18.2: Double and half angles

We can write formulas for the trigonometric functions of twice an angle and half an angle.

Proposition: Half and Double Angle Formulas

Let \(\alpha\) be an angle. Then we have the half-angle formulas:

\[\begin{aligned}
\sin \dfrac{\alpha}{2} &=\pm \sqrt{\dfrac{1-\cos \alpha}{2}} \\
\cos \dfrac{\alpha}{2} &=\pm \sqrt{\dfrac{1+\cos \alpha}{2}} \\
\tan \dfrac{\alpha}{2} &=\dfrac{1-\cos \alpha}{\sin \alpha}=\dfrac{\sin \alpha}{1+\cos \alpha}=\pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}}
\end{aligned} \nonumber \]

Here, the signs “\(\pm\)” are determined by the quadrant in which the angle \(\frac \alpha 2\) lies. (For more on the signs, see also page .)

Furthermore, we have the double angle formulas:

\[\begin{aligned}
\sin (2 \alpha) &=2 \sin \alpha \cos \alpha \\
\cos (2 \alpha) &=\cos ^{2} \alpha-\sin ^{2} \alpha=1-2 \sin ^{2} \alpha=2 \cos ^{2} \alpha-1 \\
\tan (2 \alpha) &=\frac{2 \tan \alpha}{1-\tan ^{2} \alpha}
\end{aligned} \nonumber \]

Proof

We start with the double angle formulas, which we prove using Proposition [PROP:trig-add-subt-formulas].

\[\begin{aligned} \sin(2\alpha)&= \sin(\alpha+\alpha)=\sin\alpha\cos\alpha+\cos\alpha\sin\alpha=2\sin\alpha\cos\alpha \\ \cos(2\alpha)&= \cos(\alpha+\alpha)=\cos\alpha\cos\alpha-\sin\alpha\sin\alpha=\cos^2\alpha-\sin^2\alpha \\ \tan(2\alpha)&= \tan(\alpha+\alpha) = \dfrac{\tan\alpha+\tan\alpha}{1-\tan\alpha\tan\alpha}=\dfrac{2\tan\alpha}{1-\tan^2\alpha}\end{aligned} \nonumber\]

Notice that \(\cos(2\alpha)=\cos^2\alpha-\sin^2\alpha\) can be rewritten using \(\sin^2\alpha+\cos^2\alpha=1\) as follows:

\[\begin{aligned} \cos^2\alpha-\sin^2\alpha &=& (1-\sin^2\alpha)-\sin^2\alpha=1-2\sin^2\alpha \\ \text{and } \quad\quad \cos^2\alpha-\sin^2\alpha &=& \cos^2\alpha-(1-\cos^2\alpha)=2 \cos^2\alpha-1\end{aligned}\]

This shows the double angle formulas. These formulas can now be used to prove the half-angle formulas.

\[\begin{aligned}
\cos (2 \alpha)&=1-2 \sin ^{2} \alpha \\
2 \sin ^{2} \alpha&=1-\cos (2 \alpha) \\
\sin ^{2} \alpha&=\dfrac{1-\cos (2 \alpha)}{2} \\
\sin \alpha&=\pm \sqrt{\dfrac{1-\cos (2 \alpha)}{2}} \\
\sin \dfrac{\alpha}{2}&=\pm \sqrt{\dfrac{1-\cos \alpha}{2}}\quad {\text { replace } \alpha \text { by } \dfrac{\alpha}{2}}
\end{aligned} \nonumber \]

\[\begin{aligned}
\cos (2 \alpha)&=2 \cos ^{2} \alpha -1 \\
2 \cos ^{2} \alpha&=1+\cos (2 \alpha) \\
\cos ^{2} \alpha&=\dfrac{1+\cos (2 \alpha)}{2} \\
\cos \alpha&=\pm \sqrt{\dfrac{1+\cos (2 \alpha)}{2}} \\
\cos \dfrac{\alpha}{2}&=\pm \sqrt{\dfrac{1+\cos \alpha}{2}}\quad {\text { replace } \alpha \text { by } \dfrac{\alpha}{2}}
\end{aligned} \nonumber \]

