# Power-Reduction Formulas

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Find the value of cos4α given that cos 2α = 1/3.

### Example 2

Calculate 2cos2(π/8) − 1.

### Example 3

Write an expression for $${\sin ^2}\alpha \,{\cos ^2}\alpha$$ that does not contain powers of trigonometric functions greater than $$1.$$

### Example 4

Find the value of $${\tan^4}\alpha$$ given that $$\cos 2\alpha = -\frac{1}{4}.$$

### Example 5

Calculate $${\sin ^4}\frac{\pi }{{12}} + {\cos ^4}\frac{\pi }{{12}}.$$

### Example 6

Calculate $${\tan ^5}\frac{\pi }{6}.$$

### Example 7

Find the value of $${\cot ^4}\alpha$$ given that $$\cos \alpha = \frac{2}{3}.$$

### Example 8

Write an expression for $${\sin ^3}\alpha \,{\cos ^3}\alpha$$ that does not contain powers of trigonometric functions greater than $$1.$$

### Example 1.

Find the value of $${\cos ^4}\alpha$$ given that $$\cos 2\alpha = \frac{1}{3}.$$

Solution.

We represent $${\cos ^4}\alpha$$ as $${\left( {{{\cos }^2}\alpha } \right)^2}$$ and apply the squared power reduction formula for cosine:

${\cos ^4}\alpha = {\left( {{{\cos }^2}\alpha } \right)^2} = {\left( {\frac{{1 + \cos 2\alpha }}{2}} \right)^2} = {\left( {\frac{{1 + \frac{1}{3}}}{2}} \right)^2} = {\left( {\frac{{\frac{4}{3}}}{2}} \right)^2} = {\left( {\frac{2}{3}} \right)^2} = \frac{4}{9}.$

### Example 2.

Calculate $$2\,{\cos ^2}\frac{\pi }{8} - 1.$$

Solution.

We will apply the squared power reduction rule for cosine. This yields:

$2\,{\cos ^2}\frac{\pi }{8} - 1 = 2\left( {\frac{{1 + \cos \frac{\pi }{4}}}{2}} \right) - 1 = \cancel{1} + \cos \frac{\pi }{4} - \cancel{1} = \cos \frac{\pi }{4} = \frac{{\sqrt 2 }}{2}.$

### Example 3.

Write an expression for $${\sin ^2}\alpha \,{\cos ^2}\alpha$$ that does not contain powers of trigonometric functions greater than $$1.$$

Solution.

Using the squared power reduction rules for sine and cosine, we have

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

Apply the reduction indetity for $${{{\cos }^2}2\alpha }.$$ This yields:

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

### Example 4.

Find the value of $${\tan^4}\alpha$$ given that $$\cos 2\alpha = -\frac{1}{4}.$$

Solution.

We write $${\tan^4}\alpha$$ as $${\left( {{{\tan }^2}\alpha } \right)^2}$$ and use the following power reduction rule:

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

This gives:

${\tan ^4}\alpha = {\left( {{{\tan }^2}\alpha } \right)^2} = {\left( {\frac{{1 - \cos 2\alpha }}{{1 + \cos 2\alpha }}} \right)^2} = {\left( {\frac{{1 - \left( { - \frac{1}{4}} \right)}}{{1 - \frac{1}{4}}}} \right)^2} = {\left( {\frac{{\frac{5}{4}}}{{\frac{3}{4}}}} \right)^2} = {\left( {\frac{5}{3}} \right)^2} = \frac{{25}}{9}.$

### Example 5.

Calculate $${\sin ^4}\frac{\pi }{{12}} + {\cos ^4}\frac{\pi }{{12}}.$$

Solution.

Using the power reduction formulas for $${\sin ^4}\frac{\pi }{{12}}$$ and $${\cos ^4}\frac{\pi }{{12}},$$ we can write:

${\sin ^4}\frac{\pi }{{12}} = \frac{{3 - 4\cos \frac{\pi }{6} + \cos \frac{\pi }{3}}}{8},$
${\cos ^4}\frac{\pi }{{12}} = \frac{{3 + 4\cos \frac{\pi }{6} + \cos \frac{\pi }{3}}}{8}.$

Substitute this in the original expression:

${\sin ^4}\frac{\pi }{{12}} + {\cos ^4}\frac{\pi }{{12}} = \frac{{3 - 4\cos \frac{\pi }{6} + \cos \frac{\pi }{3}}}{8} + \frac{{3 + 4\cos \frac{\pi }{6} + \cos \frac{\pi }{3}}}{8} = \frac{{3 - \cancel{{4\cos \frac{\pi }{6}}} + \cos \frac{\pi }{3} + 3 + \cancel{{4\cos \frac{\pi }{6}}} + \cos \frac{\pi }{3}}}{8} = \frac{{6 + 2\cos \frac{\pi }{3}}}{8} = \frac{{2\left( {3 + \cos \frac{\pi }{3}} \right)}}{8} = \frac{{3 + \cos \frac{\pi }{3}}}{4} = \frac{{3 + \frac{1}{2}}}{4} = \frac{{\frac{7}{2}}}{4} = \frac{7}{8}.$

### Example 6.

Calculate $${\tan ^5}\frac{\pi }{6}.$$

Solution.

