# Double-Angle and Multiple-Angle Identities

## Double-Angle Identities

The double-angle identities are a special case of the addition formulas.

To derive the double-angle formula for sine, we use the sine addition formula:

$\sin \left( {\alpha + \beta } \right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta .$

Putting $$\beta = \alpha$$ gives

$\sin \left( {\alpha + \alpha } \right) = \sin 2\alpha = \sin \alpha \cos \alpha + \cos \alpha \sin \alpha = 2\sin \alpha \cos \alpha ,$

that is,

$\sin 2\alpha = 2\sin\alpha\cos\alpha$

Similarly, substituting $$\beta = \alpha$$ in the cosine addition formula

$\cos \left( {\alpha + \beta } \right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

yields the double-angle identity for cosine:

$\cos \left( {\alpha + \alpha } \right) = \cos 2\alpha = \cos \alpha \cos \alpha - \sin \alpha \sin \alpha = {\cos ^2}\alpha - {\sin ^2}\alpha ,$

or

$\cos 2\alpha = \cos^2\alpha - \sin^2\alpha$

Using the Pythagorean trigonometric identity, we can get two more variations for the cosine of a double angle:

Since $${\cos ^2}\alpha = 1 - {\sin ^2}\alpha ,$$ we have

$\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha = 1 - {\sin ^2}\alpha - {\sin ^2}\alpha = 1 - 2\,{\sin ^2}\alpha .$

Hence, the first variation is given by

$\cos 2\alpha = 1 - 2\sin^2\alpha$

If we substitute the identity $${\sin ^2}\alpha = 1 - {\cos ^2}\alpha$$ in the original double-angle formula for cosine, we get the second variation:

$\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha = {\cos ^2}\alpha - \left( {1 - {{\cos }^2}\alpha } \right) = 2\,{\cos ^2}\alpha - 1.$

Thus,

$\cos 2\alpha = 2\cos^2\alpha - 1$

To derive the double-angle formula for tangent, recall the tangent addition formula

$\tan \left( {\alpha + \beta } \right) = \frac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}.$

Let $$\beta = \alpha.$$ Then

$\tan \left( {\alpha + \alpha } \right) = \tan 2\alpha = \frac{{\tan \alpha + \tan \alpha }}{{1 - \tan \alpha \tan \alpha }} = \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }},$

that is,

$\tan 2\alpha = \frac{2\tan\alpha}{1 - \tan^2\alpha}$

The double-angle identity for cotangent has the form

$\cot \left( {\alpha + \beta } \right) = \frac{{\cot \alpha \cot \beta - 1}}{{\cot \alpha + \cot \beta }},\;\; \Rightarrow \cot 2\alpha = \frac{{\cot \alpha \cot \alpha - 1}}{{\cot \alpha + \cot \alpha }} = \frac{{{{\cot }^2}\alpha - 1}}{{2\cot \alpha }}.$

Hence,

$\cot 2\alpha = \frac{\cot^2\alpha - 1}{2\cot\alpha}$

## Double-Angle Identities in Terms of Tangent

The double angle identities can be expressed in terms of the tangent of the single angle. To prove this, note that

$\sin 2\alpha = 2\sin \alpha \cos \alpha = \frac{{2\sin \alpha \cos \alpha }}{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}.$

Assuming $$\cos \alpha \ne 0,$$ we can divide both the numerator and denominator by $$\cos^2\alpha.$$ This yields

$\sin 2\alpha = \frac{{\frac{{2\sin \alpha \cos \alpha }}{{{{\cos }^2}\alpha }}}}{{\frac{{{{\sin }^2}\alpha + {{\cos }^2}\alpha }}{{{{\cos }^2}\alpha }}}} = \frac{{\frac{{2\sin \alpha }}{{\cos \alpha }}}}{{1 + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}} = \frac{{2\tan \alpha }}{{1 + {{\tan }^2}\alpha }}.$

We got the following identity:

$\sin 2\alpha = \frac{2\tan\alpha}{1 + \tan^2\alpha}$

Now take the double-angle formula for cosine:

$\cos 2\alpha = {\cos ^2}\alpha - {\sin ^2}\alpha = \frac{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha + {{\sin }^2}\alpha }}.$

Divide the numerator and denominator by $$\cos^2\alpha$$ again, provided $$\cos\alpha \ne 0$$ or $$\alpha \ne \frac{\pi }{2} + \pi n,$$ $$n \in \mathbb{Z}:$$

$\cos 2\alpha = \frac{{\frac{{{{\cos }^2}\alpha - {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}}{{\frac{{{{\cos }^2}\alpha + {{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}} = \frac{{1 - \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}}{{1 + \frac{{{{\sin }^2}\alpha }}{{{{\cos }^2}\alpha }}}} = \frac{{1 - {{\tan }^2}\alpha }}{{1 + {{\tan }^2}\alpha }}.$

So, we have

$\cos 2\alpha = \frac{1 - \tan^2\alpha}{1 + \tan^2\alpha}$

Finally, note that $$\cot 2\alpha = \frac{1}{{\tan 2\alpha }}.$$ Therefore,

$\cot 2\alpha = \frac{1 - \tan^2\alpha}{2\tan\alpha}$

where $$\alpha \ne \frac{{\pi n}}{2},$$ $$n \in \mathbb{Z}.$$

## Triple-Angle Identities

Using double-angle identities and addition formulas, we can derive triple-angle identities for trigonometric functions.

