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Mathematics LibreTexts

3.3: Double-Angle and Half-Angle Formulas

  • Page ID
    3287
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    A special case of the addition formulas is when the two angles being added are equal, resulting in the double-angle formulas:

    \[\begin{align} \sin\;2\theta ~&=~ 2\;\sin\;\theta ~ \cos\;\theta\label{eqn:doublesin}\\
    \cos\;2\theta ~&=~ \cos^2 \;\theta ~-~ \sin^2 \;\theta\label{eqn:doublecos}\\
    \tan\;2\theta ~&=~ \frac{2\;\tan\;\theta}{1 ~-~ \tan^2 \;\theta}\label{eqn:doubletan} \end{align}\]

    To derive the sine double-angle formula, we see that

    \[\nonumber
     \sin\;2\theta ~=~ \sin\;(\theta+\theta) ~=~ \sin\;\theta ~ \cos\;\theta ~+~ \cos\;\theta ~ \sin\;\theta ~=~ 2\;\sin\;\theta ~ \cos\;\theta~.\]

    Likewise, for the cosine double-angle formula, we have

    \[ \cos\;2\theta ~=~ \cos\;(\theta+\theta) ~=~ \cos\;\theta~\cos\;\theta ~-~ \sin\;\theta~\sin\;\theta ~=~ \cos^2 \;\theta ~-~ \sin^2 \;\theta~,\nonumber \]

    and for the tangent we get

     \[\nonumber \tan\;2\theta ~=~ \tan\;(\theta+\theta) ~=~ \frac{\tan\;\theta ~+~ \tan\;\theta}{1 ~-~ \tan\;\theta ~ \tan\;\theta} ~=~
      \frac{2\;\tan\;\theta}{1 ~-~ \tan^2 \;\theta}
    \]

    Using the identities \(\;\sin^2 \;\theta = 1 - \cos^2 \;\theta \) and \(\;\cos^2 \;\theta = 1 - \sin^2 \;\theta \), we get the following useful alternate forms for the cosine double-angle formula:

    \[\begin{align} \cos\;2\theta ~&=~ 2\;\cos^2 \;\theta ~-~ 1\label{eqn:doublecosalt1}\\
     &=~ 1 ~-~ 2\;\sin^2 \;\theta\label{eqn:doublecosalt2}
    \end{align}\]

    Example 3.13

    Prove that \(\;\sin\;3\theta ~=~ 3\;\sin\;\theta ~-~ 4\;\sin^3 \;\theta\; \).

    Solution:

    Using \(3\theta = 2\theta + \theta \), the addition Equation for sine, and the double-angle Equations \ref{eqn:doublesin} and \ref{eqn:doublecosalt2}, we get:

    \[\nonumber \begin{align*}
      \sin\;3\theta ~&=~ \sin\;(2\theta+\theta)\\ \nonumber
      &=~ \sin\;2\theta~\cos\;\theta ~+~ \cos\;2\theta~\sin\;\theta\\ \nonumber
      &=~ (2\;\sin\;\theta~\cos\;\theta)\;\cos\;\theta ~+~ (1 - 2\;\sin^2 \;\theta)\;\sin\;\theta\\ \nonumber
      &=~ 2\;\sin\;\theta~\cos^2 \;\theta ~+~ \sin\;\theta ~-~ 2\;\sin^3 \;\theta\\ \nonumber
      &=~ 2\;\sin\;\theta\;(1 - \sin^2 \;\theta) ~+~ \sin\;\theta ~-~ 2\;\sin^3 \;\theta\\ \nonumber
      &=~ 3\;\sin\;\theta ~-~ 4\;\sin^3 \;\theta
     \end{align*}\]

    Example 3.14

    Prove that \(\;\sin\;4z ~=~ \dfrac{4\;\tan\;z~(1 - \tan^2 \;z)}{(1 + \tan^2 \;z)^2}\; \).

