click to advance
Unit 5 · Analytic Trigonometry
5.5 Double Angle &
Half Angle Identities
Objective: Use multiple-angle formulas to rewrite and evaluate trigonometric functions.
sin(2α) / cos(2α) / tan(2α) Three forms of cos(2α) sin(θ/2) / cos(θ/2) / tan(θ/2) Exact values Quadrant analysis
Reference
Double-Angle Identities
sin
\[\sin(2\alpha) = 2\sin\alpha\cos\alpha\]
cos
\[\cos(2\alpha) = \cos^2\!\alpha - \sin^2\!\alpha\]
\[\phantom{\cos(2\alpha)} = 2\cos^2\!\alpha - 1\]
\[\phantom{\cos(2\alpha)} = 1 - 2\sin^2\!\alpha\]
tan
\[\tan(2\alpha) = \dfrac{2\tan\alpha}{1 - \tan^2\!\alpha}\]

Derived from the Sum Identities — let β = α in sin(α+β), cos(α+β), tan(α+β)

Example 1
Using Double-Angle Identities
Find \(\sin(2\theta)\), \(\cos(2\theta)\), and \(\tan(2\theta)\) given \(\cos\theta = \dfrac{5}{13}\) where \(\dfrac{3\pi}{2} < \theta < 2\pi\).
Setup — Find sinθ and tanθ first
\[\theta \text{ in QIV with } \cos\theta = \tfrac{5}{13} \;\Longrightarrow\; \text{5-12-13 triple} \;\Longrightarrow\; \sin\theta = -\tfrac{12}{13},\quad \tan\theta = -\tfrac{12}{5}\] QIV: cosine positive, sine negative
sin(2θ)
\[\sin(2\theta) = 2\sin\theta\cos\theta = 2\!\left(-\tfrac{12}{13}\right)\!\left(\tfrac{5}{13}\right) = -\dfrac{120}{169}\] Apply \(\sin(2\theta) = 2\sin\theta\cos\theta\)
cos(2θ)
\[\cos(2\theta) = 2\cos^2\!\theta - 1 = 2\!\left(\tfrac{5}{13}\right)^{\!2} - 1 = \tfrac{50}{169} - \tfrac{169}{169} = -\dfrac{119}{169}\] Use \(2\cos^2\theta - 1\) since cosθ is given
tan(2θ)
\[\tan(2\theta) = \dfrac{2\tan\theta}{1 - \tan^2\!\theta} = \dfrac{2\!\left(-\tfrac{12}{5}\right)}{1 - \tfrac{144}{25}} = \dfrac{-\tfrac{24}{5}}{-\tfrac{119}{25}} = \dfrac{120}{119}\] Note: \(\tan\theta = \tfrac{\sin\theta}{\cos\theta} = -\tfrac{12}{5}\)
sin(2θ) \[-\dfrac{120}{169}\]
cos(2θ) \[-\dfrac{119}{169}\]
tan(2θ) \[\dfrac{120}{119}\]
Example 2
Solving an Equation with Double Angles
Solve \(2\cos x + \sin(2x) = 0\) where \(0 \le x < 2\pi\).
Strategy
Rewrite \(\sin(2x)\) with the double-angle identity so every term involves \(x\) only, then factor.
Step 1 — Substitute the double-angle identity
\[2\cos x + 2\sin x\cos x = 0\] Replace \(\sin(2x)\) with \(2\sin x\cos x\)
Step 2 — Factor
\[2\cos x\,(1 + \sin x) = 0\] Factor out \(2\cos x\)
Step 3 — Solve each factor
\[2\cos x = 0 \;\Rightarrow\; \cos x = 0 \;\Rightarrow\; x = \tfrac{\pi}{2},\;\tfrac{3\pi}{2}\]
\[1 + \sin x = 0 \;\Rightarrow\; \sin x = -1 \;\Rightarrow\; x = \tfrac{3\pi}{2}\]
Answer
\[x = \dfrac{\pi}{2}, \quad \dfrac{3\pi}{2}\]
Example 3
Rewriting with a Double-Angle Identity
Use a double-angle identity to rewrite \(\cos^2(5\alpha) - \sin^2(5\alpha)\).
Recognize the Pattern
The expression matches \(\cos^2 u - \sin^2 u = \cos(2u)\). Let \(u = 5\alpha\)
Apply and Back-Substitute
\[\cos^2(5\alpha) - \sin^2(5\alpha) = \cos(2 \cdot 5\alpha)\] Back-substitute \(u = 5\alpha\) into \(\cos(2u)\)
Answer
\[\cos(10\alpha)\]
Reference
Half-Angle Identities
sin
\[\sin\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1-\cos\theta}{2}}\]
cos
\[\cos\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1+\cos\theta}{2}}\]
tan
\[\tan\dfrac{\theta}{2} = \pm\sqrt{\dfrac{1-\cos\theta}{1+\cos\theta}} = \dfrac{1-\cos\theta}{\sin\theta} = \dfrac{\sin\theta}{1+\cos\theta}\]

