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\tag{3.23}\\ \cos\;2\theta ~&=~ \cos^2 \;\theta ~-~ \sin^2 \;\theta\tag{3.24}\\ \tan\;2\theta ~&=~ \frac{2\;\tan\;\theta}{1 ~-~ \tan^2 \;\theta} \tag{3.25}\end{align}

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

\begin{align*} \sin\;2\theta ~&=~ \sin\;(\theta+\theta) ~=~ \sin\;\theta ~ \cos\;\theta ~+~ \cos\;\theta ~ \sin\;\theta ~=~ 2\;\sin\;\theta ~ \cos\;\theta~.\\ \intertext{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~,\\ \intertext{and for the tangent we get} \tan\;2\theta ~&=~ \tan\;(\theta+\theta) ~=~ \frac{\tan\;\theta ~+~ \tan\;\theta}{1 ~-~ \tan\;\theta ~ \tan\;\theta} ~=~ \frac{2\;\tan\;\theta}{1 ~-~ \tan^2 \;\theta} \end{align*}

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\tag{3.26}\\ &=~ 1 ~-~ 2\;\sin^2 \;\theta \tag{3.27}\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 formula for sine, and the double-angle formulas (3.23) and (3.27), we get:

\begin{align*} \sin\;3\theta ~&=~ \sin\;(2\theta+\theta)\\ &=~ \sin\;2\theta~\cos\;\theta ~+~ \cos\;2\theta~\sin\;\theta\\ &=~ (2\;\sin\;\theta~\cos\;\theta)\;\cos\;\theta ~+~ (1 - 2\;\sin^2 \;\theta)\;\sin\;\theta\\ &=~ 2\;\sin\;\theta~\cos^2 \;\theta ~+~ \sin\;\theta ~-~ 2\;\sin^3 \;\theta\\ &=~ 2\;\sin\;\theta\;(1 - \sin^2 \;\theta) ~+~ \sin\;\theta ~-~ 2\;\sin^3 \;\theta\\ &=~ 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 \(1 + \tan^2 \;z= \sec^2 \;z\,\):

\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}\\[6pt] &=~ \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 formula (\ref{eqn:doublecos}))}\\[5pt] &=~ (4\;\sin\;z~\cos\;2z)\;\cos\;z\\ &=~ 2\;(2\;\sin\;z~\cos\;z)\;\cos\;2z\\ &=~ 2\;\sin\;2z~\cos\;2z\quad\qquad\text{(by formula (\ref{eqn:doublesin}))}\\ &=~ \sin\;4z\quad\qquad\text{(by formula (\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.[1]

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}\tag{3.28}\\ \cos^2 \;\tfrac{1}{2}\theta ~&=~ \frac{1 \;+\; \cos\;\theta}{2}\tag{3.29}\\ \tan^2 \;\tfrac{1}{2}\theta ~&=~ \frac{1 \;-\; \cos\;\theta}{1 \;+\; \cos\;\theta} \tag{3.30}\end{align}

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

\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}\\ \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 formula then follows easily:

\begin{displaymath} \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} \end{displaymath}

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}}\tag{3.31}\\ \cos\;\tfrac{1}{2}\theta ~&=~ \pm\;\sqrt{\frac{1 \;+\; \cos\;\theta}{2}}\tag{3.32}\\ \tan\;\tfrac{1}{2}\theta ~&=~ \pm\;\sqrt{\frac{1 \;-\; \cos\;\theta}{1 \;+\; \cos\;\theta}} \tag{3.33}\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 \(\theta=300\Degrees\) then \(\tfrac{1}{2}\theta = 150\Degrees\) 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 formula (3.33), multiplying the numerator and denominator inside the square root by \((1 - \cos\;\theta)\) gives

\begin{displaymath} \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} ~. \end{displaymath}

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:

\begin{equation} \tan\;\tfrac{1}{2}\theta ~=~ \frac{1 \;-\; \cos\;\theta}{\sin\;\theta} \tag{3.34}\end{equation}

Multiplying the numerator and denominator in formula (3.34) by \(1 + \cos\;\theta\) gives

\begin{displaymath} \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)} ~, \end{displaymath}

so we also get:

\begin{equation} \tan\;\tfrac{1}{2}\theta ~=~ \frac{\sin\;\theta}{1 \;+\; \cos\;\theta} \tag{3.35}\end{equation}

Taking reciprocals in formulas (3.34) and (3.35) gives:

\begin{equation} \cot\;\tfrac{1}{2}\theta ~=~ \frac{\sin\;\theta}{1 \;-\; \cos\;\theta} ~=~ \frac{1 \;+\; \cos\;\theta}{\sin\;\theta} \tag{3.36}\end{equation}

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 formula (3.29) for \(\;\cos^2 \;\tfrac{1}{2}\theta\) gives us

\begin{displaymath} \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} ~. \end{displaymath}

Exercises

For Exercises 1-8, prove the given identity.