    Solution:

    Expand the right side and use \(1 + \tan^2 \;z= \sec^2 \;z\,\):

    \[\nonumber \begin{align*}
      \dfrac{4\;\tan\;z~(1 - \tan^2 \;z)}{(1 + \tan^2 \;z)^2} ~&=~
       \dfrac{4 \;\cdot\; \dfrac{\sin\;z}{\cos\;z} \;\cdot\; \left( \dfrac{\cos^2 \;z}{\cos^2 \;z} -
       \dfrac{\sin^2 \;z}{\cos^2 \;z} \right)}{( \sec^2 \;z )^2}\\ \nonumber
      &=~ \dfrac{4 \;\cdot\; \dfrac{\sin\;z}{\cos\;z} \;\cdot\; \dfrac{\cos\;2z}{\cos^2 \;z}}{\left(
       \dfrac{1}{\cos^2 \;z} \right)^2}\quad\qquad\text{(by Equation \ref{eqn:doublecos})}\\ \nonumber
      &=~ (4\;\sin\;z~\cos\;2z)\;\cos\;z\\ \nonumber
      &=~ 2\;(2\;\sin\;z~\cos\;z)\;\cos\;2z\\ \nonumber
      &=~ 2\;\sin\;2z~\cos\;2z\quad\qquad\text{(by Equation \ref{eqn:doublesin})}\\ \nonumber
      &=~ \sin\;4z\quad\qquad\text{(by Equation \ref{eqn:doublesin} with \(\theta \) replaced by \(2z\))}
     \end{align*}\]

    Note: Perhaps surprisingly, this seemingly obscure identity has found a use in physics, in the derivation of a solution of the sine-Gordon equation in the theory of nonlinear waves

    Closely related to the double-angle formulas are the half-angle formulas:

    \[\begin{align} \sin^2 \;\tfrac{1}{2}\theta ~&=~ \frac{1 \;-\; \cos\;\theta}{2}\label{eqn:halfsin}\\
    \cos^2 \;\tfrac{1}{2}\theta ~&=~ \frac{1 \;+\; \cos\;\theta}{2}\label{eqn:halfcos}\\
     \tan^2 \;\tfrac{1}{2}\theta ~&=~ \frac{1 \;-\; \cos\;\theta}{1 \;+\; \cos\;\theta}\label{eqn:halftan}\end{align}\]

    These formulas are just the double-angle formulas rewritten with \(\theta \) replaced by \(\tfrac{1}{2}\theta\):

    \[\nonumber \begin{align*}
     \cos\;2\theta \;&=\; 1 \;-\; 2\;\sin^2 \;\theta ~\Rightarrow~ \sin^2 \;\theta \;=\; \frac{1 \;-\; \cos\;2\theta}{2}
      ~\Rightarrow~ \sin^2 \;\tfrac{1}{2}\theta \;=\; \frac{1 \;-\; \cos\;2\,(\tfrac{1}{2}\theta)}{2} \;=\;
      \frac{1 \;-\; \cos\;\theta}{2}\\ \nonumber
     \cos\;2\theta \;&=\; 2\;\cos^2  \;\theta\;-\; 1 ~\Rightarrow~ \cos^2 \;\theta \;=\; \frac{1 \;+\; \cos\;2\theta}{2}
      ~\Rightarrow~ \cos^2 \;\tfrac{1}{2}\theta \;=\; \frac{1 \;+\; \cos\;2\,(\tfrac{1}{2}\theta)}{2} \;=\;
      \frac{1 \;+\; \cos\;\theta}{2}
    \end{align*}\]
    The tangent half-angle Equation then follows easily:
    \[\nonumber
     \tan^2 \;\tfrac{1}{2}\theta \;=\; \left( \dfrac{\sin\;\tfrac{1}{2}\theta}{\cos\;\tfrac{1}{2}\theta} \right)^2
      \;=\; \dfrac{\sin^2 \;\tfrac{1}{2}\theta}{\cos^2 \;\tfrac{1}{2}\theta} \;=\;
      \dfrac{\tfrac{1 \;-\; \cos\;\theta}{2}}{\tfrac{1 \;+\; \cos\;\theta}{2}} \;=\;
      \frac{1 \;-\; \cos\;\theta}{1 \;+\; \cos\;\theta}
    \]

    The half-angle formulas are often used (e.g. in calculus) to replace a squared trigonometric function by a nonsquared function, especially when \(2\theta \) is used instead of \(\theta \).