The ± sign is determined by the quadrant of θ/2 — not θ. Always identify the quadrant of θ/2 first!

Example 4
Exact Value with a Half-Angle Identity
Find the exact value of \(\cos 165°\).
Recognize as a Half Angle & Determine the Sign
\[\cos 165° = \cos\dfrac{330°}{2}\] 165° is in QII → cosine is negative → use the − sign
Apply the Half-Angle Formula
\[\cos 165° = -\sqrt{\dfrac{1 + \cos 330°}{2}}\] \(\cos 330° = \cos(-30°) = \cos 30° = \dfrac{\sqrt{3}}{2}\)
Substitute and Simplify
\[-\sqrt{\dfrac{1 + \frac{\sqrt{3}}{2}}{2}} = -\sqrt{\dfrac{\frac{2+\sqrt{3}}{2}}{2}} = -\sqrt{\dfrac{2+\sqrt{3}}{4}}\]
Answer
\[\cos 165° = -\dfrac{\sqrt{2+\sqrt{3}}}{2}\]
Example 5
Half-Angle from Given Information
Given \(\sin x = \dfrac{2}{5}\) where \(0 < x < \dfrac{\pi}{2}\), find \(\cos\dfrac{x}{2}\).
Find cos x & Determine the Sign
\[\cos x = \dfrac{\sqrt{21}}{5}\] Right triangle: opp = 2, hyp = 5 → adj = \(\sqrt{21}\). x in QI → x/2 in QI → cosine positive
Apply the Half-Angle Formula
\[\cos\dfrac{x}{2} = +\sqrt{\dfrac{1 + \cos x}{2}} = \sqrt{\dfrac{1 + \frac{\sqrt{21}}{5}}{2}}\]
Simplify
\[= \sqrt{\dfrac{\frac{5+\sqrt{21}}{5}}{2}} = \sqrt{\dfrac{5+\sqrt{21}}{10}}\]
Answer
\[\cos\dfrac{x}{2} = \sqrt{\dfrac{5+\sqrt{21}}{10}}\]
Example 6
Half-Angle — Quadrant Analysis
Given \(\cos\theta = \dfrac{5}{13}\) where \(\dfrac{3\pi}{2} < \theta < 2\pi\), find \(\sin\dfrac{\theta}{2}\).
Determine the Quadrant of θ/2
\[\dfrac{3\pi}{2} < \theta < 2\pi \;\Longrightarrow\; \dfrac{3\pi}{4} < \dfrac{\theta}{2} < \pi\] θ/2 is in QII → sine is positive → use the + sign
Apply the Half-Angle Formula
\[\sin\dfrac{\theta}{2} = +\sqrt{\dfrac{1 - \cos\theta}{2}} = \sqrt{\dfrac{1 - \frac{5}{13}}{2}} = \sqrt{\dfrac{\frac{8}{13}}{2}}\] Substitute \(\cos\theta = \tfrac{5}{13}\)
Simplify
\[\sqrt{\dfrac{8}{26}} = \sqrt{\dfrac{4}{13}} = \dfrac{2}{\sqrt{13}} = \dfrac{2\sqrt{13}}{13}\]
Answer
\[\sin\dfrac{\theta}{2} = \dfrac{2\sqrt{13}}{13}\]