2

  1. \(\cos\;3\theta ~=~ 4\;\cos^3 \;\theta ~-~ 3\;\cos\;\theta\)

  2. \(\tan\;\tfrac{1}{2}\theta ~=~ \csc\;\theta ~-~ \cot\;\theta\)

    2

  3. \(\dfrac{\sin\;2\theta}{\sin\;\theta} ~-~ \dfrac{\cos\;2\theta}{\cos\;\theta} ~=~ \sec\;\theta\)

  4. \(\dfrac{\sin\;3\theta}{\sin\;\theta} ~-~ \dfrac{\cos\;3\theta}{\cos\;\theta} ~=~ 2\)

    2

  5. \(\tan\;2\theta ~=~ \dfrac{2}{\cot\;\theta \;-\; \tan\;\theta}\)

  6. \(\tan\;3\theta ~=~ \dfrac{3\;\tan\;\theta \;-\; \tan^3 \;\theta}{1 \;-\; 3\;\tan^2 \;\theta}\)

    2

  7. \(\tan^2 \;\tfrac{1}{2}\theta ~=~ \dfrac{\tan\;\theta \;-\; \sin\;\theta}{\tan\;\theta \;+\; \sin\;\theta}\)

  8. \(\dfrac{\cos^2 \;\psi}{\cos^2 \;\theta} ~=~ \dfrac{1 \;+\; \cos\;2\psi}{1 \;+\; \cos\;2\theta}\)

  9. Some trigonometry textbooks used to claim incorrectly that \(\;\sin\;\theta ~+~ \cos\;\theta ~=~ \sqrt{1 \;+\; \sin\;2\theta}\) was an identity. Give an example of a specific angle \(\theta\) that would make that equation false. Is \(\;\sin\;\theta ~+~ \cos\;\theta ~=~ \pm\;\sqrt{1 \;+\; \sin\;2\theta}\) an identity? Justify your answer.

  10. Fill out the rest of the table below for the angles \(0\Degrees < \theta < 720\Degrees\) in increments of \(90\Degrees\), showing \(\theta\), \(\tfrac{1}{2}\theta\), and the signs (\(+\) or \(-\)) of \(\sin\;\theta\) and \(\tan\;\tfrac{1}{2}\theta\).

\(\theta\)\(\tfrac{1}{2}\theta\)\(\sin\;\theta\)\(\tan\;\tfrac{1}{2}\theta\)
\(0\Degrees - 90\Degrees\)\(0\Degrees - 45\Degrees\)\(+\)\(+\)
\(90\Degrees - 180\Degrees\)\(45\Degrees - 90\Degrees\)
\(180\Degrees - 270\Degrees\)\(90\Degrees - 135\Degrees\)
\(270\Degrees - 360\Degrees\)\(135\Degrees - 180\Degrees\)
\(\theta\)\(\tfrac{1}{2}\theta\)\(\sin\;\theta\)\(\tan\;\tfrac{1}{2}\theta\)
\(360\Degrees - 450\Degrees\)\(180\Degrees - 225\Degrees\)
\(450\Degrees - 540\Degrees\)\(225\Degrees - 270\Degrees\)
\(540\Degrees - 630\Degrees\)\(270\Degrees - 315\Degrees\)
\(630\Degrees - 720\Degrees\)\(315\Degrees - 360\Degrees\)

  1. In general, what is the largest value that \(\;\sin\;\theta~\cos\;\theta\;\) can take? Justify your answer.

For Exercises 12-17, prove the given identity for any right triangle \(\triangle\,ABC\) with \(C=90\Degrees\).

2

  1. \(\sin\;(A-B) ~=~ \cos\;2B\)

  2. \(\cos\;(A-B) ~=~ \sin\;2A\)

    2

  3. \(\sin\;2A ~=~ \dfrac{2\;ab}{c^2}\)

  4. \(\cos\;2A ~=~ \dfrac{b^2 - a^2}{c^2}\)

    2

  5. \(\tan\;2A ~=~ \dfrac{2\;ab}{b^2 - a^2}\)

  6. \(\tan\;\tfrac{1}{2}A ~=~ \dfrac{c - b}{a} ~=~ \dfrac{a}{c + b}\)

  7. Continuing Exercise 20 from Section 3.1, it can be shown that

    \begin{align*} r\;(1 \;-\; \cos\;\theta) ~&=~ a\;(1 \;+\; \epsilon)\,(1 \;-\; \cos\;\psi) ~,~\text{and}\\ r\;(1 \;+\; \cos\;\theta) ~&=~ a\;(1 \;-\; \epsilon)\,(1 \;+\; \cos\;\psi) ~, \end{align*}

    where \(\theta\) and \(\psi\) are always in the same quadrant. Show that \(\;\tan\;\tfrac{1}{2}\theta ~=~ \sqrt{\frac{1 \;+\; \epsilon}{1 \;-\; \epsilon}}~ \tan\;\tfrac{1}{2}\psi\;\).

  1. See p.331 in L.A. Ostrovsky and A.I.Potapov, Modulated Waves: Theory and Applications, Baltimore: The Johns Hopkins University Press, 1999. ↩