
# 3.3: Double-Angle and Half-Angle Formulas

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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

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