We represent $${\tan ^5}\frac{\pi }{6}$$ as the product $${\tan ^2}\frac{\pi }{6}{\tan ^3}\frac{\pi }{6}.$$ Then using the power reduction formulas, we obtain

${\tan ^5}\frac{\pi }{6} = {\tan ^2}\frac{\pi }{6}{\tan ^3}\frac{\pi }{6} = \frac{{1 - \cos \frac{\pi }{3}}}{{1 + \cos \frac{\pi }{3}}} \cdot \frac{{3\sin \frac{\pi }{6} - \sin \frac{\pi }{2}}}{{3\cos \frac{\pi }{6} + \cos \frac{\pi }{2}}} = \frac{{1 - \frac{1}{2}}}{{1 + \frac{1}{2}}} \cdot \frac{{3 \cdot \frac{1}{2} - 1}}{{3 \cdot \frac{{\sqrt 3 }}{2} + 0}} = \frac{{\frac{1}{2}}}{{\frac{3}{2}}} \cdot \frac{{\frac{1}{2}}}{{\frac{{3\sqrt 3 }}{2}}} = \frac{1}{{9\sqrt 3 }}.$

### Example 7.

Find the value of $${\cot ^4}\alpha$$ given that $$\cos \alpha = \frac{2}{3}.$$

Solution.

We will use the following identity:

${\cot ^4}\alpha = \frac{{3 + 4\cos 2\alpha + \cos 4\alpha }}{{3 - 4\cos 2\alpha + \cos 4\alpha }}.$

Calculate $${\cos 2\alpha }$$ by the double angle formula:

$\cos 2\alpha = 2\,{\cos ^2}\alpha - 1 = 2 \cdot {\left( {\frac{2}{3}} \right)^2} - 1 = 2 \cdot \frac{4}{9} - 1 = \frac{{8 - 9}}{9} = - \frac{1}{9}.$

Similarly find $${\cos 4\alpha:}$$

$\cos 4\alpha = 2{\cos ^2}2\alpha - 1 = 2 \cdot {\left( { - \frac{1}{9}} \right)^2} - 1 = 2 \cdot \frac{1}{{81}} - 1 = - \frac{{79}}{{81}}.$

Substitute these results in the expression for $${\cot ^4}\alpha:$$

${\cot ^4}\alpha = \frac{{3 + 4\cos 2\alpha + \cos 4\alpha }}{{3 - 4\cos 2\alpha + \cos 4\alpha }} = \frac{{3 + 4 \cdot \left( { - \frac{1}{9}} \right) + \left( { - \frac{{79}}{{81}}} \right)}}{{3 - 4 \cdot \left( { - \frac{1}{9}} \right) + \left( { - \frac{{79}}{{81}}} \right)}} = \frac{{3 - \frac{4}{9} - \frac{{79}}{{81}}}}{{3 + \frac{4}{9} - \frac{{79}}{{81}}}} = \frac{{\frac{{243 - 36 - 79}}{{81}}}}{{\frac{{243 + 36 - 79}}{{81}}}} = \frac{{128}}{{200}} = \frac{{16}}{{25}}.$

### Example 8.

Write an expression for $${\sin ^3}\alpha \,{\cos ^3}\alpha$$ that does not contain powers of trigonometric functions greater than $$1.$$

Solution.

Using the cubed power reduction rules, we can write:

${\sin ^3}\alpha \,{\cos ^3}\alpha = \frac{{3\sin \alpha - \sin 3\alpha }}{4} \cdot \frac{{3\cos \alpha + \cos 3\alpha }}{4} = \frac{{9\sin \alpha \cos \alpha - 3\cos \alpha \sin 3\alpha + 3\sin \alpha \cos 3\alpha - \sin 3\alpha \cos 3\alpha }}{{16}}.$

Now we apply the double angle formula for sine

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

Hence

$\sin \alpha \cos \alpha = \frac{{\sin 2\alpha }}{2},\;\sin 3\alpha \cos 3\alpha = \frac{{\sin 6\alpha }}{2}.$

By the sine subtraction formula,

$\sin \alpha \cos 3\alpha - \cos \alpha \sin 3\alpha = \sin \left( {\alpha - 3\alpha } \right) = \sin \left( { - 2\alpha } \right) = - \sin 2\alpha .$

Then

${\sin ^3}\alpha\,{\cos ^3}\alpha = \frac{{\frac{{9\sin 2\alpha }}{2} + 3\left( { - \sin 2\alpha } \right) - \frac{{\sin 6\alpha }}{2}}}{{16}} = \frac{{9\sin 2\alpha - 6\sin 2\alpha - \sin 6\alpha }}{{32}} = \frac{{3\sin 2\alpha - \sin 6\alpha }}{{32}}.$