Multiple-Angle and Half-Angle Formulas

TL;DR

Double-Angle Formulas

• \[\sin 2\alpha = 2\sin \alpha \cos \alpha\]
• \[\begin{align*} \cos 2\alpha &= \cos^{2}\alpha - \sin^{2}\alpha \\ &= 2\cos^{2}\alpha - 1 \\ &= 1 - 2\sin^{2}\alpha \end{align*}\]
• \[\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^{2}\alpha}\]

Triple-Angle Formulas

• \[\sin 3\alpha = 3\sin \alpha - 4\sin^{3}\alpha\]
• \[\cos 3\alpha = 4\cos^{3}\alpha - 3\cos \alpha\]

Half-Angle Formulas

• \[\sin^{2}\frac{\alpha}{2} = \frac{1 - \cos \alpha}{2}\]
• \[\cos^{2}\frac{\alpha}{2} = \frac{1 + \cos \alpha}{2}\]
• \[\tan^{2}\frac{\alpha}{2} = \frac{1 - \cos \alpha}{1 + \cos\alpha}\]
• \[\tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha}\]

Prerequisites

• [Trigonometric Addition Formulas]

Multiple-Angle Formulas

Double-Angle Formulas

• \[\sin 2\alpha = 2\sin \alpha \cos \alpha\]
• \[\begin{align*} \cos 2\alpha &= \cos^{2}\alpha - \sin^{2}\alpha \\ &= 2\cos^{2}\alpha - 1 \\ &= 1 - 2\sin^{2}\alpha \end{align*}\]
• \[\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^{2}\alpha}\]

Derivation

We can derive the double-angle formulas from the Trigonometric Addition Formulas.

\[\begin{gather} \sin ( \alpha + \beta ) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \label{eqn:sin_add} \\ \cos ( \alpha + \beta ) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \label{eqn:cos_add} \\ \tan ( \alpha + \beta ) = \frac { \tan \alpha + \tan \beta } { 1 - \tan \alpha \tan \beta } \label{eqn:tan_add} \end{gather}\]

If we substitute $\alpha$ for $\beta$:

From equation ($\ref{eqn:sin_add}$)

\[\sin 2\alpha = 2\sin \alpha \cos \alpha\]

From equation ($\ref{eqn:cos_add}$)

\[\begin{align*} \cos 2 \alpha &= \cos ^ { 2 } \alpha - \sin ^ { 2 } \alpha \\ &= 2 \cos ^ { 2 } \alpha - 1 \\ &= 1 - 2 \sin ^ { 2 } \alpha \end{align*}\]

From equation ($\ref{eqn:tan_add}$)

\[\tan 2\alpha = \frac{2\tan \alpha}{1 - \tan^{2} \alpha}\]

Triple-Angle Formulas

• \[\sin 3\alpha = 3\sin \alpha - 4\sin^{3}\alpha\]
• \[\cos 3\alpha = 4\cos^{3}\alpha - 3\cos \alpha\]

Derivation

Using $\sin 2\alpha = 2\sin\alpha \cos\alpha$ and $\cos 2 \alpha = 1 - 2\sin^{2}\alpha$, we get:

\[\begin{align*} \sin 3 \alpha &= \sin ( \alpha + 2 \alpha ) = \sin \alpha \cos 2 \alpha + \cos \alpha \sin 2 \alpha \\ &= \sin \alpha ( 1 - 2 \sin ^ { 2 } \alpha ) + \cos \alpha ( 2 \sin \alpha \cos \alpha ) \\ &= \sin a ( 1 - 2 \sin ^ { 2 } \alpha ) + 2 \sin \alpha ( 1 - \sin ^ { 2 } \alpha ) \\ &= 3 \sin \alpha - 4 \sin ^ { 3 } \alpha . \end{align*}\]

Similarly, using $\sin 2\alpha = 2\sin\alpha \cos\alpha$ and $\cos 2 \alpha = 2\cos^{2}\alpha - 1$, we get:

\[\begin{align*} \cos 3 \alpha &= \cos ( \alpha + 2 \alpha ) = \cos \alpha \cos 2 \alpha - \sin \alpha \sin 2 \alpha \\ &= \cos \alpha ( 2 \cos ^ { 2 } \alpha - 1 ) - \sin \alpha ( 2 \sin \alpha \cos \alpha ) \\ &= \cos \alpha ( 2 \cos ^ { 2 } \alpha - 1 ) - 2 \cos \alpha ( 1 - \cos ^ { 2 } \alpha ) \\ &= 4 \cos ^ { 3 } \alpha - 3 \cos \alpha \end{align*}\]

Half-Angle Formulas

• \[\sin^{2}\frac{\alpha}{2} = \frac{1 - \cos \alpha}{2}\]
• \[\cos^{2}\frac{\alpha}{2} = \frac{1 + \cos \alpha}{2}\]
• \[\tan^{2}\frac{\alpha}{2} = \frac{1 - \cos \alpha}{1 + \cos\alpha}\]
• \[\tan \frac{\alpha}{2} = \frac{\sin \alpha}{1 + \cos \alpha}\]

Derivation

From the double-angle formula $\cos 2\alpha = 2\cos^{2}\alpha - 1 = 1 - 2\sin^{2}\alpha$, if we substitute $\frac{\alpha}{2}$ for $\alpha$, we get:

\[\cos \alpha = 1 - 2\sin^{2}\frac{\alpha}{2} = 2 \cos^{2}\frac{\alpha}{2} - 1 .\]

From $ \cos \alpha = 1 - 2\sin^{2}\frac{\alpha}{2} $, we get:

\[\sin^{2}\frac{\alpha}{2}=\frac{1-\cos \alpha}{2} .\]

From $ \cos \alpha = 2 \cos^{2}\frac{\alpha}{2} - 1 $, we get:

\[\cos^{2}\frac{\alpha}{2}=\frac{1+\cos \alpha}{2} .\]

From these, we can show that:

\[\tan ^ { 2 } \frac { \alpha } { 2 } = \left . \left( \sin ^ { 2 } \frac{\alpha}{2}\right) \middle/ \left( \cos ^ { 2 } \frac { \alpha } { 2 } \right) \right . = \frac { 1 - \cos \alpha } { 1 + \cos \alpha }\]

And also:

\[\tan \frac { \alpha } { 2 } = \frac { \sin \frac { \alpha } { 2 } } { \cos \frac { \alpha } { 2 } } = \frac { 2 \sin \frac { \alpha } { 2 } \cos \frac { \alpha } { 2 } } { 2 \cos ^ { 2 } \frac { \alpha } { 2 } } = \frac { \sin \alpha } { 1 + \cos \alpha }\]