Grade 12 Maths · Trigonometry

Use a compound angle identity

Expand the trig ratio of a sum or difference of angles, for example cos(A − B) = cos A cos B + sin A sin B.

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Question 7.3 · March 2021 · Provincial · 4 marks

7.3 Determine the maximum value of

3 sin x + cos x

, without the use of a calculator.(4)

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More past paper questions: use a compound angle identity

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3. Question 6.2.2 · May/June 2025 · Paper 2 · National (DBE) · 2 marks
6.2 Consider the expression: $sin (A - B) - sin (A + B)$
6.2.2 Simplify the following expression to a single term: $sin 4 x - sin 10 x$ (2)
Show the memo (2 ticks)Memo
1. ✓
  Rewrite the expression to match the identity from 6.2.1We need to find angles A and B such that $A - B = 4 x$ and $A + B = 10 x$ .
  $A = 7 x$ and $B = 3 x$ .
  So, $sin 4 x - sin 10 x = sin (7 x - 3 x) - sin (7 x + 3 x)$
  1 mark · Use a compound angle identity
2. ✓
  Apply the proven identityUsing the result from 6.2.1: $sin (A - B) - sin (A + B) = - 2 cos A sin B$
  $= - 2 cos 7 x sin 3 x$
  1 mark · Use a compound angle identity
4. Question 5.1.3 · May/June 2025 · Paper 2 · National (DBE) · 4 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:
5.1.3 $cos (θ - 13 5^{\circ})$ (4)
Show the memo (4 ticks)Memo
1. ✓
  Expand using the compound angle identity for cosine $cos (θ - 13 5^{\circ}) = cos θ cos 13 5^{\circ} + sin θ sin 13 5^{\circ}$
  1 mark · Use a compound angle identity
2. ✓
  Reduce angles to special angles $= cos θ (- cos 4 5^{\circ}) + sin θ (sin 4 5^{\circ})$
  1 mark · Use a reduction formula
3. ✓
  Substitute trigonometric values $= (- \frac{5}{13}) (- \frac{2}{2}) + (- \frac{12}{13}) (\frac{2}{2})$
  
  OR
  
  $= (- \frac{5}{13}) (- \frac{1}{2}) + (- \frac{12}{13}) (\frac{1}{2})$
  1 mark · Find trig ratios from a sketch
4. ✓
  Simplify to final answer $= \frac{5 2}{26} - \frac{12 2}{26}$
  $= - \frac{7 2}{26}$
  1 mark · Find trig ratios from a sketch
5. Question 3.3 · March 2021 · Provincial · 3 marks
3.3 Given: $cos (A - B) = cos A cos B + sin A sin B$ . Use the identity for $cos (A - B)$ to derive an identity for $sin (A + B)$ .(3)
Show the memo (3 ticks)Memo
1. ✓
  Use co-function identity $sin (A + B) = cos (9 0^{\circ} - (A + B))$
  1 mark · Use co-ratio identities
2. ✓
  Regroup angles $= cos ((9 0^{\circ} - A) - B)$
  1 mark · Use a compound angle identity
3. ✓
  Expand and simplify $= cos (9 0^{\circ} - A) cos B + sin (9 0^{\circ} - A) sin B$
  $= sin A cos B + cos A sin B$
  1 mark · Use a compound angle identity
6. Question 5.2 · March 2025 · Provincial · 6 marks
5.2 Show that: $\frac{- 2 s i n 1 5 ^{\circ} c o s 1 5 ^{\circ}}{c o s ( 4 5 ^{\circ} - x ) c o s x - s i n ( 4 5 ^{\circ} - x ) s i n x} = - \frac{2}{2}$ (6)
Show the memo (5 ticks)Memo
1. ✓
  $- sin 3 0^{\circ}$ Numerator $= - 2 sin 1 5^{\circ} cos 1 5^{\circ} = - sin 3 0^{\circ}$
  1 mark · Use a double angle identity
2. ✓
  $- \frac{1}{2}$ $- sin 3 0^{\circ} = - \frac{1}{2}$
  1 mark · Use special angle values
3. ✓
  Compound angle: $cos [(4 5^{\circ} - x) + x]$ Denominator $= cos [(4 5^{\circ} - x) + x]$
  2 marks · Use a compound angle identity
4. ✓
  $cos 4 5^{\circ}$ $= cos 4 5^{\circ}$
  1 mark · Use a compound angle identity
5. ✓
  $\frac{1}{2}$ $cos 4 5^{\circ} = \frac{1}{2}$ , so $\frac{- \frac{1}{2}}{\frac{1}{2}} = - \frac{2}{2}$
  1 mark · Use special angle values

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