Trig Identities

Double-Angle Formulas

Derive and apply sin(2θ) = 2 sin θ cos θ and cos(2θ) = cos²θ − sin²θ using exact special-angle values.

Example: θ = π/6 (30°), so 2θ = π/3 (60°). sin(π/3) = 2·(1/2)·(√3/2) = √3/2 ✓ cos(π/3) = (√3/2)² − (1/2)² = 3/4 − 1/4 = 1/2 ✓

double-angle The double-angle formulas follow from the angle-sum formulas with a = b = θ: sin(2θ) = sin(θ+θ) = 2 sin θ cos θ cos(2θ) = cos(θ+θ) = cos²θ − sin²θ

Example

Apply double-angle identities using exact special-angle values.

highlighted = computed this step

Step 1 — Set up

Set up the expression.

θ=π62θ=π3\theta= \frac{\pi}{6} \quad 2 \theta= \frac{\pi}{3}

Step 2 — Sine formula

Apply the sine double-angle formula.

sin⁡(2θ)=2sin⁡θcos⁡θ\sin( \hl{2} \theta)= \hl{2} \sin\theta\cos\theta

Step 3 — Sine value

Compute 2 times sine theta times cosine theta to get root 3 over 2.

2⋅12⋅32=322 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \hlmath{\frac{\sqrt{3}}{2}}

Step 4 — Cosine formula

Apply the cosine double-angle formula.

cos⁡(2θ)=cos⁡2θ−sin⁡2θ\cos( \hl{2} \theta)=\cos^{ \hl{2} }\theta-\sin^{ \hl{2} }\theta

Step 5 — Cosine value

Square the exact values to get 1 over 2.

(32)2−(12)2=12( \frac{\sqrt{3}}{2} )^{ 2 }-( \frac{1}{2} )^{ 2 }= \hlmath{\frac{1}{2}}