Multiple and sub-multiple angles

Angles such as $2A$ and $3A$ are called mul­ti­ple angles of $A$; angles such as $\displaystyle \frac{A}{2}$ and $\displaystyle \frac{A}{3}$ are sub-mul­ti­ple angles. Putting $B = A$ in the com­pound angle for­mu­lae gives for­mu­lae for $\sin 2A$, $\cos 2A$ and $\tan 2A$, and from these come the half-angle and triple-angle for­mu­lae. They also give exact val­ues for awk­ward angles such as $18°$, $36°$ and $\displaystyle 22\frac{1}{2}°$. This les­son sets out all of them.

1. Dou­ble angle for­mu­lae#

Put $B = A$ in $\sin(A + B)$, $\cos(A + B)$ and $\tan(A + B)$:

$$\displaystyle \sin 2A = 2\sin A\cos A = \frac{2\tan A}{1 + \tan^2 A}$$

$$\displaystyle \cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A = 2\cos^2 A - 1 = \frac{1 - \tan^2 A}{1 + \tan^2 A}$$

$$\displaystyle \tan 2A = \frac{2\tan A}{1 - \tan^2 A}$$

The other forms of $\cos 2A$ come from replac­ing $\cos^2 A$ by $1 - \sin^2 A$, or $\sin^2 A$ by $1 - \cos^2 A$. The forms with $\tan A$ come from divid­ing top and bot­tom by $\cos^2 A$ and using $\sec^2 A = 1 + \tan^2 A$.

The for­mula $\sin 2A = 2\sin A\cos A$ works for any angle and its dou­ble. Learn to see it in both direc­tions:

$$\sin 4A = 2\sin 2A\cos 2A, \quad \sin 8A = 2\sin 4A\cos 4A, \quad \sin 20° = 2\sin 10°\cos 10°$$

$$\displaystyle \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}, \quad 2\sin 2y\cos 2y = \sin 4y, \quad 2\sin 8x\cos 8x = \sin 16x$$

2. Squares in terms of $\cos 2A$#

From the sec­ond and third forms of $\cos 2A$:

$$\displaystyle 1 - \cos 2A = 2\sin^2 A, \qquad 1 + \cos 2A = 2\cos^2 A, \qquad \frac{1 - \cos 2A}{1 + \cos 2A} = \tan^2 A$$

So

$$\displaystyle \sin^2 A = \frac{1 - \cos 2A}{2}, \qquad \cos^2 A = \frac{1 + \cos 2A}{2}, \qquad \tan^2 A = \frac{1 - \cos 2A}{1 + \cos 2A}$$

and tak­ing square roots,

$$\displaystyle \sin A = \pm\sqrt{\frac{1 - \cos 2A}{2}}, \quad \cos A = \pm\sqrt{\frac{1 + \cos 2A}{2}}, \quad \tan A = \pm\sqrt{\frac{1 - \cos 2A}{1 + \cos 2A}}$$

The sign is cho­sen by the quad­rant of $A$.

3. Triple angle for­mu­lae#

$$\sin 3A = 3\sin A - 4\sin^3 A$$

$$\cos 3A = 4\cos^3 A - 3\cos A$$

$$\displaystyle \tan 3A = \frac{3\tan A - \tan^3 A}{1 - 3\tan^2 A}$$

For exam­ple, $\sin 3A = \sin(2A + A) = \sin 2A\cos A + \cos 2A\sin A = 2\sin A\cos^2 A + (1 - 2\sin^2 A)\sin A$, and replac­ing $\cos^2 A$ by $1 - \sin^2 A$ gives $3\sin A - 4\sin^3 A$.

4. Half angles#

Replac­ing $A$ by $\displaystyle \frac{A}{2}$ in the dou­ble angle for­mu­lae:

$$\displaystyle \sin A = 2\sin\frac{A}{2}\cos\frac{A}{2} = \frac{2\tan\frac{A}{2}}{1 + \tan^2\frac{A}{2}}$$

$$\displaystyle \cos A = \cos^2\frac{A}{2} - \sin^2\frac{A}{2} = 1 - 2\sin^2\frac{A}{2} = 2\cos^2\frac{A}{2} - 1 = \frac{1 - \tan^2\frac{A}{2}}{1 + \tan^2\frac{A}{2}}$$

$$\displaystyle \tan A = \frac{2\tan\frac{A}{2}}{1 - \tan^2\frac{A}{2}}$$

5. Squares in terms of $\cos A$#

$$\displaystyle 1 - \cos A = 2\sin^2\frac{A}{2}, \qquad 1 + \cos A = 2\cos^2\frac{A}{2}$$

so

$$\displaystyle \sin\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{2}}, \quad \cos\frac{A}{2} = \pm\sqrt{\frac{1 + \cos A}{2}}, \quad \tan\frac{A}{2} = \pm\sqrt{\frac{1 - \cos A}{1 + \cos A}}$$

6. Third angles#

Replac­ing $A$ by $\displaystyle \frac{A}{3}$ in the triple angle for­mu­lae:

$$\displaystyle \sin A = 3\sin\frac{A}{3} - 4\sin^3\frac{A}{3}, \qquad \cos A = 4\cos^3\frac{A}{3} - 3\cos\frac{A}{3}$$

$$\displaystyle \tan A = \frac{3\tan\frac{A}{3} - \tan^3\frac{A}{3}}{1 - 3\tan^2\frac{A}{3}}$$

