Power Reduction Formulas/Sine to 4th

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Theorem

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

where $\sin$ and $\cos$ denote sine and cosine respectively.


Proof

\(\displaystyle \sin^4 x\) \(=\) \(\displaystyle \paren {\sin^2 x}^2\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {\frac {1 - \cos 2 x} 2}^2\) Square of Sine
\(\displaystyle \) \(=\) \(\displaystyle \frac {1 - 2 \cos 2 x + \cos^2 2 x} 4\) 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

$\blacksquare$


Sources