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Mathematics LibreTexts

3.3: Double-Angle and Half-Angle Formulas

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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