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Grade 12 · Trigonometry
Compound & Double Angle Identities (Grade 12)
MARKING GUIDELINE
Marks
35
Duration
55 minutes
Questions
3
Name:
Class:
Date:
Mark
/ 35
Instructions and Information
- Answer ALL the questions in this question paper.
- Answer QUESTION 1 by circling the letter (A–D) in the answer grid at the end of that section.
- Show ALL calculations clearly.
- Show all units where applicable.
- Number the answers correctly according to the numbering system used in this question paper.
- A non-programmable calculator may be used, unless stated otherwise.
- Write neatly and legibly.
Question 1
[10 MARKS]Four options are given as possible answers to the following questions. Choose the correct answer and circle the letter (A–D) in the grid at the end of this section. If you want to change your choice, put a cross through the wrong letter and circle your new choice.
- 1.1sin(A + B) = …(1)A)sin A cos B + cos A sin B✓B)sin A cos B − cos A sin BC)cos A cos B − sin A sin BD)sin A + sin BAnswer: A — That is the compound-angle formula on the information sheet.
- B — that is sin(A − B)
- C — that is cos(A + B)
- D — sine does not distribute over addition
- 1.2cos(A + B) = …(1)A)cos A cos B − sin A sin B✓B)cos A cos B + sin A sin BC)sin A cos B + cos A sin BD)cos A + cos BAnswer: A — The cosine compound formula carries the OPPOSITE sign to the one inside the bracket.
- B — that is cos(A − B)
- C — that is sin(A + B)
- D — cosine does not distribute over addition
- 1.3sin 2θ = …(1)A)2 sin θ cos θ✓B)2 sin θC)sin²θ + cos²θD)sin θ + cos θAnswer: A — Put A = B = θ into sin(A + B).
- B — doubling the ANGLE is not the same as doubling the ratio
- C — that expression equals 1, not sin 2θ
- D — the two ratios are multiplied, not added
- 1.4Which is NOT one of the three forms of cos 2θ?(1)A)2 sin²θ − 1✓B)cos²θ − sin²θC)1 − 2 sin²θD)2 cos²θ − 1Answer: A — The sine form is 1 − 2sin²θ; reversing it changes the sign of the whole expression.
- B — that is the basic form
- C — that is the sine form
- D — that is the cosine form
- 1.5Using the double-angle formula on 30°, sin 60° = …(1)A)√32✓B)12C)√34D)1Answer: A — sin 60° = 2 sin 30° cos 30° = 2(½)(√3/2) = √3/2.
- B — that is sin 30°, not sin 60°
- C — forgot the factor of 2
- D — sin 60° is less than 1
- 1.6cos 2θ written in terms of cos θ only is …(1)A)2 cos²θ − 1✓B)1 − 2 cos²θC)cos²θ − 1D)2 cos²θ + 1Answer: A — Substitute sin²θ = 1 − cos²θ into cos²θ − sin²θ.
- B — the signs have been reversed — that is −cos 2θ
- C — the coefficient of cos²θ is 2, not 1
- D — the constant is −1, not +1
- 1.7If sin θ = 35 and θ is acute, then sin 2θ = …(1)A)2425✓B)65C)725D)1225Answer: A — cos θ = 4/5, so sin 2θ = 2(3/5)(4/5) = 24/25.
- B — doubled sin θ instead of using the formula
- C — that is cos 2θ, not sin 2θ
- D — forgot the factor of 2
- 1.8For that same θ, cos 2θ = …(1)A)725✓B)2425C)1625D)−725Answer: A — cos 2θ = 1 − 2sin²θ = 1 − 18/25 = 7/25.
- B — that is sin 2θ
- C — that is cos²θ, not cos 2θ
- D — θ is acute and small here, so cos 2θ comes out positive
- 1.9sin 75° written as a compound angle is …(1)A)sin 45° cos 30° + cos 45° sin 30°✓B)sin 45° cos 30° − cos 45° sin 30°C)sin 45° + sin 30°D)cos 45° cos 30° − sin 45° sin 30°Answer: A — 75° = 45° + 30°, and sin(A + B) = sin A cos B + cos A sin B.
- B — that is sin(45° − 30°) = sin 15°
- C — sine does not distribute over addition
- D — that is cos 75°, not sin 75°
- 1.10cos 15° cos 45° + sin 15° sin 45° = …(1)A)√32✓B)12C)√22D)cos 75°Answer: A — This is cos(45° − 15°) = cos 30° = √3/2.
- B — that is cos 60°, which would come from 45° + 15°
- C — that is cos 45°
- D — the PLUS in the expansion means a MINUS inside the bracket
Answer grid — marking guideline
| 1.1 | A | B | C | D |
| 1.2 | A | B | C | D |
| 1.3 | A | B | C | D |
| 1.4 | A | B | C | D |
| 1.5 | A | B | C | D |
| 1.6 | A | B | C | D |
| 1.7 | A | B | C | D |
| 1.8 | A | B | C | D |
| 1.9 | A | B | C | D |
| 1.10 | A | B | C | D |
Question 2
[13 MARKS]Answer the questions below WITHOUT the use of a calculator.
- 2.1Determine the exact value of cos 15° using a compound-angle formula.(5)cos 15° = cos(45° − 30°) (1)
= cos 45°cos 30° + sin 45°sin 30° (2)
= 22·32 + 22·12 (1)
= 6 + 24 (1) - 2.2Simplify fully: sin 2A1 + cos 2A(4)sin 2A = 2 sin A cos A (1)
1 + cos 2A = 2cos2A (2)
= tan A (1) - 2.3Prove the identity: sin(A + B) − sin(A − B) = 2 cos A sin B(4)LHS = (sin A cos B + cos A sin B) − (sin A cos B − cos A sin B) (2)
= 2 cos A sin B (1) = RHS (1)
Question 3
[12 MARKS]Answer the questions below.
- 3.1Write down the THREE forms of the identity for cos 2A.(3)cos2A − sin2A (1); 2cos2A − 1 (1); 1 − 2sin2A (1)
- 3.2If sin 25° = p, express sin 50° in terms of p.(4)sin 50° = 2 sin 25°cos 25° (2)
cos 25° = 1 − p2 (1)
= 2p1 − p2 (1) - 3.3Solve for x ∈ [0° ; 360°]: sin 2x = cos x(5)2 sin x cos x − cos x = 0 (1)
cos x(2 sin x − 1) = 0 (1)
cos x = 0 gives x = 90° or 270° (1)
sin x = 12 gives x = 30° or 150° (2)
TOTAL: 35 marks
This question paper consists of 3 questions.