The addition formulas expand a trigonometric function of a sum or difference of angles. The double-angle formulas are the special case where both angles are equal.
This lesson is part of trigonometric functions. It builds on the identities in using identities to simplify expressions.
What are the formulas?
| Formula | Double-angle form (B = A) |
|---|---|
| sin(A ± B) = sin A cos B ± cos A sin B | sin 2A = 2 sin A cos A |
| cos(A ± B) = cos A cos B ∓ sin A sin B | cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A |
| tan(A ± B) = (tan A ± tan B) ÷ (1 ∓ tan A tan B) | tan 2A = 2 tan A ÷ (1 − tan²A) |
Notice the sign flip in cosine: cos(A + B) has a minus between the products.
Worked example 1: an exact value
Find the exact value of sin 75°.
Write 75° = 45° + 30°. Then sin 75° = sin 45° cos 30° + cos 45° sin 30° = (√2 ÷ 2)(√3 ÷ 2) + (√2 ÷ 2)(1 ÷ 2) = (√6 + √2) ÷ 4.
Check numerically: (2.449 + 1.414) ÷ 4 ≈ 0.966, and sin 75° ≈ 0.966.
Worked example 2: from a given ratio
Given sin A = 3/5 with A acute, find sin 2A and cos 2A.
Since A is acute, cos A = 4/5 (from the 3-4-5 triangle). Then sin 2A = 2 sin A cos A = 2 × 3/5 × 4/5 = 24/25, and cos 2A = 1 − 2sin²A = 1 − 2(9/25) = 7/25.
Check: (24/25)² + (7/25)² = (576 + 49) ÷ 625 = 1.
Worked example 3: an equation
Solve sin 2x = sin x for 0° ≤ x ≤ 360°.
Replace sin 2x with 2 sin x cos x: 2 sin x cos x − sin x = 0, so sin x(2cos x − 1) = 0. Do not divide by sin x, because that loses the solutions where sin x = 0.
Sin x = 0 gives x = 0°, 180°, 360°. Cos x = ½ gives x = 60°, 300°. The full set is 0°, 60°, 180°, 300°, 360°.
The mistake that costs marks
The common slip is to treat the formulas as if they distribute, writing sin(A + B) = sin A + sin B or sin 2A = 2 sin A.
| Step | Wrong | Right |
|---|---|---|
| Expand sin 2A | 2 sin A | 2 sin A cos A |
| Test with A = 30° | 2 sin 30° = 1, but sin 60° = 0.866 | 2 × 0.5 × 0.866 = 0.866, which matches sin 60° |
| Verdict | Fails the test | Passes the test |
Always test a new formula with a simple angle before you trust it in a long question.
Check yourself
Given cos A = 5/13 with A acute, find sin 2A and cos 2A.
Answer
Since A is acute, sin A = 12/13 (from the 5-12-13 triangle).
sin 2A = 2 × 12/13 × 5/13 = 120/169, and cos 2A = 2cos²A − 1 = 2(25/169) − 1 = −119/169.
Check: 120² + 119² = 14 400 + 14 161 = 28 561 = 169².
What to study next
Next, see how these functions look as graphs in sketching transformed trigonometric functions. Then test the whole chapter with the trigonometric functions practice set.
If you want a teacher to work through these formulas with you, see online one-to-one Additional Mathematics tuition.