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Lesson 3: Double-Angle and Half-Angle Formulas

Estimated time: 35-40 minutes

Learning Objectives

Double-Angle Formulas

sin 2θ = 2 sin θ cos θ

cos 2θ = cos²θ − sin²θ = 2cos²θ − 1 = 1 − 2sin²θ

tan 2θ = 2 tan θ / (1 − tan²θ)

These follow directly from the sum formulas with A = B = θ.

Example 1: Finding sin 2θ

If sin θ = 3/5 and θ in QI, find sin 2θ.

cos θ = 4/5. sin 2θ = 2(3/5)(4/5) = 24/25.

Example 2: cos 2θ Three Ways

If sin θ = 3/5, find cos 2θ using all three forms.

cos²−sin² = 16/25 − 9/25 = 7/25. 2cos²−1 = 32/25−1 = 7/25. 1−2sin² = 1−18/25 = 7/25.

Power-Reducing Formulas

sin²θ = (1 − cos 2θ) / 2

cos²θ = (1 + cos 2θ) / 2

These are rearrangements of the cos 2θ formulas and are very useful in calculus.

Half-Angle Formulas

sin(θ/2) = ±√[(1 − cos θ) / 2]

cos(θ/2) = ±√[(1 + cos θ) / 2]

tan(θ/2) = (1 − cos θ) / sin θ = sin θ / (1 + cos θ)

Example 3: Exact Value Using Half-Angle

Find cos 15° using the half-angle formula (15 = 30/2).

cos 15° = √[(1 + cos 30)/2] = √[(1 + √3/2)/2] = √[(2+√3)/4] = √(2+√3)/2

Example 4: Half-Angle with Given Info

If cos θ = −3/5 and π < θ < 3π/2, find sin(θ/2).

θ/2 is in (π/2, 3π/4), so QII where sin is positive.

sin(θ/2) = +√[(1−(−3/5))/2] = √[(8/5)/2] = √(4/5) = 2/√5 = 2√5/5

Check Your Understanding

1. If cos θ = 12/13 (θ in QI), find sin 2θ.

sin θ = 5/13. sin 2θ = 2(5/13)(12/13) = 120/169.

2. Express cos²(3x) using a power-reducing formula.

(1 + cos 6x)/2.

3. Find tan 22.5° using a half-angle formula (22.5 = 45/2).

tan(45/2) = (1 − cos 45)/sin 45 = (1 − √2/2)/(√2/2) = (2 − √2)/√2 = √2 − 1.

Key Takeaways

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