# Power Reduction Formulas/Cosine to 4th

(Redirected from Fourth Power of Cosine)

## Theorem

$\cos^4 x = \dfrac {3 + 4 \cos 2 x + \cos 4 x} 8$

where $\cos$ denotes cosine.

## Proof 1

 $\displaystyle \cos^4 x$ $=$ $\displaystyle \left({\cos^2 x}\right)^2$ $\displaystyle$ $=$ $\displaystyle \left({\frac {1 + \cos 2 x} 2}\right)^2$ Square of Cosine $\displaystyle$ $=$ $\displaystyle \frac {1 + 2 \cos 2 x + \cos^2 2 x} 4$ multiplying out $\displaystyle$ $=$ $\displaystyle \frac {1 + 2 \cos 2 x + \frac {1 + \cos 4 x} 2} 4$ Square of Cosine $\displaystyle$ $=$ $\displaystyle \frac {2 + 4 \cos 2 x + 1 + \cos 4 x} 8$ multiplying top and bottom by $2$ $\displaystyle$ $=$ $\displaystyle \frac {3 + 4 \cos 2 x + \cos 4 x} 8$ rearrangement

$\blacksquare$

## Proof 2

 $\displaystyle \cos 4 x$ $=$ $\displaystyle \paren {\frac {e^{i x} + e^{-i x} } 2}^4$ Cosine Exponential Formulation $\displaystyle$ $=$ $\displaystyle \frac {\paren {e^{i x} + e^{-i x} }^4} {16}$ rearranging $\displaystyle$ $=$ $\displaystyle \frac {\paren {e^{i x} }^4 + 4 \paren {e^{i x} }^3 \paren {e^{-i x} } + 6 \paren {e^{i x} }^2 \paren {e^{-i x} }^2 + 4 \paren {e^{i x} } \paren {e^{-i x} }^3 + \paren {e^{-i x} }^4} {16}$ multiplying out $\displaystyle$ $=$ $\displaystyle \frac {e^{4 i x} + 4 e^{2 i x} + 6 + 4 e^{-2 i x} + e^{-4 i x} } {16}$ multiplying out $\displaystyle$ $=$ $\displaystyle \frac {3 + 4 \paren {\dfrac {e^{2 i x} + e^{-2 i x} } 2} + \paren {\dfrac {e^{4 i x} + e^{-4 i x} } 2} } 8$ gathering terms $\displaystyle$ $=$ $\displaystyle \frac {3 + 4 \cos 2 x + \cos 4 x} 8$ Cosine Exponential Formulation

$\blacksquare$

## Also defined as

This result can often be seen as:

$\cos^4 x = \dfrac 3 8 + \dfrac {\cos 2 x} 2 + \dfrac {\cos 4 x} 8$