In particular,

\[\tan \dfrac{\alpha}{2}=\dfrac{\sin \left(\frac{\alpha}{2}\right)}{\cos \left(\frac{\alpha}{2}\right)}=\dfrac{\pm \sqrt{\frac{1-\cos \alpha}{2}}}{\pm \sqrt{\frac{1+\cos \alpha}{2}}}=\pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}} \nonumber \]

For the first two formulas for \(\tan \dfrac \alpha 2\) we simplify \(\sin(2\alpha)\cdot \tan(\alpha)\) and \((1+\cos(2\alpha))\cdot \tan(\alpha)\) as follows.

\[\begin{aligned}
\sin (2 \alpha) \cdot \tan (\alpha)&= 2 \sin \alpha \cos \alpha \cdot \dfrac{\sin \alpha}{\cos \alpha}\\
&=2 \sin ^{2} \alpha\\
&=1-\cos (2 \alpha) \\
\tan (\alpha)&=\dfrac{1-\cos (2 \alpha)}{\sin (2 \alpha)} \\
\tan \left(\frac{\alpha}{2}\right)&=\dfrac{1-\cos (\alpha)}{\sin (\alpha)} \quad \text { replace } \alpha \text { by } \frac{\alpha}{2}\\
(1+\cos (2 \alpha)) \cdot \tan (\alpha) &= 2 \cos ^{2} \alpha \cdot \dfrac{\sin \alpha}{\cos \alpha}\\
&=2 \sin \alpha \cos \alpha\\
&=\sin (2 \alpha) \\
\tan (\alpha)&=\dfrac{\sin (2 \alpha)}{1+\cos (2 \alpha)}\\
\tan \left(\dfrac{\alpha}{2}\right)&=\dfrac{\sin (\alpha)}{1+\cos (\alpha)} \quad \text { replace } \alpha \text { by } \dfrac{\alpha}{2}
\end{aligned} \nonumber \]

This completes the proof of the proposition.

Here is an example involving the half-angle identities.

Example \(\PageIndex{1}\)

Find the trigonometric functions using the half-angle formulas.

1. \(\sin\left(\dfrac{\pi}{8}\right)\)
2. \(\cos\left(\dfrac{9\pi}{8}\right)\)
3. \(\tan\left(\dfrac{\pi}{24}\right)\)

Solution

1. Since \(d\frac{\pi}{8}=\dfrac{\frac{\pi}{4}}{2}\), we use the half-angle formula with \(\alpha=\dfrac{\pi}{4}\).

\[\begin{aligned}
\sin\left(\dfrac{\pi}{8}\right) &= \sin\left(\dfrac{\frac{\pi}{4}}{2}\right)\\
&=\pm\sqrt{\dfrac{1-\cos\frac{\pi}{4}}{2}}\\
&=\pm\sqrt{\dfrac{1-\frac{\sqrt{2}}{2}}{2}}\\
&=\pm\sqrt{\dfrac{\frac{2-\sqrt{2}}{2}}{2}}\\
&=\pm\sqrt{\dfrac{2-\sqrt{2}}{4}}\\
&=\pm\dfrac{\sqrt{2-\sqrt{2}}}{2}
\end{aligned} \nonumber \]

Since \(\dfrac{\pi}{8}=\dfrac{180^\circ}{8}=22.5^\circ\) is in the first quadrant, the sine is positive, so that \(\sin(\dfrac{\pi}{8})=\dfrac{\sqrt{2-\sqrt{2}}}{2}\).