$\sin 3\alpha = \sin \left( {2\alpha + \alpha } \right) = \sin 2\alpha \cos \alpha + \cos 2\alpha \sin \alpha = 2\sin \alpha \cos \alpha \cos \alpha + \left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)\sin \alpha = 2\sin \alpha \,{\cos ^2}\alpha + \left( {1 - 2{{\sin }^2}\alpha } \right)\sin \alpha = 2\sin \alpha \left( {1 - {{\sin }^2}\alpha } \right) + \left( {1 - 2\,{{\sin }^2}\alpha } \right)\sin \alpha = 2\sin \alpha - 2\,{\sin ^3}\alpha + \sin \alpha - 2\,{\sin ^3}\alpha = 3\sin \alpha - 4\,{\sin ^3}\alpha .$

Thus,

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

The triple-angle formula for cosine can be proved in a similar manner:

$\cos 3\alpha = \cos \left( {2\alpha + \alpha } \right) = \cos 2\alpha \cos \alpha - \sin 2\alpha \sin \alpha = \left( {{{\cos }^2}\alpha - {{\sin }^2}\alpha } \right)\cos \alpha - 2\sin \alpha \cos \alpha \sin \alpha = \left( {2\,{{\cos }^2}\alpha - 1} \right)\cos \alpha - 2\cos \alpha\,{\sin ^2}\alpha = \left( {2\,{{\cos }^2}\alpha - 1} \right)\cos \alpha - 2\cos \alpha \left( {1 - {{\cos }^2}\alpha } \right) = 2\,{\cos ^3}\alpha - \cos \alpha - 2\cos \alpha + 2\,{\cos ^3}\alpha = 4\,{\cos ^3}\alpha - 3\cos \alpha .$

Hence,

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

The triple-angle identity for tangent is given by

$\require{cancel} \tan 3\alpha = \tan \left( {2\alpha + \alpha } \right) = \frac{{\tan 2\alpha + \tan \alpha }}{{1 - \tan 2\alpha \tan \alpha }} = \frac{{\frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} + \tan \alpha }}{{1 - \frac{{2\tan \alpha }}{{1 - {{\tan }^2}\alpha }} \cdot \tan \alpha }} = \frac{{\frac{{2\tan \alpha + \tan \alpha \left( {1 - {{\tan }^2}\alpha } \right)}}{\cancel{1 - {{\tan }^2}\alpha }}}}{{\frac{{1 - {{\tan }^2}\alpha - 2\,{{\tan }^2}\alpha }}{\cancel{1 - {{\tan }^2}\alpha }}}} = \frac{{2\tan \alpha + \tan \alpha - {{\tan }^3}\alpha }}{{1 - {{\tan }^2}\alpha - 2\,{{\tan }^2}\alpha }} = \frac{{3\tan \alpha - {{\tan }^3}\alpha }}{{1 - 3\,{{\tan }^2}\alpha }},$

that is,

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

Let's also consider the triple-angle formula for cotangent:

$\cot 3\alpha = \cot \left( {2\alpha + \alpha } \right) = \frac{{\cot 2\alpha \cot \alpha - 1}}{{\cot 2\alpha + \cot \alpha }} = \frac{{\frac{{{{\cot }^2}\alpha - 1}}{{2\cot \alpha }} \cdot \cot \alpha - 1}}{{\frac{{{{\cot }^2}\alpha - 1}}{{2\cot \alpha }} + \cot \alpha }} = \frac{{\frac{{{{\cot }^3}\alpha - \cot \alpha - 2\cot \alpha }}{\cancel{2\cot \alpha }}}}{{\frac{{{{\cot }^2}\alpha - 1 + 2\,{{\cot }^2}\alpha }}{\cancel{2\cot \alpha }}}} = \frac{{{{\cot }^3}\alpha - \cot \alpha - 2\cot \alpha }}{{{{\cot }^2}\alpha - 1 + 2\,{{\cot }^2}\alpha }} = \frac{{{{\cot }^3}\alpha - 3\cot \alpha }}{{3\,{{\cot }^2}\alpha - 1}},$

or

$\cot 3\alpha = \frac{\cot^3\alpha - 3\cot\alpha}{3\cot^2\alpha - 1}$

See solved problems on Page 2.