7. The ratios of $18°$ and $36°$#

$$\displaystyle \sin 18° = \frac{\sqrt{5} - 1}{4} = \cos 72°, \qquad \cos 18° = \frac{\sqrt{10 + 2\sqrt{5}}}{4} = \sin 72°$$

$$\displaystyle \sin 36° = \frac{\sqrt{10 - 2\sqrt{5}}}{4} = \cos 54°, \qquad \cos 36° = \frac{\sqrt{5} + 1}{4} = \sin 54°$$

Where $\sin 18°$ comes from. Let $A = 18°$. Then $5A = 90°$, so $2A = 90° - 3A$ and $\sin 2A = \cos 3A$:

$$2\sin A\cos A = 4\cos^3 A - 3\cos A$$

Divide by $\cos A$ (which is not zero) and write $\cos^2 A = 1 - \sin^2 A$:

$$2\sin A = 4(1 - \sin^2 A) - 3 \implies 4\sin^2 A + 2\sin A - 1 = 0$$

$$\displaystyle \sin A = \frac{-2 \pm \sqrt{4 + 16}}{8} = \frac{-1 \pm \sqrt{5}}{4}$$

Since $\sin 18° > 0$, $\displaystyle \sin 18° = \frac{\sqrt{5} - 1}{4}$. Then $\displaystyle \cos 36° = 1 - 2\sin^2 18° = \frac{\sqrt{5} + 1}{4}$.

8. The ratios of $\displaystyle 22\frac{1}{2}°$#

$$\displaystyle \sin 22\tfrac{1}{2}° = \sqrt{\frac{\sqrt{2} - 1}{2\sqrt{2}}} = \cos 67\tfrac{1}{2}°, \qquad \cos 22\tfrac{1}{2}° = \sqrt{\frac{\sqrt{2} + 1}{2\sqrt{2}}} = \sin 67\tfrac{1}{2}°$$

$$\displaystyle \tan 22\tfrac{1}{2}° = \sqrt{2} - 1 = \cot 67\tfrac{1}{2}°, \qquad \cot 22\tfrac{1}{2}° = \sqrt{2} + 1 = \tan 67\tfrac{1}{2}°$$

Hint. Put $A = 45°$ in $\displaystyle \sin\frac{A}{2} = \sqrt{\frac{1 - \cos A}{2}}$: $\displaystyle \sin 22\frac{1}{2}° = \sqrt{\frac{1 - \frac{1}{\sqrt{2}}}{2}} = \sqrt{\frac{\sqrt{2} - 1}{2\sqrt{2}}}$.

Exam­ple 1#

If $\displaystyle \tan A = \frac{3}{4}$ and $A$ is acute, find $\sin 2A$, $\cos 2A$ and $\tan 2A$.

$$\displaystyle \sin 2A = \frac{2 \cdot \frac{3}{4}}{1 + \frac{9}{16}} = \frac{3/2}{25/16} = \frac{24}{25}, \qquad \cos 2A = \frac{1 - \frac{9}{16}}{1 + \frac{9}{16}} = \frac{7}{25}$$

$$\displaystyle \tan 2A = \frac{2 \cdot \frac{3}{4}}{1 - \frac{9}{16}} = \frac{3/2}{7/16} = \frac{24}{7}$$

Exam­ple 2#

If $\displaystyle \cos A = \frac{1}{3}$, find $\cos 3A$.

$$\displaystyle \cos 3A = 4\left(\frac{1}{3}\right)^3 - 3\left(\frac{1}{3}\right) = \frac{4}{27} - 1 = -\frac{23}{27}$$

Exam­ple 3#

Show that $\displaystyle \tan 22\frac{1}{2}° = \sqrt{2} - 1$.

Use $\displaystyle \tan\frac{A}{2} = \frac{2\sin^2\frac{A}{2}}{2\sin\frac{A}{2}\cos\frac{A}{2}} = \frac{1 - \cos A}{\sin A}$ with $A = 45°$:

$$\displaystyle \tan 22\tfrac{1}{2}° = \frac{1 - \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = \sqrt{2} - 1$$

Prac­tice#

1. If $\displaystyle \tan A = \frac{1}{2}$, find $\sin 2A$, $\cos 2A$ and $\tan 2A$.
2. If $\displaystyle \cos A = \frac{3}{5}$ and $A$ is acute, find $\sin 2A$.
3. If $\displaystyle \sin A = \frac{1}{3}$, find $\cos 2A$.
4. If $\displaystyle \cos 2A = \frac{1}{9}$ and $A$ is acute, find $\sin A$ and $\cos A$.
5. Find $2\sin 15°\cos 15°$.
6. Find $\cos^2 15° - \sin^2 15°$.
7. If $\displaystyle \sin A = \frac{3}{5}$, find $\sin 3A$.
8. If $\displaystyle \cos A = \frac{1}{2}$, find $\cos 3A$.
9. If $\displaystyle \tan A = \frac{1}{3}$, find $\tan 3A$.
10. Find $\cos 36° - \sin 18°$.
11. Find $\sin 18°\cos 36°$.
12. Using $\displaystyle \tan\frac{A}{2} = \frac{\sin A}{1 + \cos A}$, find $\tan 15°$.

Answers#