    By taking square roots, we can write the above formulas in an alternate form:

    \[\begin{align}
     \sin\;\tfrac{1}{2}\theta ~&=~ \pm\;\sqrt{\frac{1 \;-\; \cos\;\theta}{2}}\label{eqn:halfsinsq}\\
     \cos\;\tfrac{1}{2}\theta ~&=~ \pm\;\sqrt{\frac{1 \;+\; \cos\;\theta}{2}}\label{eqn:halfcossq}\\
     \tan\;\tfrac{1}{2}\theta ~&=~ \pm\;\sqrt{\frac{1 \;-\; \cos\;\theta}{1 \;+\; \cos\;\theta}}\label{eqn:halftansq}
    \end{align}\]

    In the above form, the sign in front of the square root is determined by the quadrant in which the angle \(\tfrac{1}{2}\theta \) is located. For example, if \(\theta=300^\circ \) then \(\tfrac{1}{2}\theta = 150^\circ \) is in QII. So in this case \(\cos\;\tfrac{1}{2}\theta < 0 \) and hence we would have \(\cos\;\tfrac{1}{2}\theta = -\;\sqrt{\frac{1 \;+\; \cos\;\theta}{2}} \).

    In Equation \ref{eqn:halftansq}, multiplying the numerator and denominator inside the square root by \((1 - \cos\;\theta) \) gives

    \[\nonumber
     \tan\;\tfrac{1}{2}\theta ~=~ \pm\;\sqrt{\frac{1 - \cos\;\theta}{1 + \cos\;\theta} \,\cdot\,
      \frac{1 - \cos\;\theta}{1 - \cos\;\theta}} ~=~
       \pm\;\sqrt{\frac{(1 - \cos\;\theta)^2}{1 - \cos^2 \;\theta}} ~=~
       \pm\;\sqrt{\frac{(1 - \cos\;\theta)^2}{\sin^2 \;\theta}} ~=~ \pm\;\frac{1 - \cos\;\theta}{\sin\;\theta} ~.
    \]

    But \(1 - \cos\;\theta \ge 0 \), and it turns out (see Exercise 10) that \(\tan\;\tfrac{1}{2}\theta \) and \(\sin\;\theta \) always have the same sign. Thus, the minus sign in front of the last expression is not possible (since that would switch the signs of \(\tan\;\tfrac{1}{2}\theta \) and \(\sin\;\theta\)), so we have:

    \[\tan\;\tfrac{1}{2}\theta ~=~ \frac{1 \;-\; \cos\;\theta}{\sin\;\theta}\label{eqn:halftanalt1}\]

    Multiplying the numerator and denominator in Equation \ref{eqn:halftanalt1} by \(1 + \cos\;\theta \) gives

    \[\nonumber
     \tan\;\tfrac{1}{2}\theta ~=~ \frac{1 \;-\; \cos\;\theta}{\sin\;\theta} \;\cdot\;
      \frac{1 \;+\; \cos\;\theta}{1 \;+\; \cos\;\theta} ~=~ \frac{1 \;-\; \cos^2 \;\theta}{\sin\;\theta\;(1 \;+\; \cos\;\theta)}
      ~=~ \frac{\sin^2 \;\theta}{\sin\;\theta\;(1 \;+\; \cos\;\theta)} ~,
    \]

    so we also get:

    \[ \tan\;\tfrac{1}{2}\theta ~=~ \frac{\sin\;\theta}{1 \;+\; \cos\;\theta}\label{eqn:halftanalt2}\]

    Taking reciprocals in Equations \ref{eqn:halftanalt1} and \ref{eqn:halftanalt2} gives:

    \[ \cot\;\tfrac{1}{2}\theta ~=~ \frac{\sin\;\theta}{1 \;-\; \cos\;\theta} ~=~
      \frac{1 \;+\; \cos\;\theta}{\sin\;\theta}\label{eqn:halfcot}\]

    Example 3.15

    Prove the identity \(\;\sec^2 \;\tfrac{1}{2}\theta ~=~\dfrac{2\;\sec\;\theta}{\sec\;\theta \;+\; 1}\; \).

    Solution:

    Since secant is the reciprocal of cosine, taking the reciprocal of Equation \ref{eqn:halfcos} for \(\;\cos^2 \;\tfrac{1}{2}\theta \) gives us

    \[\sec^2 \;\tfrac{1}{2}\theta ~=~ \frac{2}{1 \;+\; \cos\;\theta}
      ~=~ \frac{2}{1 \;+\; \cos\;\theta} \;\cdot\; \frac{\sec\;\theta}{\sec\;\theta}
      ~=~ \frac{2\;\sec\;\theta}{\sec\;\theta \;+\; 1} ~. \nonumber \]

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