Compound & Double Angle Identities — Grade 12 Maths (CAPS)

The formulae for sin(A ± B) and cos(A ± B), the three forms of cos 2A, using them to find exact values without a calculator, and proving identities.

1The compound angle formulae

cos(A − B) = cos A cos B + sin A sin B
cos(A + B) = cos A cos B − sin A sin B

sin(A + B) = sin A cos B + cos A sin B
sin(A − B) = sin A cos B − cos A sin B

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Remember the sign rule: cos changes the sign, sin keeps it. cos(A + B) has a minus in the middle; sin(A + B) has a plus.

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sin(A + B) is not sin A + sin B. Test it: sin(30° + 60°) = sin 90° = 1, but sin 30° + sin 60° = 0,5 + 0,866 = 1,366.

2Exact values without a calculator

Worked ExampleDetermine the exact value of cos 75°

1
Write 75° using two special angles: 75° = 45° + 30°.
Both are on the special-angle table.
2
cos 75° = cos 45° cos 30° − sin 45° sin 30°.
Compound formula for cos(A + B).
3
= 1/(√2) × (√3)/2 − 1/(√2) × 1/2.
4
= (√3 − 1)/(2√2).
Common denominator.
5
= (√6 − √2)/4.
Rationalise. Check on a calculator: 0,2588. ✓

3The double angle formulae

sin 2A = 2 sin A cos A

cos 2A = cos²A − sin²A
cos 2A = 2 cos²A − 1
cos 2A = 1 − 2 sin²A

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cos 2A has three forms and choosing the right one is the whole skill. If the question also mentions sinA, use 1 − 2 sin²A. If it mentions cosA, use 2 cos²A − 1.

Worked ExampleIf sin θ = 3/5 and θ is acute, determine cos 2θ

1
Use cos 2θ = 1 − 2 sin²θ.
We were given sin θ, so pick the form with only sin in it.
2
= 1 − 2(3/5)².
3
= 1 − 2(9/25) = 1 − 18/25.
4
= 7/25.
No need to find cos θ at all.

4Proving an identity

Work down one side only until it equals the other. Never move terms across the equals sign — you are not solving an equation.

Worked ExampleProve that (sin 2A)/(1 + cos 2A) = tan A

1
LHS = (2 sin A cos A)/(1 + (2 cos²A − 1)).
Expand sin 2A, and choose the cos 2A form that will cancel the 1.
2
= (2 sin A cos A)/(2 cos²A).
The +1 and −1 cancel — that is why this form was chosen.
3
= (sin A)/(cos A).
Cancel 2 cos A top and bottom — allowed only while cos A ≠ 0 (see the restriction below).
4
= tan A = RHS. ✓
5
Restriction: 1 + cos 2A = 2 cos²A, which is zero when cos A = 0. So the identity holds only for cos A ≠ 0, i.e. A ≠ 90° + k·180°, k ∈ ℤ.
Take the restriction from the ORIGINAL expression, before anything cancels. tan A is undefined at exactly those angles too.

Practice exercises

Work each one out, then click to reveal the answer.

📄 Worksheet ✅ Memo 📝 Exam paper 🗝️ Marking guide

1
Expand cos(A − B).
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cos A cos B + sin A sin B
2
Expand sin(A + B).
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sin A cos B + cos A sin B
3
Determine the exact value of cos 75° without a calculator.
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cos(45°+30°) = (√6 − √2)/4
4
Determine the exact value of sin 15° without a calculator.
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sin(45°−30°) = (√6 − √2)/4
5
Write down the three forms of cos 2A.
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cos²A − sin²A, 2 cos²A − 1, 1 − 2 sin²A
6
If sin θ = 3/5 and θ is acute, determine cos 2θ.
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1 − 2(9/25) = 7/25
7
If sin θ = 3/5 and θ is acute, determine sin 2θ.
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cos θ = 4/5, so 2(3/5)(4/5) = 24/25
8
Prove that (sin 2A)/(1 + cos 2A) = tan A, and state the restriction on A.
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(2 sin A cos A)/(1 + (2 cos²A − 1)) ✓ = (2 sin A cos A)/(2 cos²A) ✓ = (sin A)/(cos A) ✓ = tan A ✓
The original denominator 1 + cos 2A = 2 cos²A is zero when cos A = 0, so cos A ≠ 0, i.e. A ≠ 90° + k·180°, k ∈ ℤ ✓
9
Simplify 2 sin 15° cos 15°.
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= sin 30° = 1/2
10
Why is sin(A + B) ≠ sin A + sin B?
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Test A = 30°, B = 60°: sin 90° = 1 but sin 30° + sin 60° ≈ 1,37, so they are not equal.