Practice Maths

Compound and Double Angle Formulas

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Key Terms

sin(A ± B)
sin A cos B ± cos A sin B.
cos(A ± B)
cos A cos B ∓ sin A sin B (NOTE: cos uses the OPPOSITE sign pattern to sin).
tan(A ± B)
(tan A ± tan B) / (1 ∓ tan A tan B).
Double angle sin
sin 2A = 2 sin A cos A.
Double angle cos
cos 2A = cos²A − sin²A = 2cos²A − 1 = 1 − 2sin²A (three equivalent forms).
Exact values using compound formulas
Use 75° = 45° + 30°, 15° = 45° − 30°, π/12 = π/4 − π/6 to find exact values.

Compound and Double Angle Formulas — Key Facts

Compound angle formulas:

sin(A ± B) = sin A cos B ± cos A sin B

cos(A ± B) = cos A cos B ∓ sin A sin B   (note: signs are reversed for cos)

tan(A ± B) = (tan A ± tan B) / (1 ∓ tan A tan B)

Double angle formulas (set B = A in the compound formulas):

sin 2A = 2 sin A cos A

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

tan 2A = 2 tan A / (1 − tan²A)

Useful for exact values: 75° = 45° + 30°; 15° = 45° − 30°; π/12 = π/4 − π/6

Worked Example 1 — Exact Value Using Compound Angle

Find the exact value of sin 75°.

sin 75° = sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°

= (1/√2)(√3/2) + (1/√2)(1/2) = √3/(2√2) + 1/(2√2) = (√3 + 1)/(2√2)

Rationalising: = (√3 + 1)√2/4 = (√6 + √2)/4

Worked Example 2 — Double Angle from Given Ratio

Given sinθ = 3/5 (θ acute), find sin 2θ, cos 2θ, and tan 2θ.

cosθ = 4/5 (positive, since θ is acute).

sin 2θ = 2 sinθ cosθ = 2(3/5)(4/5) = 24/25

cos 2θ = cos²θ − sin²θ = 16/25 − 9/25 = 7/25

tan 2θ = sin 2θ / cos 2θ = (24/25)/(7/25) = 24/7

Hot Tip: The sign rule for cos(A±B) is the OPPOSITE of what you might expect: cos(A+B) uses minus, and cos(A−B) uses plus. A mnemonic: for cosine, the compound angle formula “crosses over” the sign.

Why Compound Angle Formulas Matter

Trigonometry tables (and calculators) give us values for standard angles — 30°, 45°, 60°, 90° and so on. But in mathematics, we frequently encounter angles like 75°, 105°, or 15°. The compound angle formulas allow us to compute exact values for these angles by decomposing them into sums or differences of standard angles. This is far more than a computational trick — these formulas are fundamental identities used throughout calculus (especially in integration of trigonometric functions) and Fourier analysis.

The Compound Angle Formulas

The sine compound angle formula is: sin(A + B) = sin A cos B + cos A sin B. Setting B negative: sin(A − B) = sin A cos B − cos A sin B. For cosine, the signs reverse: cos(A + B) = cos A cos B − sin A sin B. This sign reversal for cosine is the most commonly confused point. The tangent formula follows by dividing: tan(A + B) = (tan A + tan B)/(1 − tan A tan B).

These formulas can be proved geometrically (using the unit circle and distance formula) or algebraically. The key point for exam work is to memorise and apply them correctly.

Deriving the Double Angle Formulas

Setting B = A in the compound formulas immediately gives the double angle formulas. For cosine, there are three equivalent forms of cos 2A, and knowing which to use in context is an important skill:

cos 2A = cos²A − sin²A   (directly from cos(A+A))

cos 2A = 2cos²A − 1   (apply cos²A = 1 − sin²A to the first form)

cos 2A = 1 − 2sin²A   (apply sin²A = 1 − cos²A to the first form)

The second form is used to express cos²A = (1 + cos 2A)/2, and the third gives sin²A = (1 − cos 2A)/2. These “power-reducing” identities are essential in integration.

Finding Exact Values

To find an exact value like sin 15°, write it as sin(45° − 30°) and apply the compound formula. A good strategy is to recognise which standard angles sum or differ to give the target. Angles that appear frequently: 75° = 45°+30°, 105° = 60°+45°, 15° = 45°−30°, π/12 = π/4−π/6.

Given One Ratio, Finding Double Angle Values

If you are given sinθ = k (and told the quadrant of θ), first find cosθ using the Pythagorean identity, then apply the double angle formulas. The quadrant of θ determines the sign of cosθ, which in turn determines the signs of all double-angle values (remembering that 2θ may be in a different quadrant from θ).

Applications in Proving Identities

Many identity proofs require substituting double angle formulas strategically. For example, to prove that sin 2θ / (1 + cos 2θ) = tanθ, substitute sin 2θ = 2 sinθ cosθ and 1 + cos 2θ = 1 + 2cos²θ − 1 = 2cos²θ. The ratio becomes 2sinθcosθ / (2cos²θ) = sinθ/cosθ = tanθ.

Mastery Practice

  1. Fluency

    Q1 — Expanding Compound Angles

    Expand, then simplify if possible:

    (a) sin(A + 30°)    (b) cos(45° − B)    (c) tan(π/4 + x)

  2. Fluency

    Q2 — Exact Values Using Compound Angles

    Find exact values for: (a) cos 75°    (b) sin(π/12)    (c) tan 15°

  3. Fluency

    Q3 — Double Angle Values

    Given sinθ = 5/13 with θ acute, find: (a) sin 2θ    (b) cos 2θ    (c) tan 2θ

  4. Fluency

    Q4 — Three Forms of cos 2A

    Given cosθ = 3/5 (θ acute), evaluate cos 2θ using all three forms and verify they agree.

  5. Understanding

    Q5 — Using sin(A + B) to Simplify

    Prove that sin(x + π) = −sin x.

  6. Understanding

    Q6 — Proving a Double Angle Identity

    Prove that sin 2θ / (1 + cos 2θ) = tanθ.

  7. Understanding

    Q7 — Exact Value for a Non-Standard Angle

    Find the exact value of cos 105°.

  8. Understanding

    Q8 — Double Angle with Obtuse Angle

    Given that cosα = −3/5 and π/2 < α < π, find sin 2α and cos 2α.

  9. Problem Solving

    Q9 — Compound Angle Identity Proof

    Prove that cos 2A / (1 − sin 2A) = (cos A − sin A) / (cos A + sin A) [hint: factorise numerator and denominator on the LHS].

  10. Problem Solving

    Q10 — Finding Exact Value with Half-Angle Strategy

    Given that sinθ = 7/25 and θ is acute, find the exact value of sin(θ + 45°).

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