Power-Reduction Formulas
Solved Problems
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
\[{\sin ^3}\alpha\,{\cos ^3}\alpha = \frac{{\frac{{9\sin 2\alpha }}{2} + 3\left( { - \sin 2\alpha } \right) - \frac{{\sin 6\alpha }}{2}}}{{16}} = \frac{{9\sin 2\alpha - 6\sin 2\alpha - \sin 6\alpha }}{{32}} = \frac{{3\sin 2\alpha - \sin 6\alpha }}{{32}}.\]