2. Note that \(\dfrac{9\pi}{8}=\dfrac{\frac{9\pi}{4}}{2}\). So we use \(\alpha=\dfrac{9\pi}{4}\). Now, \(\dfrac{9\pi}{8}=\dfrac{9\cdot 180^\circ}{8}=202.5^\circ\) is in the third quadrant, so that the cosine is negative. We have:

\[\cos\left(\dfrac{9\pi}{8}\right) = \cos\left(\dfrac{\frac{9\pi}{4}}{2}\right)=-\sqrt{\dfrac{1+\cos\frac{9\pi}{4}}{2}} \nonumber \]

Now, \(\cos(\dfrac{9\pi}{4})=\cos(\dfrac{8\pi+\pi}{4})=\cos(2\pi+\dfrac{\pi}{4})=\cos(\dfrac{\pi}{4})=\dfrac{\sqrt{2}}{2}\), so that

\[\cos\left(\dfrac{9\pi}{8}\right) =-\sqrt{\dfrac{1+\frac{\sqrt{2}}{2}}{2}}= -\sqrt{\dfrac{\frac{2+\sqrt{2}}{2}}{2}}=-\sqrt{\dfrac{2+\sqrt{2}}{4}}=-\dfrac{\sqrt{2+\sqrt{2}}}{2} \nonumber \]

3. Note that \(\dfrac{\pi}{24}=\dfrac{\frac{\pi}{12}}{2}\), and we already calculated the trigonometric function values of \(\alpha=\dfrac{\pi}{12}\) in Example 18.1.1 (c). So that we obtain:

\[\begin{aligned}
\tan\left(\dfrac{\pi}{24}\right) &= \tan\left(\dfrac{\frac{\pi}{12}}{2}\right)\\
&=\dfrac{1-\cos\frac{\pi}{12}}{\sin\frac{\pi}{12}}\\
&=\dfrac{1-\frac{\sqrt{2}+\sqrt{6}}{4}}{\frac{\sqrt{6}-\sqrt{2}}{4}}\\
&=\dfrac{\frac{4-\sqrt{2}-\sqrt{6}}{4}}{\frac{\sqrt{6}-\sqrt{2}}{4}} \\
&=\dfrac{4-\sqrt{2}-\sqrt{6}}{4}\cdot \dfrac{4}{\sqrt{6}-\sqrt{2}}\\
&=\dfrac{4-\sqrt{2}-\sqrt{6}}{\sqrt{6}-\sqrt{2}}
\end{aligned} \nonumber \]

We can rationalize the denominator by multiplying numerator and denominator by \((\sqrt{6}+\sqrt{2})\).

\[\begin{aligned}
\tan\left(\dfrac{\pi}{24}\right) &= \dfrac{4-\sqrt{2}-\sqrt{6}}{\sqrt{6}-\sqrt{2}} \cdot \dfrac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}} \\
&= \dfrac{4\sqrt{6}+4\sqrt{2}-\sqrt{12}-\sqrt{4}-\sqrt{36}-\sqrt{12}}{6-2} \\
&= \dfrac{4\sqrt{6}+4\sqrt{2}-2\sqrt{12}-2-6}{4} \\
&= \dfrac{4\sqrt{6}+4\sqrt{2}-4\sqrt{3}-8}{4}\\
&= \sqrt{6}+\sqrt{2}-\sqrt{3}-2
\end{aligned} \nonumber \]

Although we used the first formula for \(\tan\dfrac{\alpha}{2}\) from the proposition, we could as well have used the other two formulas.

Example \(\PageIndex{2}\)

Find the trigonometric functions of \(2\alpha\) when \(\alpha\) has the properties below.

1. \(\sin(\alpha)=\dfrac{3}{5}\), and \(\alpha\) is in quadrant II
2. \(\tan(\alpha)=\dfrac{12}{5}\), and \(\alpha\) is in quadrant III

Solution

1. From \(\sin^2(\alpha)+\cos^2(\alpha)=1\), we find that \(\cos^2(\alpha)=1-\sin^2(\alpha)\), and since \(\alpha\) is in the second quadrant, \(\cos(\alpha)\) is negative, so that

\[\begin{aligned} \cos(\alpha)&=-\sqrt{1-\sin^2(\alpha)}\\&=-\sqrt{1-\Big(\dfrac{3}{5}\Big)^2}\\&=-\sqrt{1-\dfrac{9}{25}} \\ &= -\sqrt{\dfrac{25-9}{25}}\\&=-\sqrt{\dfrac{16}{25}}\\&=-\dfrac{4}{5},\end{aligned} \nonumber \]

and

\[\tan(\alpha)=\dfrac{\sin\alpha}{\cos\alpha}=\dfrac{\frac{3}{5}}{\frac{-4}{5}}=\dfrac{3}{5}\cdot\dfrac{5}{-4}=-\dfrac{3}{4} \nonumber \]

