Skip to main content
Mathematics LibreTexts

9.3: Double-Angle Identities

  1. Last updated
  2. Save as PDF
  • Page ID
    203495

    • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \( \newcommand{\dsum}{\displaystyle\sum\limits} \)

    \( \newcommand{\dint}{\displaystyle\int\limits} \)

    \( \newcommand{\dlim}{\displaystyle\lim\limits} \)

    \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)

    ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\id}{\mathrm{id}}\)

    \( \newcommand{\Span}{\mathrm{span}}\)

    \( \newcommand{\kernel}{\mathrm{null}\,}\)

    \( \newcommand{\range}{\mathrm{range}\,}\)

    \( \newcommand{\RealPart}{\mathrm{Re}}\)

    \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)

    \( \newcommand{\Argument}{\mathrm{Arg}}\)

    \( \newcommand{\norm}[1]{\| #1 \|}\)

    \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)

    \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    \( \newcommand{\vectorA}[1]{\vec{#1}}      % arrow\)

    \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}}      % arrow\)

    \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vectorC}[1]{\textbf{#1}} \)

    \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)

    \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)

    \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    \(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)

    The Double Angle Identities

    Theorem: Double-Angle Identities

    \[ \begin{array}{rcl}
    \sin\left( 2\theta \right) & = & 2 \sin\left( \theta \right) \cos\left( \theta \right) \\[6pt] \cos\left( 2\theta \right) & = & \cos^2\left( \theta \right) - \sin^2\left( \theta \right) \\[6pt] & = & 2\cos^2\left( \theta \right) - 1 \\[6pt] & = & 1 - 2 \sin^2\left( \theta \right) \\[6pt] \tan\left( 2\theta \right) & = & \dfrac{2 \tan\left( \theta \right)}{1 - \tan^2\left( \theta \right)} \\[6pt] \end{array} \nonumber \]

    Proof
    Deriving the Double-Angle Identity for sine begins with the Sum Identity,\[\sin(\alpha+\beta)=\sin \left(\alpha\right) \cos \left(\beta\right)+\cos \left(\alpha\right) \sin \left(\beta\right). \nonumber \]If we let \(\alpha=\beta=\theta\), then we have\[\begin{array}{rrcl}
    & \sin(\theta+\theta) & = & \sin \left(\theta\right) \cos \left(\theta\right)+\cos \left(\theta\right) \sin \left(\theta\right) \\[6pt] \implies & \sin(2\theta) & = & 2\sin \left(\theta\right) \cos \left(\theta\right) \\[6pt] \end{array} \nonumber \]Deriving the Double-Angle Identity for cosine gives us three options. First, starting from the Sum Identity,\[\cos(\alpha+\beta)=\cos \left(\alpha\right) \cos \left(\beta\right) - \sin \left(\alpha\right) \sin \left(\beta\right),\nonumber \]and letting \(\alpha=\beta=\theta\), we have\[\begin{array}{rrcl}
    & \cos(\theta+\theta) & = & \cos \left(\theta\right) \cos \left(\theta\right)-\sin \left(\theta\right) \sin \left(\theta\right) \\[6pt] \implies & \cos(2\theta) & = & \cos^2 \left(\theta\right) - \sin^2 \left(\theta\right) \\[6pt] \end{array} \nonumber \]Using the Pythagorean Identities, we can expand this Double-Angle Identity for cosine and get two more variations. The first variation is\[\begin{array}{rcl}
    \cos(2\theta) & = & \cos^2 \left(\theta\right) - \sin^2 \left(\theta\right) \\[6pt] & = & \left(1-\sin^2 \left(\theta\right)\right)-\sin^2 \left(\theta\right) \\[6pt] & = & 1 - 2 \sin^2\left( \theta \right) \\[6pt] \end{array}\nonumber \]The second variation is\[\begin{array}{rcl}
    \cos(2\theta) & = & \cos^2 \left(\theta\right)-\sin^2 \left(\theta\right) \\[6pt] & = & \cos^2 \left(\theta\right)-\left(1-\cos^2 \left(\theta\right)\right) \\[6pt] & = & 2 \cos^2 \left(\theta\right)-1 \\[6pt] \end{array} \nonumber \]Similarly, to derive the Double-Angle Identity for the tangent, replacing \(\alpha=\beta=\theta\) in the sum formula gives\[\begin{array}{rrcl}
    & \tan(\alpha+\beta) & = & \dfrac{\tan \left(\alpha\right)+\tan \left(\beta\right)}{1-\tan \left(\alpha\right) \tan \left(\beta\right)} \\[6pt] \implies & \tan(\theta+\theta) & = & \dfrac{\tan \left(\theta\right)+\tan \left(\theta\right)}{1-\tan \left(\theta\right) \tan \left(\theta\right)} \\[6pt] \implies & \tan(2\theta) & = & \dfrac{2\tan \left(\theta\right)}{1-\tan^2 \left(\theta\right)} \\[6pt] \end{array} \nonumber \]
    Caution: Don't Factor Out of Functions!

    \[ \text{trig}(2 \theta ) \neq 2 \text{trig}(\theta) \nonumber \]

    Finding Exact Values of Trigonometric Functions Involving Double Angles

    Example \( \PageIndex{ 1 } \): Using double angles with triangles

    Let's consider a right triangle with an interior unkown angle of \(\theta\), where we are given two sides. Find the values of \( \sin(2 \theta) \), \( \cos(2\theta)\), and \(\tan(2\theta)\).

    Example \(\PageIndex{2}\): A popular style of problem revisited.

    If \(\alpha\) is a Quadrant III angle with \(\sin(\alpha) = -\frac{12}{13}\), and \(\beta\) is a Quadrant IV angle with \(\tan(\beta) = -5\), find the following

    1. \(\sin(2 \alpha)\)
    2. \(\cos(2 \alpha)\)
    3. \(\cos(2 \beta)\)

    Revisiting Proofs of Identities

    Example \(\PageIndex{3}\): We can now do even more proofs!

    Prove each identity.

    1. \(\cos^4 \left(\theta\right)-\sin^4 \left(\theta\right) =\cos\left(2\theta\right)\)
    2. \(\cos(2\theta)\cos \theta=\cos^3 \left(\theta\right)−\cos \left(\theta\right) \sin^2 \left(\theta\right)\)

    Revisiting Graphs of Trigonometric Functions

    Example \(\PageIndex{4}\): We can graph more interesting functions!

    Graph the function.\[ g(x) = \dfrac{6\tan\left( \frac{\pi}{4}x \right)}{1 - \tan^2\left( \frac{\pi}{4} x \right)} \nonumber \]

    This page titled 9.3: Double-Angle Identities is shared under a CC BY-SA 12 license and was authored, remixed, and/or curated by Roy Simpson.