# 18.2: Double and half angles

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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

\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 $$\sin(\dfrac{\pi}{8})=\dfrac{\sqrt{2-\sqrt{2}}}{2}$$.

1. 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

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

1. 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

1. 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

This page titled 18.2: Double and half angles is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Thomas Tradler and Holly Carley (New York City College of Technology at CUNY Academic Works) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.