From this we can calculate the solution by plugging these values into the double angle formulas.

\[\begin{aligned} \sin(2\alpha)&= 2\sin\alpha\cos\alpha =2\cdot \dfrac{3}{5}\cdot \dfrac{(-4)}{5}=\dfrac{-24}{25}\\ \cos(2\alpha)&= \cos^2(\alpha)-\sin^2(\alpha) =\Big(\dfrac{-4}{5}\Big)^2-\Big(\dfrac{3}{5}\Big)^2=\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25}\\ \tan(2\alpha)&=\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2\cdot\left(\frac{-3}{4}\right)}{1-\left(\frac{-3}{4}\right)^2}=\dfrac{\frac{-3}{2}}{1-\frac{9}{16}}=\dfrac{\frac{-3}{2}}{\frac{16-9}{16}}=\dfrac{-3}{2}\cdot \dfrac{16}{7}=\dfrac{-24}{7}\end{aligned} \nonumber \]

2. Similar to the calculation in part (a), we first calculate \(\sin(\alpha)\) and \(\cos(\alpha)\), which are both negative in the third quadrant. Recall from equation [EQU:sin2+cos2=1] that \(\sec^2\alpha=1+\tan^2\alpha\), where \(\sec\alpha=\dfrac{1}{\cos\alpha}\). Therefore,

\[\sec^2\alpha=1+\Big(\dfrac{12}{5}\Big)^2=1+\dfrac{144}{25}=\dfrac{25+144}{25}=\dfrac{169}{25} \implies \sec\alpha=\pm\dfrac{13}{5} \nonumber \]

Since \(\cos(\alpha)\) is negative (in quadrant III), so is \(\sec(\alpha)\), so that we get,

\[\cos\alpha=\dfrac{1}{\sec\alpha}=\dfrac{1}{-\frac{13}{5}}=-\dfrac{5}{13} \nonumber \]

Furthermore, \(\sin^2\alpha=1-\cos^2\alpha\), and \(\sin\alpha\) is negative (in quadrant III), we have

\[\begin{aligned} \sin\alpha &=-\sqrt{1-\cos^2\alpha}\\&=-\sqrt{1-\Big(-\dfrac{5}{13}\Big)^2} \\&= -\sqrt{1-\dfrac{25}{169}} \\ &= -\sqrt{\dfrac{169-25}{169}}\\&=-\sqrt{\dfrac{144}{169}}\\&=-\dfrac{12}{13}\end{aligned} \nonumber \]

Thus, we obtain the solution as follows:

\[\begin{aligned} \sin(2\alpha)&= 2\sin\alpha\cos\alpha =2\cdot \dfrac{(-12)}{13}\cdot \dfrac{(-5)}{13} =\dfrac{120}{169}\\ \cos(2\alpha)&= \cos^2(\alpha)-\sin^2(\alpha) =\Big(\dfrac{-5}{13}\Big)^2-\Big(\dfrac{-12}{13}\Big)^2=\dfrac{25}{169}-\dfrac{144}{169}=\dfrac{-119}{169}\\ \tan(2\alpha)&=\dfrac{2\tan\alpha}{1-\tan^2\alpha}=\dfrac{2\cdot\frac{12}{5}}{1-\left(\frac{12}{5}\right)^2}=\dfrac{\frac{24}{5}}{1-\frac{144}{25}}=\dfrac{\frac{24}{5}}{\frac{25-144}{25}}\\ &= \dfrac{24}{5}\cdot \dfrac{25}{-119}=\dfrac{120}{-119}\end{aligned} \nonumber \]