# Power-Reduction Formulas

Power-reduction identities allow to simplify expressions involving powers of trigonometric functions. These formulas are quite useful in calculus. In particular, using these formulas one can integrate powers of trigonometric expressions.

The power-reduction formulas can be derived through the use of double-angle and half-angle identities as well as the Pythagorean identities.

## Power-Reduction Formulas for Squares

Recall the double angle identity for cosine. It can be written in two forms - in terms of sine and in terms of cosine:

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

Solve the first equation for $${\sin ^2}\alpha:$$

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

Similarly, take the second equation and solve it for $${\cos ^2}\alpha:$$

$\cos 2\alpha = 2\,{\cos ^2}\alpha - 1, \;\Rightarrow 2\,{\cos ^2}\alpha = 1 + \cos 2\alpha , \;\Rightarrow {\cos ^2}\alpha = \frac{{1 + \cos 2\alpha }}{2}.$

Next it's easy to find reduction formulas for tangent and cotangent squared:

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

Let's put these formulas together:

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

## Power-Reduction Formulas for Cubes

To derive the formula for $${\sin ^3}\alpha,$$ we use the triple angle identity for sine

$\sin 3\alpha = 3\sin \alpha - 4\,{\sin ^3}\alpha .$

After rearranging and dividing both sides of the equation by $$4,$$ we obtain

$4\,{\sin ^3}\alpha = 3\sin \alpha - \sin 3\alpha , \;\Rightarrow {\sin ^3}\alpha = \frac{{3\sin \alpha - \sin 3\alpha }}{4}.$

Find the similar identity for cosine cubed. Using the triple angle formula

$\cos 3\alpha = 4{\cos ^3}\alpha - 3\cos \alpha ,$

we have

$4\,{\cos ^3}\alpha = 3\cos \alpha + \cos 3\alpha , \;\Rightarrow {\cos ^3}\alpha = \frac{{3\cos \alpha + \cos 3\alpha }}{4}.$

From these formulas, we can easily derive the expressions for tangent and cotangent cubed:

${\tan ^3}\alpha = \frac{{{{\sin }^3}\alpha }}{{{{\cos }^3}\alpha }} = \frac{{\frac{{3\sin \alpha - \sin 3\alpha }}{4}}}{{\frac{{3\cos \alpha + \cos 3\alpha }}{4}}} = \frac{{3\sin \alpha - \sin 3\alpha }}{{3\cos \alpha + \cos 3\alpha }}.$
${\cot ^3}\alpha = \frac{1}{{{{\tan }^3}\alpha }} = \frac{{3\cos \alpha + \cos 3\alpha }}{{3\sin \alpha - \sin 3\alpha }}.$

Hence,

$\sin^3\alpha = \frac{3\sin\alpha - \sin 3\alpha}{4}$
$\cos^3\alpha = \frac{3\cos\alpha + \cos 3\alpha}{4}$
$\tan^3\alpha = \frac{3\sin\alpha - \sin 3\alpha}{3\cos\alpha + \cos 3\alpha}$
$\cot^3\alpha = \frac{3\cos\alpha + \cos 3\alpha}{3\sin\alpha - \sin 3\alpha}$

## Power-Reduction Formulas for Fourth Powers

To derive the reduction identity for $${\sin ^4}\alpha$$ we apply the formula for $${\sin ^2}\alpha$$ twice.

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

Substituting the reduction formula for $${\cos ^2}2\alpha$$ gives

${\sin ^4}\alpha = \frac{{1 - 2\cos 2\alpha + \frac{{1 + \cos 4\alpha }}{2}}}{4} = \frac{{\frac{{2 - 4\cos 2\alpha + 1 + \cos 4\alpha }}{2}}}{4} = \frac{{3 - 4\cos 2\alpha + \cos 4\alpha }}{8}.$

Similarly we can find the fourth power of cosine:

${\cos ^4}\alpha = {\left( {{{\cos }^2}\alpha } \right)^2} = {\left( {\frac{{1 + \cos 2\alpha }}{2}} \right)^2} = \frac{{1 + 2\cos 2\alpha + {{\cos }^2}2\alpha }}{4} = \frac{{1 + 2\cos 2\alpha + \frac{{1 + \cos 4\alpha }}{2}}}{4} = \frac{{3 + 4\cos 2\alpha + \cos 4\alpha }}{8}.$

The reduction identities for $${\tan ^4}\alpha$$ and $${\cot ^4}\alpha$$ are given by

${\tan ^4}\alpha = \frac{{{{\sin }^4}\alpha }}{{{{\cos }^4}\alpha }} = \frac{{\frac{{3 - 4\cos 2\alpha + \cos 4\alpha }}{8}}}{{\frac{{3 + 4\cos 2\alpha + \cos 4\alpha }}{8}}} = \frac{{3 - 4\cos 2\alpha + \cos 4\alpha }}{{3 + 4\cos 2\alpha + \cos 4\alpha }};$
${\cot ^4}\alpha = \frac{1}{{{{\tan }^4}\alpha }} = \frac{{3 + 4\cos 2\alpha + \cos 4\alpha }}{{3 - 4\cos 2\alpha + \cos 4\alpha }}.$

These formulas are summarized below:

$\sin^4\alpha = \frac{3 - 4\cos 2\alpha + \cos 4\alpha}{8}$
$\cos^4\alpha = \frac{3 + 4\cos 2\alpha + \cos 4\alpha}{8}$
$\tan^4\alpha = \frac{3 - 4\cos 2\alpha + \cos 4\alpha}{3 + 4\cos 2\alpha + \cos 4\alpha}$
$\cot^4\alpha = \frac{3 + 4\cos 2\alpha + \cos 4\alpha}{3 - 4\cos 2\alpha + \cos 4\alpha}$

See solved problems on Page 2.