8 cos^{4} x - 8 cos^{2} x + 2 - 1$
  $= 8 cos^{4} x - 8 cos^{2} x + 1$
  1 mark · Collect like terms
8. Question 3.2 · March 2021 · Provincial · 6 marks
3.2 Simplify the following expression into a single trigonometric ratio: $\frac{s i n ( x - 18 0 ^{\circ} ) \cdot c o s ( x - 9 0 ^{\circ} )}{c o s ( - x - 360 ) \cdot s i n ( 9 0 ^{\circ} + x )}$ (6)
Show the memo (3 ticks)Memo
1. ✓
  Reduce numerator $sin (x - 18 0^{\circ}) = - sin (18 0^{\circ} - x) = - sin x$
  $cos (x - 9 0^{\circ}) = cos (9 0^{\circ} - x) = sin x$
  2 marks · Use a reduction formula
2. ✓
  Reduce denominator $cos (- x - 36 0^{\circ}) = cos (- (x + 36 0^{\circ})) = cos (x + 36 0^{\circ}) = cos x$
  $sin (9 0^{\circ} + x) = cos x$
  2 marks · Use a reduction formula
3. ✓
  Simplify fraction $\frac{- s i n x \cdot s i n x}{c o s x \cdot c o s x} = - \frac{s i n ^{2} x}{c o s ^{2} x} = - tan^{2} x$
  2 marks · Write tan as sin over cos
9. Question 6.2 · March 2021 · Provincial · 5 marks
6.2 Simplify without using a calculator: $\frac{c o s ( - 18 0 ^{\circ} ) \cdot t a n θ \cdot c o s 69 0 ^{\circ} \cdot s i n ( θ - 18 0 ^{\circ} )}{c o s ^{2} ( θ - 9 0 ^{\circ} )}$ (5)
Show the memo (2 ticks)Memo
1. ✓
  Simplify individual terms $cos (- 18 0^{\circ}) = - 1$
  $cos 69 0^{\circ} = cos (33 0^{\circ}) = cos 3 0^{\circ} = \frac{3}{2}$
  $sin (θ - 18 0^{\circ}) = - sin (18 0^{\circ} - θ) = - sin θ$
  $cos (θ - 9 0^{\circ}) = cos (9 0^{\circ} - θ) = sin θ ⟹ cos^{2} (\dots) = sin^{2} θ$
  3 marks · Use a reduction formula
2. ✓
  Substitute and Simplify $= \frac{( - 1 ) ( \frac{s i n θ}{c o s θ} ) ( \frac{3}{2} ) ( - s i n θ )}{s i n ^{2} θ}$
  $= \frac{\frac{3}{2} s i n ^{2} θ}{s i n ^{2} θ c o s θ}$
  $= \frac{3}{2 c o s θ}$
  2 marks · Simplify an algebraic fraction
10. Question 7.1 · March 2021 · Provincial · 3 marks
7.1 Prove the identity: $\frac{1 - s i n 2 x}{s i n x - c o s x} = sin x - cos x$ (3)
Show the memo (2 ticks)Memo
1. ✓
  Expand numerator $1 - sin 2 x = (sin^{2} x + cos^{2} x) - 2 sin x cos x = (sin x - cos x)^{2}$
  2 marks · Use the square identity
2. ✓
  Simplify $\frac{( s i n x - c o s x ) ^{2}}{s i n x - c o s x} = sin x - cos x$
  1 mark · Simplify an algebraic fraction

Trigonometric Identities Grade 12 videos

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Grade 11 Trigonometry Identities IntroductionMiss Martins Maths and Science

Compound Angles Grade 12Kevinmathscience

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